text:
"use strict";

var chai = require("chai"),
    sortedArray = require("../lib/sortedArray.js"),
    expect = chai.expect;

describe("sortedArray()", function () {
    var arr;

    function descending(a, b) {
        return b - a;
    }

    function ascending(a, b) {
        return a - b;
    }

    function firstLetterLexically(a, b) {
        return a.charAt(0).localeCompare(b.charAt(0));
    }

    function byLengthDesc(a, b) {
        return b.length - a.length;
    }

    beforeEach(function () {
        arr = sortedArray([1, 2, 4]);
    });

    describe("when passing no arguments", function () {

        it("should return a new array", function () {
            // we need to use .slice() for comparison because the arr is extended (and thus can't be compared via .eql())
            expect(sortedArray().slice()).to.eql([]);
        });

    });

    describe("when passing an array", function () {

        it("should return the passed array", function () {
            arr = [];
            expect(sortedArray(arr)).to.equal(arr);
        });

        describe("with length > 0 and a comparator", function () {

            it("should sort the array by the given comparator", function () {
                arr = [5, 1, 2];
                arr.comparator = descending;
                arr = sortedArray(arr);
                expect(arr.slice()).to.eql([5, 2, 1]);
            });

        });

        describe("with length > 0 and no comparator", function () {

            it("should sort the array by the default comparator", function () {
                arr = sortedArray([5, 2, 1]);
                expect(arr.slice()).to.eql([1, 2, 5]);
            });

        });

    });

    describe("when passing an array and multiple comparators", function () {

        it("should sort the array by default", function () {
            arr = sortedArray(["bb", "a", "cc", "b", "aa", "c"], firstLetterLexically, byLengthDesc);
            expect(arr.slice()).to.eql(["aa", "a", "bb", "b", "cc", "c"]);
        });

    });

    describe(".push()", function () {

        it("should insert the new value respecting the given order", function () {
            arr.push(3);
            expect(arr.slice()).to.eql([1, 2, 3, 4]);
        });

        describe("when passing several arguments", function () {

            it("should insert all new values at their sorted index", function () {
                arr.push(3, 0, 5);
                expect(arr.slice()).to.eql([0, 1, 2, 3, 4, 5]);
            });

        });

    });

    describe(".unshift()", function () {

        it("should be an alias for .push()", function () {
            expect(arr.unshift).to.equal(arr.push);
        });

    });

    describe(".splice()", function () {

        it("should remove the elements at the given index", function () {
            arr.push(3, 5, 6);
            arr.splice(1, 3);
            expect(arr.slice()).to.eql([1, 5, 6]);
        });

        it("should return removed elements just like Array.prototype.splice", function () {
            arr.push(3, 5, 6);
            expect(arr.splice(1, 3)).to.eql([2, 3, 4]);
            expect(arr.splice(1, 0)).to.eql([]);
        });

        it("should add the additional elements at the index specified by the current order", function () {
            arr.splice(0, 0, 3, 5, 6);
            expect(arr.slice()).to.eql([1, 2, 3, 4, 5, 6]);
        });

    });

    describe(".indexOf()", function () {

        it("should work like Array.prototype.indexOf", function () {
            expect(arr.indexOf(2)).to.equal(1);
            expect(arr.indexOf(100)).to.equal(-1);
            expect(arr.indexOf(2, 2)).to.equal(-1);
            expect(arr.indexOf(2, -100)).to.equal(1);
        });

    });

    describe(".sort()", function () {

        it("should sort the array by the default comparator", function () {
            arr[3] = 2;
            arr.sort();
            expect(arr.slice()).to.eql([1, 2, 2, 4]);
        });

        describe("when passing comparators", function () {

            beforeEach(function () {
                arr = sortedArray(["bb", "a", "cc", "b", "aa", "c"]);
            });

            it("should sort the array by all comparators in order", function () {
                arr = sortedArray(["bb", "a", "cc", "b", "aa", "c"]);
                arr.sort(firstLetterLexically, byLengthDesc);
                expect(arr.slice()).to.eql(["aa", "a", "bb", "b", "cc", "c"]);
            });

            it("should order all subsequent new elements by the passed comparator", function () {
                arr.sort(firstLetterLexically, byLengthDesc);
                arr.push("aaa");
                expect(arr.slice()).to.eql(["aaa", "aa", "a", "bb", "b", "cc", "c"]);
            });

        });

        describe("when comparators have been set", function () {

            beforeEach(function () {
                arr = sortedArray(["bb", "a", "cc", "b", "aa", "c"]);
                arr.sort(firstLetterLexically, byLengthDesc);
                arr[arr.length] = "aaa";
            });

            it("should sort the array by the previously set comparator", function () {
                arr.sort();
                expect(arr.slice()).to.eql(["aaa", "aa", "a", "bb", "b", "cc", "c"]);
            });

        });

    });

    describe(".reverse()", function () {

        function evenOdd(a, b) {
            var compared = a % 2 - b % 2;

            if (compared === 0) { // both are even or odd
                return a - b;
            } else {
                return compared;
            }
        }

        it("should work like Array.prototype.reverse()", function () {
            arr.reverse();
            expect(arr.slice()).to.eql([4, 2, 1]);
        });

        it("should invert the comparator so the original order is preserved", function () {
            arr.reverse();
            arr.push(3);
            expect(arr.slice()).to.eql([4, 3, 2, 1]);
            arr.reverse();
            arr.push(0);
            expect(arr.slice()).to.eql([0, 1, 2, 3, 4]);
        });

        it("should reset if a different comparator is used", function () {
            arr.reverse();
            arr.sort(evenOdd);
            arr.push(3, 5, 6, 7);
            expect(arr.slice()).to.eql([2, 4, 6, 1, 3, 5, 7]);
            arr.reverse();
            expect(arr.slice()).to.eql([7, 5, 3, 1, 6, 4, 2]);
            arr.push(0);
            expect(arr.slice()).to.eql([7, 5, 3, 1, 6, 4, 2, 0]);
        });

    });

    describe(".sortedIndex()", function () {

        it("should return the insertion index for the given element", function () {
            // arr = [1, 2, 4]
            expect(arr.sortedIndex(-1)).to.equal(0);
            expect(arr.sortedIndex(0)).to.equal(0);
            expect(arr.sortedIndex(1)).to.equal(0);
            expect(arr.sortedIndex(3)).to.equal(2);
            expect(arr.sortedIndex(4)).to.equal(2);
            expect(arr.sortedIndex(5)).to.equal(3);
            expect(arr.sortedIndex(6)).to.equal(3);
        });

    });

    describe(".toString()", function () {

        it("should return the same result as the original array", function () {
            expect(arr.toString()).to.equal([1, 2, 4].toString());
        });

    });

});

redpajama_set_name: RedPajamaGithub


Prior to joining bookmaker and funding your account in order to view Union Douala – Eding Sport, or any other particular event via the bookmaker live stream, you are strongly advised to check with bookmaker if, depending on your place of residence, it is possible to view this live stream.
This time we will watch a game between teams from the middle of a tournament table (ranked 9 and 8). One and the other teams get a very uncertain game. In this match the chances to win of both teams are almost equal.
Last 4 head-to-head matches Union Douala won 1 matches, drawn 2 matches, lost 1 matches and goals 6-6. Including matches at home between the teams Union Douala won 1 matches, drawn 0 matches, lost 1 matches and goals 3-3.


text:
\section{Introduction}
In the study of Riemnnian geometry, the notion of curvature plays an significant role. As the average of sectional curvature, the scalar curvature is one of the simplest curvature invariants on a Riemannian manifold. In general, the scalar curvature can be regarded as a kind of weak measure of the local geometry. It is tempting to ask which manifolds can admit metric of positive scalar curvature. When the underlying manifold is torus $\mathbb{T}^n$, the Geroch Conjecture predicted that metrics with non-negative scalar curvature must be flat. The problem was solved by Schoen-Yau \cite{SchoenYau1979,SchoenYau1979-2} for $n\leq 7$ using minimal surface method  and Gromov-Lawson \cite{GromovLawson1980} for general $n$ using Atiyah-Singer index theorem for a twisted spinor bundle on a spin manifold.

In \cite{Gromov2014},  Gromov  conjectured a stability for the torus rigidity. Namely, a sequence of Riemannian manifolds with almost non-negative scalar curvature, which are diffeomorphic to tori, combined with appropriate compactness conditions should converge to a flat torus in some weak sense. In \cite{Sormani2017}, Sormani had formulated the conjecture more concretely using the notion of intrinsic flat distance which is a distance between integral current spaces and was introduced by Sormani-Wenger \cite{SormaniWenger2011}. This is believed to be the suitable notion for taking limits of manifolds with lower scalar curvature bounds, see also the recent work of the second named author, Naber and Neumayer in \cite{LeeNaberNeumayer2020} which suggested that the geodesic distance should be replaced by the $L^p$ version of distance function $d_p$ at least when $n>3$. For general dimension, it is unclear what conditions should serve as the non-collapsing assumption. While in $n=3$, Sormani \cite{Sormani2017} gave a precise prediction of the non-collapsing conditions called the MinA condition in order to avoid bubbling occurring.

The first result in this direction is given by Gromov in \cite{Gromov2014}  where if one assumes that a sequence of tori with almost non-negative scalar curvature converges in the $C^0$ sense to a $C^2$ metric, then one can show that the $C^2$ limit is a flat Riemannian metric. In \cite{Bamler2016}, Bamler gave an alternative proof using the Ricci flow to perform regularization. It was recently generalized by Burkhardt-Guim \cite{Burkhardt2019} to the case when the limiting metrics $g_\infty$ are only $C^0$ metric and it was shown that $g_\infty$ is isometric to the flat torus as a metric space. Further progresses toward the conjecture formulated by Sormani \cite{Sormani2017} has been made in various cases. In \cite{AHPPS2019}, Allen, Hernandez-Vazquez, Parise, Payne, and Wang studied the warped product case. In \cite{CAP2020}, Cabrera Pacheco, Ketterer, and Perales studied the case of graphical tori. In \cite{Allen2020}, Allen studied the case when the sequence is conformal to the flat tori. In \cite{LeeNaberNeumayer2020}, the second named author with Naber and Neumayer considered the case when the entropy is small and proved the stability in the sense of $d_p$ convergence.

Motivated by the work of Allen in \cite{Allen2020}, we study the case when the metrics are uniformly conformal to some metrics with better controlled geometry. We show that under the natural non-collapsing assumption, if in addition the conformal factor is bounded in $L^{p_{0}}$ for some sufficiently large $p_{0}$, then the sequence will converge to a flat metric in the volume preserving intrinsic flat sense.

\begin{thm}\label{Sec1: maintheorem}
Let $(M_{i},g_{i})$ be a sequence of $n$-dimensional Riemannian manifolds ($n>2$). Suppose $M_{i}$ is diffeomorphic to $\mathbb{T}^n$ and $g_i=u_i^\frac4{n-2}h_i$ for some metric $h_i$ such that
\begin{enumerate}\setlength{\itemsep}{1mm}
\item[(i)]  $|\text{\rm Ric}(h_i)|_{h_i}+(\mathrm{inj}(M_i,h_i))^{-1}+\mathrm{diam}(M_i,h_i)\leq \Lambda$;
\item[(ii)] $\mathrm{diam}(M_i,g_i)+\|u_i\|_{L^{p_0}(M_i,h_i)}\leq \Lambda $;
\item[(iii)] $\mathrm{Vol}(M_i,g_i)\geq \Lambda^{-1}$;
\item[(iv)] $R(g_i)\geq -\delta_i$
\end{enumerate}
for some $\Lambda>0$, $p_0>\frac{n}{2}+\frac{2n(n+4)}{(n-2)(n+2)}$ and $\delta_i\rightarrow 0$. Then after passing to a subsequence, $(M_i,g_i)$ converges to a flat torus in the volume preserving intrinsic flat sense.
\end{thm}
\begin{rem}
By the H\"older inequality, condition (ii) and (iii) will imply a lower bound of $\mathrm{Vol}(M_i,h_i)$. In this case, the lower bound of injectivity radius of $h_i$ will follow from the work of Cheeger-Gromov-Taylor \cite{CheegerGromovTaylor1982} if we strengthen the curvature bound of $h_i$ from $\text{\rm Ric}(h_i)$ to $\text{\rm Rm}(h_i)$.
\end{rem}

In Theorem~\ref{Sec1: maintheorem}, $h_i$ serves as a sequence of well-behaved reference metrics within the conformal classes. In certain sense, we show that if the metric within the conformal has almost non-negative scalar curvature, then it is almost the solution to the Yamabe problem. The volume condition (iii) is necessary in order to prevent collapsing. Condition (ii) is stronger than volume non-expanding which plays the role to rule out bubbling. If we strengthen condition (ii) from $L^{p_{0}}$ to $L^\infty$, better convergence can be obtained. Note that in this case, the upper bound of $\mathrm{diam}(M_i,g_i)$ follows from condition (i).

\begin{thm}\label{Sec1: maintheorem2}
Let $(M_{i},g_{i})$ be a sequence of $n$-dimensional Riemannian manifolds ($n>2$). Suppose $M_{i}$ is diffeomorphic to $\mathbb{T}^n$ and $g_i=u_i^\frac4{n-2}h_i$ for some metric $h_i$ such that
\begin{enumerate}\setlength{\itemsep}{1mm}
\item[(i)]  $|\text{\rm Ric}(h_i)|_{h_i}+(\mathrm{inj}(M_i,h_i))^{-1}+\mathrm{diam}(M_i,h_i)\leq \Lambda$;
\item[(ii)] $ \sup_{M_i}u_i\leq\Lambda$;
\item[(iii)] $\mathrm{Vol}(M_i,g_i)\geq \Lambda^{-1}$;
\item[(iv)] $R(g_i)\geq -\delta_i$
\end{enumerate}
for some $\Lambda>0$ and $\delta_i\to 0$. Then after passing to a subsequence, $(M_i,g_i)$ converges to a flat torus in $L^p$ for all $p>0$ modulo diffeomorphism. Moreover, $(M_i,g_i)$ converges to a flat torus in the measured Gromov-Hausdorff sense.
\end{thm}

\begin{rem}
It will  be clear from the proof that the torus structure doesn't play any important role. Indeed, the same stability will be true if $\mathbb{T}^n$ is replaced by any closed manifolds $M^n$ with non-positive Yamabe invariant. We refer readers to Section~\ref{stability-T} for the detailed statement.
\end{rem}

The main technique is motivated by that of \cite{Bamler2016,Burkhardt2019,ShiTam2018} which used the Ricci flow to regularize the metrics using the estimates from \cite{KochLamm2012, Simon2002}. In our case, we do not a-priori assume any convergence of $g_i$ or $C^0$ closeness of metrics and therefore Ricci flow's method do not apply directly. Instead,  we make use of the Yamabe flow which is a geometric heat flow evolving inside the conformal class to (partially) regularize the metric $g_i$. Unlike the Ricci flow, its regularization ability is  relatively limited and highly depends on the uniform geometry of $h_i$. In our case, although the corresponding Yamabe flow $g_i(t)$ is not uniformly regular in $C^\infty$, we are able to show that it converges to a fixed flat metric away from $t=0$ weakly modulo diffeomorphism. This reduces the problem to establishing uniform weak convergence of $g_i(t)$ as $t\rightarrow 0$. This will be done in Section~\ref{stability-T}.

{\it Acknowledgement}:
We would like to thank Brian Allen and Shengwen Wang for useful communication. We would also like to thank Davi Maximo for pointing out a discrepancy in the earlier version of the paper. M.-C. Lee was partially supported by NSF grant DMS-1709894 and EPSRC  grant number P/T019824/1.

\section{A-priori estimates along the Yamabe flow}\label{estimateYamaflow}
In the following, we will consider a Riemannian manifold $(M,h)$ which satisfies
\begin{equation*}
{\bf (A)}:\quad\;\;
\left\{
\begin{array}{ll}
|\text{\rm Ric}(h)|_{h} \leq \Lambda;\\[1mm]
\mathrm{diam}(M,h)\leq \Lambda;\\[1mm]
\mathrm{inj}(M,h)\geq \Lambda^{-1}
\end{array}
\right.
\end{equation*}
for some $\Lambda>0$. For later use, here we recall the Sobolev inequality and the Poincar\'e inequality: for $f\in W^{1,2}(M)$,
\begin{equation}
\left(\int_{M}|f|^{\frac{2n}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}}
\leq C_{S}\int_{M}|\nabla^{h}f|_{h}^{2}d\mu_{h}+C_{S}\int_{M}f^{2}d\mu_{h},
\end{equation}
and
\begin{equation}
\int_{M}|f-\underline{f}|^{2}d\mu_{h}
\leq C_{P}\int_{M}|\nabla^{h}f|_{h}^{2}d\mu_{h}, \quad \underline{f} = \frac{1}{\mathrm{Vol}(M,h)}\int_{M}fd\mu_{h}.
\end{equation}
Note that the lower bound of injective radius implies the lower bound of volume (see \cite{BergerKazdan80}), and then constants $C_{S}$ (see \cite{Ilias83}) and $C_{P}$ (see \cite{LiYau80}) depend only on $n$ and $\Lambda$.

Let $g_0=u_0^\frac4{n-2}h$ be a metric inside the conformal class of $h$. Suppose that
\begin{equation*}
{\bf (B)}:\quad\;\; \left\{
\begin{array}{ll}
\|u_0\|_{L^{p_0}(M,h)}\leq \Lambda;\\[1.5mm]
\mathrm{Vol}(M,g_{0})\geq \Lambda^{-1};\\[1mm]
R(g_0)\geq -\delta\geq -1
\end{array}
\right.
\end{equation*}
for some $p_0,\Lambda,\delta>0$. In this section, we will use the Yamabe flow to regularize $g_0$ slightly. This is the family of metric $g(t)$ which solves
\begin{equation}\label{Yamabe flow}
\left\{
\begin{array}{ll}
\frac{\partial g}{\partial t} = -R(g) g,\\[1mm]
g(0) = g_0.
\end{array}
\right.
\end{equation}
Equivalently, if we write $g(t)=u(t)^{\frac{4}{n-2}}h$, then the function $u(t)$ solves
\begin{equation}\label{evo-u}
\begin{cases}
\ \frac{\partial u^N}{\partial t} = \frac{n+2}{4}\left(\frac{4(n-1)}{n-2}\Delta_{h}u-R_{h}u \right), \\
\ u(0) = u_{0},
\end{cases}
\end{equation}
where $N=\frac{n+2}{n-2}$ and $R_{h}$ denotes the scalar curvature of $h$. We will establish a-priori estimates of $g(t)$ or equivalently $u(t)$ along the Yamabe flow.

\subsection{Lower bound of the Yamabe flow}
We first show that if $p_{0}$ is sufficiently large, then assumption {\bf (B)} implies the positive lower bound of $u_{0}$. We begin with the following lemma, which is a global version of \cite[(4.8)]{HanLin1997}.
\begin{lma}\label{Sec2: inequality}
If $p_{0}>\frac{2n}{n-2}$, then there is a constant $\varepsilon_{0}(n,\Lambda,p_{0})>0$ such that
\begin{equation}
\left(\int_{M}u_{0}^{\varepsilon_{0}}d\mu_{h}\right)\left(\int_{M}u_{0}^{-\varepsilon_{0}}d\mu_{h}\right) \leq 4e^{2}.
\end{equation}
\end{lma}
\begin{proof}
The proof is similar to that of the local version \cite[(4.8)]{HanLin1997}. For the sake of completeness, we include the proof here. Since $g_{0}=u_{0}^{\frac{4}{n-2}}h$, then the scalar curvature of $g_{0}$ is
\begin{equation}\label{Sec2: inequality eqn 5}
R_{g_{0}} = u_{0}^{-\frac{n+2}{n-2}}\left(R_{h}u_{0}-\frac{4(n-1)}{n-2}\Delta_{h} u_{0}\right),
\end{equation}
Define
\begin{equation}\label{def of w}
w = \log u_{0}-\frac{1}{\mathrm{Vol}(M,h)}\int_{M}(\log u_{0})d\mu_{h}.
\end{equation}
Direct calculation shows
\begin{equation}
\begin{split}
R_{g_{0}} = {} & u_{0}^{-\frac{4}{n-2}}\left(R_{h}-\frac{4(n-1)}{n-2}\frac{\Delta_{h}u_{0}}{u_{0}}\right) \\
= {} & u_{0}^{-\frac{4}{n-2}}\left(R_{h}-\frac{4(n-1)}{n-2}\left(\Delta_{h}w+|\nabla^{h}w|_{h}^{2}\right)\right).
\end{split}
\end{equation}
Recalling $R_{g_{0}}\geq-1$,
\begin{equation}\label{Sec2: inequality eqn 1}
|\nabla^{h}w|_{h}^{2} = -\Delta_{h}w+\frac{n-2}{4(n-1)}\left(R_{h}-R_{g_{0}}u_{0}^{\frac{4}{n-2}}\right)
\leq -\Delta_{h}w+Cu_{0}^{\frac{4}{n-2}}+C
\end{equation}
for some $C(n,\Lambda)$. Since $p_{0}>\frac{2n}{n-2}>\frac{4}{n-2}$, we obtain
\begin{equation}\label{Sec2: inequality eqn 4}
\int_{M}|\nabla^{h}w|_{h}^{2}d\mu_{h} \leq \int_{M}(Cu_{0}^{\frac{4}{n-2}}+C)d\mu_{h}
\leq C\|u_{0}\|_{p_{0}}^{p_{0}}+C \leq C(n,\Lambda_{0},p_{0}).
\end{equation}
Combining this with the Sobolev inequality and the Poincar\'{e} inequality,
\begin{equation}\label{Sec2: inequality eqn 2}
\begin{split}
\left(\int_{M}|w|^{\frac{2n}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}}
\leq {} & C_{S}\int_{M}|\nabla^{h}w|_{h}^{2}d\mu_{h}+C_{S}\int_{M}|w|^{2}d\mu_{h} \\
\leq {} & C_{S}(1+C_{P})\int_{M}|\nabla^{h}w|_{h}^{2}d\mu_{h} \leq C(n,\Lambda,p_{0}).
\end{split}
\end{equation}
Define two constants $\beta$ and $\beta^{*}$ by
\begin{equation}
\frac{4\beta}{n-2} = p_{0}, \quad \frac{1}{\beta}+\frac{1}{\beta^{*}} = 1.
\end{equation}
Then $p_{0}>\frac{2n}{n-2}$ implies $\beta>\frac{n}{2}$ and $\beta^{*}<\frac{n}{n-2}$. Using \eqref{Sec2: inequality eqn 1}, \eqref{Sec2: inequality eqn 4} and the H\"older inequality, for $p>1$, we compute
\begin{equation}
\begin{split}
& \int_{M}|w|^{p}|\nabla^{h}w|_{h}^{2}d\mu_{h} \\
\leq {} & p\int_{M}|w|^{p-1}|\nabla^{h}w|_{h}^{2}d\mu_{h}+C\int_{M}|w|^{p}u_{0}^{\frac{4}{n-2}}d\mu_{h}+C\int_{M}|w|^{p}d\mu_{h} \\
\leq {} & \frac{1}{2}\int_{M}|w|^{p}|\nabla^{h}w|_{h}^{2}d\mu_{h}+(Cp)^{p}\int_{M}|\nabla^{h}w|_{h}^{2}d\mu_{h} \\
&+C\left(\int_{M}u_{0}^{p_{0}}d\mu_{h}\right)^{\frac{1}{\beta}}\left(\int_{M}|w|^{p\beta^{*}}d\mu_{h}\right)^{\frac{1}{\beta^{*}}}
+C\left(\int_{M}|w|^{p\beta^{*}}d\mu_{h}\right)^{\frac{1}{\beta^{*}}} \\
\leq {} & \frac{1}{2}\int_{M}|w|^{p}|\nabla^{h}w|_{h}^{2}d\mu_{h}+(Cp)^{p}+C\left(\int_{M}|w|^{p\beta^{*}}d\mu_{h}\right)^{\frac{1}{\beta^{*}}},
\end{split}
\end{equation}
which implies
\begin{equation}
\int_{M}|\nabla(|w|^{\frac{p+2}{2}})|_{h}^{2}d\mu_{h} \leq C(p+2)^{2}\left(\int_{M}|w|^{p\beta^{*}}d\mu_{h}\right)^{\frac{1}{\beta^{*}}}+(C(p+2))^{p+2},
\end{equation}
Combining this with the Sobolev inequality and the H\"older inequality,
\begin{equation}
\begin{split}
& \left(\int_{M}|w|^{\frac{n(p+2)}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}} \\
\leq {} & C(p+2)^{2}\left(\int_{M}|w|^{p\beta^{*}}d\mu_{h}\right)^{\frac{1}{\beta^{*}}}+C\int_{M}|w|^{p+2}d\mu_{h}+(C(p+2))^{p+2} \\
\leq {} & C(p+2)^{2}\left(\int_{M}|w|^{(p+2)\beta^{*}}d\mu_{h}\right)^{\frac{1}{\beta^{*}}}+(C(p+2))^{p+2}.
\end{split}
\end{equation}
It then follows that
\begin{equation}
\|w\|_{L^{\frac{n(p+2)}{n-2}}} \leq (C(p+2))^{\frac{2}{p+2}}\|w\|_{L^{(p+2)\beta^{*}}}+C(p+2), \quad \text{for $p>1$}.
\end{equation}
Replacing $(p+2)\beta^{*}$ by $p$, and writing $\gamma=\frac{n}{\beta^{*}(n-2)}$, we obtain
\begin{equation}\label{Sec2: inequality eqn 3}
\|w\|_{L^{\gamma p}} \leq (Cp)^{\frac{C}{p}}\|w\|_{L^{p}}+Cp, \quad \text{for $p>3\beta^{*}$}.
\end{equation}
Recalling $\beta^{*}<\frac{n}{n-2}$, we have $\gamma>1$. Using \eqref{Sec2: inequality eqn 3} repeatedly,
\begin{equation}\label{Sec2: inequality eqn 6}
\|w\|_{L^{6\beta^{*}}} \leq C\|w\|_{L^{4\beta^{*}}}+C.
\end{equation}
By \eqref{Sec2: inequality eqn 2}, we have $\|w\|_{L^{2\beta^{*}}}\leq C$. Using the Cauchy-Schwarz inequality,
\begin{equation}\label{Sec2: inequality eqn 7}
\|w\|_{L^{4\beta^{*}}} = \left(\int_{M}|w|^{\beta^{*}}|w|^{3\beta^{*}}d\mu_{h}\right)^{\frac{1}{4\beta^{*}}}
\leq  \|w\|_{L^{2\beta^{*}}}^{\frac{1}{4}}\|w\|_{L^{6\beta^{*}}}^{\frac{3}{4}}
\leq  C\|w\|_{L^{6\beta^{*}}}^{\frac{3}{4}}.
\end{equation}
Then \eqref{Sec2: inequality eqn 6} and \eqref{Sec2: inequality eqn 7} show $\|w\|_{L^{6\beta^{*}}}\leq C$. Combining this with \eqref{Sec2: inequality eqn 3}, we obtain
\begin{equation}
\|w\|_{L^{\gamma^{k}}} \leq \left(C\gamma^{k-1}\right)^{\frac{C}{\gamma^{k-1}}}\|w\|_{L^{\gamma^{k-1}}}+C\gamma^{k-1}, \quad \text{for $k\in\mathbb{N}$},
\end{equation}
and so
\begin{equation}
\|w\|_{L^{\gamma^{k}}} \leq C(n,\Lambda,p_{0})\gamma^{k}, \quad \text{for $k\in\mathbb{N}$}.
\end{equation}
This shows
\begin{equation}
\|w\|_{L^{p}} \leq C_{0}(n,\Lambda,p_{0})p,\quad \text{for $p\geq1$}.
\end{equation}
Choosing $\varepsilon_{0}(n,\Lambda,p_{0})$ such that $C_{0}\varepsilon_{0}\leq\frac{1}{2}$, then
\begin{equation}
\int_{M}e^{\varepsilon_{0}|w|}d\mu_{h}
= \sum_{k=0}^{\infty}\int_{M}\frac{\varepsilon_{0}^{k}|w|^{k}}{k!}d\mu_{h}
\leq \sum_{k=0}^{\infty}\frac{(C_{0}\varepsilon_{0})^{k}k^{k}}{k!}
\leq e\sum_{k=0}^{\infty}(C_{0}\varepsilon_{0})^{k} \leq 2e.
\end{equation}
Recalling the definition of $w$ \eqref{def of w}, we obtain
\begin{equation}
\left(\int_{M}u_{0}^{\varepsilon_{0}}d\mu_{h}\right)\left(\int_{M}u_{0}^{-\varepsilon_{0}}d\mu_{h}\right)
= \left(\int_{M}e^{\varepsilon_{0}w}d\mu_{h}\right)\left(\int_{M}e^{-\varepsilon_{0}w}d\mu_{h}\right) \leq 4e^{2}.
\end{equation}
\end{proof}

\begin{lma}\label{Sec2: low}
If $p_0>\frac{2n}{n-2}$, then there is a constant $c(n,\Lambda,p_0)>0$ such that
\begin{equation}
\inf_M u_0(x) \geq c(n,\Lambda,p_0).
\end{equation}
\end{lma}

\begin{proof}
By \eqref{Sec2: inequality eqn 5}, we have
\begin{equation}
\frac{4(n-1)}{n-2}\Delta_{h}u_{0} = R_{h}u_{0}-R_{g_0}u_{0}^\frac{n+2}{n-2}.
\end{equation}
Using $R_{g_{0}}\geq-1$, the function $v=u_{0}^{-1}$ satisfies
\begin{equation}\label{Sec2: low eqn 1}
\begin{split}
\Delta_{h} v = {} & -v^2\Delta_{h}u_{0}+2v^3 |\nabla^{h}u|_{h}^2 \\[1mm]
\geq {} & -\frac{(n-2)v}{4(n-1)} \left(R_h -R_{g_0}u_{0}^\frac{4}{n-2}\right) \\
\geq {} & -C(n)\Lambda v-C(n)u_{0}^\frac{4}{n-2}.
\end{split}
\end{equation}
Define two constants $\beta$ and $\beta^{*}$ by
\begin{equation}
\frac{4\beta}{n-2} = p_{0}, \quad \frac{1}{\beta}+\frac{1}{\beta^{*}} = 1.
\end{equation}
Then $p_{0}>\frac{2n}{n-2}$ implies $\beta>\frac{n}{2}$ and $\beta^{*}<\frac{n}{n-2}$. For $p>1$, multiplying $-v^{p-1}$ on both sides of \eqref{Sec2: low eqn 1} and integrating on $(M,h)$, we have
\begin{equation}\label{Sec2:low lemma}
\begin{split}
 \frac{4(p-1)}{p^2}\int_M | \nabla^{h}v^{\frac{p}{2}}|_h^2 d\mu_h = {} & -\int_M (v^{p-1} \Delta_h v) d\mu_h  \\[1mm]
\leq {} & C \int_M v^p d\mu_h +C \int_M (u_{0}^\frac4{n-2} v^p) d\mu_h\\
\leq {} & C \int_M v^p d\mu_h +C \left(\int_M u_{0}^{p_{0}} d\mu_h\right)^{\frac{1}{\beta}} \left(\int_M v^{p\beta^*}d\mu_h \right)^{\frac{1}{\beta^{*}}}
\end{split}
\end{equation}
for some $C(n,\Lambda)$. Combining this with $\|u_{0}\|_{L^{p_{0}}}\leq\Lambda$, the Sobolev inequality and the H\"older inequality, for $p>1$,
\begin{equation}
\begin{split}
\frac{p-1}{p^2} \left(\int_M v^\frac{pn}{n-2} d\mu_h \right)^\frac{n-2}{n}
&\leq C(n,\Lambda,p_0) \left(\int_M v^{p\beta^*}d\mu_h\right)^{1/\beta^*}.
\end{split}
\end{equation}
Recalling $\beta^*<\frac{n}{n-2}$, we may apply the iteration method to conclude that
\begin{equation}\label{Sec2: low-MV}
\|v\|_{L^\infty} \leq C(n,\Lambda,p_0) \|v\|_{L^{\varepsilon_{0}}},
\end{equation}
where $\varepsilon_{0}$ is the constant in Lemma \ref{Sec2: inequality}.

It remains to control $L^{\varepsilon_{0}}$ bound of $v=u_{0}^{-1}$. Thanks to Lemma \ref{Sec2: inequality}, it suffices to establish the positive lower bound of $\|u_{0}\|_{L^{\varepsilon_{0}}}$. We may assume $\varepsilon_0=\frac{2n\sigma }{n-2}$ for some $\sigma<1$. Otherwise, the required estimate follows from $\mathrm{Vol}(M,g_{0})\geq\Lambda^{-1}$ and the H\"older inequality. For any $\gamma\in(0,\sigma)$,
\begin{equation}\label{Sec2: rev--low}
\begin{split}
\Lambda^{-1} \leq {} & \mathrm{Vol}(M,g_{0}) = \int_M u_{0}^\frac{2n}{n-2} d\mu_h
=  \int_M u_{0}^{\frac{\varepsilon_{0}\gamma}{\sigma}+\frac{2n(1-\gamma)}{n-2}} d\mu_h \\[1mm]
\leq {} & \left(\int_M u_{0}^{\varepsilon_0}d\mu_h \right)^\frac{\gamma}{\sigma}
\left(\int_M u_{0}^\frac{2n\sigma(1-\gamma)}{(n-2)(\sigma-\gamma)}d\mu_{h}\right)^\frac{\sigma-\gamma}{\sigma}.
\end{split}
\end{equation}
Since $p_0>\frac{2n}{n-2}$, we may choose $\gamma<\sigma$ such that
\begin{equation}
\frac{2n\sigma(1-\gamma)}{(n-2)(\sigma-\gamma)} = p_0.
\end{equation}
Using $\|u_{0}\|_{L^{p_{0}}}\leq\Lambda$, we obtain
\begin{equation}
\int_M u_{0}^{\varepsilon_0}d\mu_h \geq c(n,\Lambda,p_{0}) > 0,
\end{equation}
which completes the proof.
\end{proof}

Using Lemma \ref{Sec2: low}, we establish the lower bound of the Yamabe flow:
\begin{prop}\label{Sec2: low YF}
Let $g(t)=u(t)^{\frac{4}{n-2}}h$ be a solution of the Yamabe flow \eqref{Yamabe flow}. Suppose that $g(t)$ exists on $M\times[0,T]$. If $p_{0}>\frac{2n}{n-2}$, then there are constants $\hat{T}$ and $c$ depending only on $(n,\Lambda,p_{0})$ such that
\[
\inf_{M\times[0,\hat{T}\wedge T]}u \geq c > 0.
\]
\end{prop}

\begin{proof}
Applying the minimum principle to \eqref{evo-u} and using Lemma \ref{Sec2: low}, for $(x,t)\in M\times[0,T]$,
\begin{equation}
u(x,t) \geq \left(-C(n)\Lambda t+\inf_M u_0^\frac{4}{n-2}\right)^\frac{n-2}{4}
\geq \left(-C(n)\Lambda t+c_0(n,\Lambda,p_0)\right)^\frac{n-2}{4}.
\end{equation}
Choosing $\hat{T}=\frac{c_{0}}{2C(n)\Lambda}$, we complete the proof.
\end{proof}

\subsection{Upper bound of the Yamabe flow}
Next, we will show that if $p_0$ is sufficiently large, then along the Yamabe flow, $u(t)$ will be bounded from above instantaneously.
\begin{prop}\label{Sec2: up YF}
Let $g(t)=u(t)^{\frac{4}{n-2}}h$ be a solution of the Yamabe flow \eqref{Yamabe flow}. Suppose that $g(t)$ exists on $M\times[0,T]$ for some $T\leq1$. If
$\frac{n}{2}+\frac{2n(n+4)}{(n-2)(n+2)}<p_0<\infty$, then for any $\left(\frac{2p_{0}}{n}-\frac{4(n+4)}{(n-2)(n+2)}\right)^{-1}<\alpha<1$, there is a constant $C(\alpha,n,\Lambda,p_{0})$ such that
\[
\sup_{M}u(\cdot,t) \leq Ct^{-\alpha}, \quad \text{for $t\in[0,T]$}.
\]
If $p_{0}=\infty$, then there is a constant $C(n,\Lambda)$ such that
\[
\sup_{M\times[0,T]}u \leq C.
\]
\end{prop}
\begin{proof}
When $\frac{n}{2}+\frac{2n(n+4)}{(n-2)(n+2)}<p_0<\infty$, we split the proof into two steps:

\bigskip
\noindent
{\bf Step 1.} $\int_{0}^{T}\int_{M}u^{\frac{(n+2)p_{0}}{n}-N+1}d\mu_{h}dt\leq C(n,\Lambda,p_{0})$.
\bigskip

It is clear that $p_{0}>N=\frac{n+2}{n-2}$. Multiplying both sides of \eqref{evo-u} by $u^{p_{0}-N}$ and integrating by parts,
\begin{equation}
\begin{split}
& \frac{N}{p_{0}}\frac{\partial}{\partial t}\left(\int_{M}u^{p_{0}}d\mu_{h}\right)
+\frac{4N(n-1)(p_{0}-N)}{(p_{0}-N+1)^{2}}\int_{M}|\nabla^{h}u^{\frac{p_{0}-N+1}{2}}|_{h}^{2}d\mu_{h} \\
= {} & -\frac{n+2}{4}\int_{M}R_{h}u^{p_{0}-N+1}d\mu_{h}
\leq C(n)\Lambda\int_{M}u^{p_{0}-N+1}d\mu_{h}.
\end{split}
\end{equation}
Integrating this on $[0,T]$ and using $\|u_{0}\|_{L^{p_{0}}}\leq\Lambda$, we obtain
\begin{equation}
\begin{split}
& \sup_{t\in[0,T]}\left(\int_{M}u(t)^{p_{0}}d\mu_{h}\right)+\int_{0}^{T}\int_{M}|\nabla^{h}u^{\frac{p_{0}-N+1}{2}}|_{h}^{2}d\mu_{h}dt \\
\leq {} & C(n)\Lambda\int_{0}^{T}\int_{M}u^{p_{0}-N+1}d\mu_{h}dt+C(n,\Lambda,p_{0}).
\end{split}
\end{equation}
Combining this with the Sobolev inequality,
\begin{equation}
\begin{split}
& \sup_{t\in[0,T]}\left(\int_{M}u(t)^{p_{0}}d\mu_{h}\right)+\int_{0}^{T}\left(\int_{M}u^{\frac{n(p_{0}-N+1)}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}}dt \\
\leq {} & C(n)\Lambda\int_{0}^{T}\int_{M}u^{p_{0}-N+1}d\mu_{h}dt+C(n,\Lambda,p_{0}).
\end{split}
\end{equation}
Using $N>1$ and the Young inequality,
\begin{equation}
C(n)\Lambda\int_{0}^{T}\int_{M}u^{p_{0}-N+1}d\mu_{h}dt \leq \frac{1}{2}\sup_{t\in[0,T]}\left(\int_{M}u(t)^{p_{0}}d\mu_{h}\right)+C(n,\Lambda,p_{0}).
\end{equation}
Then
\begin{equation}\label{Sec2: up YF eqn 2}
\sup_{t\in[0,T]}\left(\int_{M}u(t)^{p_{0}}d\mu_{h}\right)+\int_{0}^{T}\left(\int_{M}u^{\frac{n(p_{0}-N+1)}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}}dt
\leq C(n,\Lambda,p_{0}).
\end{equation}
By the H\"older inequality,
\begin{equation}
\begin{split}
& \int_{0}^{T}\int_{M}u^{\frac{(n+2)p_{0}}{n}-N+1}d\mu_{h}dt
= \int_{0}^{T}\int_{M}u^{(p_{0}-N+1)+\frac{2p_{0}}{n}}d\mu_{h}dt \\
\leq {} & \left(\int_{0}^{T}\left(\int_{M}u^{\frac{n(p_{0}-N+1)}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}}dt\right)
\sup_{t\in[0,T]}\left(\int_{M}u^{p_{0}}(t)d\mu_{h}\right)^{\frac{2}{n}}
\leq C(n,\Lambda,p_{0}).
\end{split}
\end{equation}

\bigskip
\noindent
{\bf Step 2.} $\sup_{M}u(\cdot,t) \leq C(\alpha,n,\Lambda,p_{0})t^{-\alpha}$.
\bigskip

Define $v=t^{\alpha}u$. For $p>\max(\alpha^{-1},2)$, multiplying both sides of \eqref{evo-u} by $t^{\alpha}v^{p-1}$,
\begin{equation}
t^{\alpha}v^{p-1}\frac{\partial u^{N}}{\partial t}-N(n-1)v^{p-1}\Delta_{h}v = -\frac{n+2}{4}R_{h}v^{p}.
\end{equation}
We compute the first term of the left hand side:
\begin{equation}
\begin{split}
t^{\alpha}v^{p-1}\frac{\partial u^{N}}{\partial t}
= {} &  t^{\alpha p}u^{p-1}\frac{\partial u^{N}}{\partial t} = \frac{Nt^{\alpha p}}{N+p-1}\frac{\partial u^{N+p-1}}{\partial t} \\
= {} & \frac{N}{N+p-1}\frac{\partial}{\partial t}\left(t^{\alpha p}u^{N+p-1}\right)-\frac{\alpha Np}{N+p-1}t^{\alpha p-1}u^{N+p-1} \\
= {} & \frac{N}{N+p-1}\frac{\partial}{\partial t}\left(v^{p}u^{N-1}\right)-\frac{\alpha Np}{N+p-1}v^{p-\frac{1}{\alpha}}u^{N+\frac{1}{\alpha}-1}.
\end{split}
\end{equation}
Then
\begin{equation}
\begin{split}
& \frac{N}{N+p-1}\frac{\partial}{\partial t}\left(v^{p}u^{N-1}\right)-N(n-1)v^{p-1}\Delta_{h}v \\[1mm]
= {} & -\frac{n+2}{4}R_{h}v^{p}+\frac{\alpha Np}{N+p-1}v^{p-\frac{1}{\alpha}}u^{N+\frac{1}{\alpha}-1} \\[1mm]
\leq {} & C(n)\Lambda v^{p}+C(\alpha,n)v^{p-\frac{1}{\alpha}}u^{N+\frac{1}{\alpha}-1}.
\end{split}
\end{equation}
Integrating both sides on $(M,h)$,
\begin{equation}
\begin{split}
& \frac{N}{N+p-1}\frac{\partial}{\partial t}\left(\int_{M}v^{p}u^{N-1}d\mu_{h}\right)+\frac{4N(n-1)(p-1)}{p^{2}}\int_{M}|\nabla^{h}v^{\frac{p}{2}}|_{h}^{2}d\mu_{h} \\
\leq {} & C(\alpha,n,\Lambda)\int_{M}\left(v^{p}+v^{p-\frac{1}{\alpha}}u^{N+\frac{1}{\alpha}-1}\right)d\mu_{h}.
\end{split}
\end{equation}
Recalling that $v(0)=0$, the above inequality shows
\begin{equation}
\begin{split}
& \sup_{t\in[0,T]}\left(\int_{M}v^{p}(t)u^{N-1}(t)d\mu_{h}\right)+\int_{0}^{T}\int_{M}|\nabla^{h}v^{\frac{p}{2}}|_{h}^{2}d\mu_{h}dt \\
\leq {} & C(\alpha,n,\Lambda)p\int_{0}^{T}\int_{M}\left(v^{p}+v^{p-\frac{1}{\alpha}}u^{N+\frac{1}{\alpha}-1}\right)d\mu_{h}dt.
\end{split}
\end{equation}
By the Sobolev inequality,
\begin{equation}
\begin{split}
& \sup_{t\in[0,T]}\left(\int_{M}v^{p}(t)u^{N-1}(t)d\mu_{h}\right)+\int_{0}^{T}\left(\int_{M}v^{\frac{np}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}}dt \\
\leq {} & C(\alpha,n,\Lambda)p\int_{0}^{T}\int_{M}\left(v^{p}+v^{p-\frac{1}{\alpha}}u^{N+\frac{1}{\alpha}-1}\right)d\mu_{h}dt.
\end{split}
\end{equation}
Since $T\leq1$, then $v=t^{\alpha}u\leq u$, and so
\begin{equation}
\begin{split}
& \sup_{t\in[0,T]}\left(\int_{M}v^{p+N-1}(t)d\mu_{h}\right)+\int_{0}^{T}\left(\int_{M}v^{\frac{np}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}}dt \\
\leq {} & C(\alpha,n,\Lambda)p\int_{0}^{T}\int_{M}\left(v^{p}+v^{p-\frac{1}{\alpha}}u^{N+\frac{1}{\alpha}-1}\right)d\mu_{h}dt.
\end{split}
\end{equation}
Using $N>1$, we obtain
\begin{equation}
\begin{split}
& \sup_{t\in[0,T]}\left(\int_{M}v^{p}(t)d\mu_{h}\right)+\int_{0}^{T}\left(\int_{M}v^{\frac{np}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}}dt \\
\leq {} & Cp\int_{0}^{T}\int_{M}\left(v^{p}+v^{p-\frac{1}{\alpha}}u^{N+\frac{1}{\alpha}-1}\right)d\mu_{h}dt+C
\end{split}
\end{equation}
for some $C(\alpha,n,\Lambda)$. Combining this with the H\"older inequality,
\begin{equation}
\begin{split}
& \int_{0}^{T}\int_{M}v^{\frac{(n+2)p}{n}}d\mu_{h}dt
= \int_{0}^{T}\int_{M}v^{p+\frac{2p}{n}}d\mu_{h}dt \\
\leq {} & \left(\int_{0}^{T}\left(\int_{M}v^{\frac{np}{n-2}}d\mu_{h}\right)^{\frac{n-2}{n}}dt\right)
\sup_{t\in[0,T]}\left(\int_{M}v^{p}(t)d\mu_{h}\right)^{\frac{2}{n}} \\
\leq {} & \left(Cp\int_{0}^{T}\int_{M}\left(v^{p}+v^{p-\frac{1}{\alpha}}u^{N+\frac{1}{\alpha}-1}\right)d\mu_{h}dt+C\right)^{\frac{n+2}{n}}.
\end{split}
\end{equation}
Then
\begin{equation}\label{Sec2: up YF eqn 1}
\begin{split}
& \left(\int_{0}^{T}\int_{M}v^{\frac{(n+2)p}{n}}d\mu_{h}dt\right)^{\frac{n}{n+2}}
\leq Cp\int_{0}^{T}\int_{M}(v^{p}+1)(u^{N+\frac{1}{\alpha}-1}+1)d\mu_{h}dt+C \\
\leq {} & Cp\left(\left(\int_{0}^{T}\int_{M}v^{\beta p}d\mu_{h}dt\right)^{\frac{1}{\beta}}+1\right)
\left(\left(\int_{0}^{T}\int_{M}u^{\frac{(n+2)p_{0}}{n}-N+1}d\mu_{g}dt\right)^{\frac{1}{\beta^{*}}}+1\right),
\end{split}
\end{equation}
for some $C(\alpha,n,\Lambda)$, where
\begin{equation}
\frac{1}{\beta}+\frac{1}{\beta^{*}} = 1, \quad \beta^{*} = \frac{\frac{(n+2)p_{0}}{n}-N+1}{N+\frac{1}{\alpha}-1}.
\end{equation}
Combining \eqref{Sec2: up YF eqn 1} with Step 1,
\begin{equation}
\left(\int_{0}^{T}\int_{M}v^{\frac{(n+2)p}{n}}d\mu_{h}dt\right)^{\frac{n}{n+2}}
\leq  Cp\left(\int_{0}^{T}\int_{M}v^{\beta p}d\mu_{h}dt\right)^{\frac{1}{\beta}}+Cp
\end{equation}
for some $C(\alpha,n,\Lambda,p_{0})$. Recalling $\alpha>\left(\frac{2p_{0}}{n}-\frac{4(n+4)}{(n-2)(n+2)}\right)^{-1}$ and $N=\frac{n+2}{n-2}$, we obtain
\begin{equation}
\beta^{*} = \frac{\frac{(n+2)p_{0}}{n}-N+1}{N+\frac{1}{\alpha}-1}
= \frac{\frac{(n+2)p_{0}}{n}-\frac{4}{n-2}}{\frac{1}{\alpha}+\frac{4}{n-2}} > \frac{n+2}{2},
\end{equation}
which implies $\beta <\frac{n+2}{n}$. Applying the iteration method and Step 1, we obtain
\begin{equation}
\sup_{M\times[0,T]}(t^{\alpha}u) = \sup_{M\times[0,T]}v
\leq C(\alpha,n,\Lambda,p_{0})\left(\int_{0}^{T}\int_{M}v^{p_{0}}d\mu_{h}dt\right)^{\frac{1}{p_{0}}}.
\end{equation}
Combining this with \eqref{Sec2: up YF eqn 2}, $v\leq u$ and $T\leq1$,
\begin{equation}
\sup_{M\times[0,T]}(t^{\alpha}u) \leq C(\alpha,n,\Lambda,p_{0})\left(\int_{0}^{T}\int_{M}u^{p_{0}}d\mu_{h}dt\right)^{\frac{1}{p_{0}}}
\leq C(\alpha,n,\Lambda,p_{0}).
\end{equation}

\bigskip

When $p_{0}=\infty$, applying the maximum principle to \eqref{evo-u} and using $T\leq1$, for any $(x,t)\in M\times[0,T]$,
\begin{equation}
u(x,t) \leq \left(C(n)\Lambda t+\sup_M u_0^\frac{4}{n-2}\right)^\frac{n-2}{4}
\leq C(n,\Lambda).
\end{equation}
\end{proof}

\subsection{More estimates along the Yamabe flow}
\begin{prop}\label{Sec2: YFC1-estimates}
Let $g(t)=u(t)^{\frac{4}{n-2}}h$ be a solution of the Yamabe flow \eqref{Yamabe flow}. If $p_0>\frac{n}{2}+\frac{2n(n+4)}{(n-2)(n+2)}$, then there is a constant $T_{0}(n,\Lambda,p_{0})$ such that $g(t)$ exists on $[0,T_{0}]$. Moreover, for all $[a,T_{0}]\subset(0,T_{0}]$, there is a constant $\lambda(a,n,\Lambda,p_0)>1$ such that
\begin{equation}\label{Sec2: YFC1-estimates eqn 3}
\left\{
\begin{array}{ll}
\lambda^{-1}h \leq g(t)\leq \lambda h;\\[1mm]
|\nabla^h g(t)|_{h} \leq \lambda;\\[1mm]
-\delta \leq R_{g(t)} \leq \lambda
\end{array}
\right.
\end{equation}
on $M\times [a,T_{0}]$.
\end{prop}
\begin{proof}
In the following, all norms are with respect to metric $h$. Denote the maximal existence time of the Yamabe flow \eqref{Yamabe flow} by $T_{\textrm{max}}$. First, we establish the lower bound of $T_{\textrm{max}}$. By \cite[Lemma 2.2]{Anderson1990}, for any $\beta\in(0,1)$, the $C^{1,\beta}$ harmonic radius of $h$ is bounded from below by $r(\beta,n,\Lambda)$. Within harmonic radius, the Laplacian operator of $h$ can be expressed as $\Delta_{h} = h^{ij}\partial_i \partial_j$, and then \eqref{evo-u} can be written as
\begin{equation}\label{harmonic-YF}
\frac{\partial u}{\partial t} -(n-1)u^{-\frac{4}{n-2}}h^{ij}\partial_i \partial_j u = -\frac{n-2}{4} R_h u^\frac{n-6}{n-2}.
\end{equation}
Write $T_{0}=\min(\hat{T},1)$, where $\hat{T}$ is the constant in Proposition \ref{Sec2: low YF}. We will show $T_{\textrm{max}}>T_{0}$. If $T_{\textrm{max}}\leq T_{0}$, then by Proposition \ref{Sec2: low YF} and \ref{Sec2: up YF}, for any $a\in(0,T_{\textrm{max}})$,
\begin{equation}\label{Sec2: YFC1-estimates eqn 4}
C^{-1}(a,n,\Lambda,p_0) \leq \inf_{M\times[\frac{a}{3},T_{\textrm{max}})} u \leq \sup_{M\times[\frac{a}{3},T_{\textrm{max}})} u \leq C(a,n,\Lambda,p_0).
\end{equation}
This shows the equation \eqref{harmonic-YF} is uniformly parabolic on $[\frac{a}{3},T_{\textrm{max}})$ with $L^\infty$ inhomogeneous term. Since $h$ is uniformly equivalent to the Euclidean metric $g_{\mathrm{euc}}$ in the harmonic coordinate system, we obtain uniform $C^{\delta,\frac{\delta}{2}}$ estimate on a slightly smaller ball by \cite{KrylovSafonov1980} for some $\delta(a,n,\Lambda,p_0)$. The covering argument shows
\begin{equation}\label{Sec2: YFC1-estimates eqn 1}
\|u\|_{C^{\delta,\frac{\delta}{2}}(M\times[\frac{a}{2},T_{\textrm{max}}))} \leq C(a,n,\Lambda,p_0).
\end{equation}
Combining this with the parabolic Schauder estimates \cite{Friedman1964}, we obtain the higher order estimates of $u$ on $[a,T_{max})$, which contradicts with the definition of $T_{\textrm{max}}$. Then we obtain $T_{\textrm{max}}>T_{0}$.

Next, we establish the required estimates. The estimate $\lambda^{-1}h\leq g(t)\leq\lambda h$ follows from \eqref{Sec2: YFC1-estimates eqn 4}. Within $C^{1,\beta}$ harmonic radius, $h$ and $g_{\mathrm{euc}}$ are uniformly comparable in $C^{1,\beta}$. Using \eqref{Sec2: YFC1-estimates eqn 1}, $|R_{h}|\leq n\Lambda$, the parabolic $L^{p}$ estimate \cite{Friedman1964} and Sobolev embedding theorem, for $p>1$,
\begin{equation}\label{Sec2: YFC1-estimates eqn 2}
\|u\|_{C^{1}} \leq C(a,n,\Lambda,\delta,p_0), \quad \|u\|_{W^{2,p}} \leq C(p,a,n,\Lambda,\delta,p_0),
\end{equation}
on $[\frac{a}{2},T_{0}]$. This shows $|\nabla^h g(t)|_{h}\leq\lambda$.

The scalar curvature $R_{g(t)}$ satisfies (see \cite[Lemma 2.2]{Chow1992}),
\begin{equation}\label{evo-R}
\left(\frac{\partial}{\partial t} -(n-1)\Delta_{g(t)}\right) R_{g(t)}=R_{g(t)}^2.
\end{equation}
The lower bound $R_{g(t)}\geq-\delta$ follows from the minimum principle. For the upper bound of $R_{g(t)}$ on $[a,T_{0}]$, \eqref{Sec2: YFC1-estimates eqn 2} shows for $p>1$ and $t\in [\frac{a}{2},T_0]$,
\begin{equation}
\|R_{g(t)}\|_{L^{p}} \leq C(p,a,n,\Lambda,\delta,p_0).
\end{equation}
Combining this with \eqref{Sec2: YFC1-estimates eqn 1}, the equation \eqref{evo-R} has $C^{\delta}$ parabolic coefficient and $L^{\frac{p}{2}}$ inhomogeneous term for any $p>1$ in the local coordinate system. Then $R_{g(t)}\leq\lambda$ on $M\times [a,T_0]$ follows from the parabolic $L^{p}$ estimate \cite{Friedman1964} and Sobolev embedding theorem.
\end{proof}

Lastly, we will establish the weak convergence of $g(t)$ as $t\to 0$.
\begin{prop}\label{Prop: Lp}
Let $g(t)=u(t)^{\frac{4}{n-2}}h$ be a solution of the Yamabe flow \eqref{Yamabe flow} on $[0,T_{0}]$, where $T_{0}$ is the constant in Proposition \ref{Sec2: YFC1-estimates}. If
$\frac{n}{2}+\frac{2n(n+4)}{(n-2)(n+2)}<p_{0}<\infty$, then for any $\left(\frac{2p_{0}}{n}-\frac{4(n+4)}{(n-2)(n+2)}\right)^{-1}<\alpha<1$, there is a constant $C(\alpha,n,\Lambda,p_0)$ such that
\begin{equation}
\big|\mathrm{Vol}(M,g(t))-\mathrm{Vol}(M,g_0)\big| \leq Ct^{1-\alpha}.
\end{equation}
If $p_{0}=\infty$, then for any $p>0$, there is a constant $C(p,n,\Lambda)$ such that
\begin{equation}
\int_M \big|g(t)-g(0)\big|_h^p d\mu_h \leq C(t+t^{p}).
\end{equation}
\end{prop}
\begin{proof}
When $\frac{n}{2}+\frac{2n(n+4)}{(n-2)(n+2)}<p_{0}<\infty$, along the Yamabe flow \eqref{Yamabe flow}, using $R_{g(t)}\geq-\delta\geq-1$ (see \eqref{Sec2: YFC1-estimates eqn 3}), we obtain
\begin{equation}
\frac{d}{dt}\mathrm{Vol}(M,g(t)) = -\frac{n}{2}\int_{M}R_{g(t)}d\mu_{g(t)} \leq \frac{n}{2}\mathrm{Vol}(M,g(t)).
\end{equation}
It then follows that
\begin{equation}\label{Prop3: vol-up}
\mathrm{Vol}(M,g(t)) \leq e^{\frac{nt}{2}}\mathrm{Vol}(M,g_0)
\leq \mathrm{Vol}(M,g_0)+C(n)t\mathrm{Vol}(M,g_0).
\end{equation}
By volume comparison theorem, we have $\mathrm{Vol}(M,h)\leq C(n,\Lambda)$. Combining this with $\|u_{0}\|_{L^{p_{0}}}\leq\Lambda$ and $p_{0}>\frac{2n}{n-2}$, we obtain
\begin{equation}\label{Prop3: vol-up 1}
\mathrm{Vol}(M,g_0) = \int_{M}u_{0}^{\frac{2n}{n-2}}d\mu_{h} \leq C(n,\Lambda,p_{0})
\end{equation}
and so
\begin{equation}\label{Prop3: vol-up 2}
\mathrm{Vol}(M,g(t)) \leq \mathrm{Vol}(M,g_0)+C(n,\Lambda,p_{0})t .
\end{equation}
It suffices to estimate the lower bound. By Proposition \ref{Sec2: low YF}, \ref{Sec2: up YF} and $R_{g(t)}\geq-\delta\geq-1$ (see \eqref{Sec2: YFC1-estimates eqn 3}),
\begin{equation}
\begin{split}
\frac{d}{dt}\mathrm{Vol}(M,g(t))
&=-\frac{n}{2}\int_M  u^{N+1}R_{g(t)}d\mu_{h}\\
&\geq -\frac{C}{t^{\alpha}} \int_M u^N(R_{g(t)}+1) d\mu_h\\
&=-\frac{C}{t^{\alpha}} \int_M u^N R_{g(t)} d\mu_h-\frac{C}{t^{\alpha}}\int_M u^{-1}d\mu_{g(t)}\\
&\geq -\frac{C}{t^{\alpha}} \int_M \left(R_hu-C(n)\Delta_h u \right) d\mu_h-\frac{C}{t^{\alpha}}\mathrm{Vol}(M,g(t))\\
&\geq -\frac{C}{t^{\alpha}} \left(\mathrm{Vol}(M,g(t))\right)^\frac{n-2}{2n}-\frac{C}{t^{\alpha}}\mathrm{Vol}(M,g(t)),
\end{split}
\end{equation}
where $C(\alpha,n,\Lambda,p_0)>0$. Using \eqref{Prop3: vol-up 1} and \eqref{Prop3: vol-up 2}, we obtain $\mathrm{Vol}(M,g(t))\leq C(n,\Lambda,p_{0})$, and then
\begin{equation}
\frac{d}{dt}\mathrm{Vol}(M,g(t)) \geq -\frac{C(\alpha,n,\Lambda,p_{0})}{t^{\alpha}}.
\end{equation}
This implies
\begin{equation}\label{Prop3: vol-low}
\mathrm{Vol}(M,g(t)) \geq \mathrm{Vol}(M,g_0) -C(\alpha,n,\Lambda,p_0)t^{1-\alpha}
\end{equation}
Combining \eqref{Prop3: vol-up 2} and \eqref{Prop3: vol-low}, we obtain
\begin{equation}
\big|\mathrm{Vol}(M,g(t))-\mathrm{Vol}(M,g_0)\big| \leq (\alpha,n,\Lambda,p_{0})t^{1-\alpha}.
\end{equation}

\bigskip

When $p_{0}=\infty$, for $t\in[0,T_{0}]$, Proposition \ref{Sec2: low YF} and \ref{Sec2: up YF} show
\begin{equation}\label{Prop3: eqn 1}
C(n,\Lambda)^{-1} \leq u(t) \leq C(n,\Lambda), \quad C(n,\Lambda)^{-1}h \leq g(t) \leq C(n,\Lambda)h.
\end{equation}
Using $R_{g(t)}\geq-\delta\geq-1$ (see \eqref{Sec2: YFC1-estimates eqn 3}), we compute
\begin{equation}
\begin{split}
\int_M|R_{g(t)}| d\mu_h \leq {} & \int_M(R_{g(t)}+1) d\mu_h+C \\
\leq {} & C\int_{M}u^{N}(R_{g(t)}+1) d\mu_h+C \\
\leq {} & C\int_{M}\left(R_{h}u-C(n)\Delta_{h}u\right)d\mu_h+C \\[1.5mm]
\leq {} & C(n,\Lambda).
\end{split}
\end{equation}
Therefore,
\begin{equation}\label{L1}
\begin{split}
\int_M \big|g(t)-g(0)\big|_h d\mu_h =& \int_M\left| \int^t_0 \frac{\partial g}{\partial s} ds  \right|_{h} d\mu_h \\
\leq & C\int^t_0\int_M  |R_{g(s)}|d\mu_hds \\
\leq & C(n,\Lambda)t.
\end{split}
\end{equation}
For $p\in(0,1)$, the Holder inequality shows
\begin{equation}
\int_M \big|g(t)-g(0)\big|_h^p d\mu_h \leq C\left(\int_M \big|g(t)-g(0)\big|_h d\mu_h\right)^{p} = C(p,n,\Lambda)t^{p}.
\end{equation}
For $p>1$, using \eqref{Prop3: eqn 1},
\begin{equation}
\int_M \big|g(t)-g(0)\big|_h^p d\mu_h
\leq C\int_M \big|g(t)-g(0)\big|_h d\mu_h \leq C(p,n,\Lambda)t.
\end{equation}
\end{proof}

\section{Stability on torus}\label{stability-T}
In this section, we will give the proof of Theorem~\ref{Sec1: maintheorem} and \ref{Sec1: maintheorem2}. Indeed, we will consider a slightly more general case where $M$ is a closed manifold with non-positive Yamabe invariant. Recall that
\begin{equation}
\sigma(M)=\sup\left\{ \mathcal{Y}(M,[g]): [g] \textit{ is a conformal class of metrics on } M\right\},
\end{equation}
where
\begin{equation}
\mathcal{Y}(M,[g_0])=\inf\left\{ \int_M R(g)\;d\mu_g: g\in[g_0],\; \mathrm{Vol}(M,g)=1\right\}.
\end{equation}

It is well known that if a smooth metric on a
compact manifold attains the Yamabe invariant and if the invariant is nonpositive, then the metric is Einstein \cite{Schoen1989}. In particular, if $g$ is a smooth metric with $R_g\geq 0$ on $M$, then $\text{\rm Ric}(g)\equiv 0$.
\begin{rem}\label{torus-Yama}
For $n\geq 3$, it is in general difficult to compute the Yamabe invariant. One can usually show that $\sigma(M)=0$ by proving the non-existence of metrics with positive scalar curvature. For example, Schoen-Yau \cite{SchoenYau1979,SchoenYau1979-2} and Gromov-Lawson \cite{GromovLawson1980} showed that torus cannot admit metric with positive scalar curvature and hence $\sigma(\mathbb{T}^n)=0$.
\end{rem}
 In the following, we will consider a sequence of $g_{i,0}=u_i^\frac4{n-2}h_i$ on $M^n$ with $\sigma(M)\leq 0$  where $h_i$ and $g_{i,0}$ satisfy assumptions {\bf (A)} and {\bf (B)} with $\delta=i^{-1}\rightarrow 0$ and $p_0$ sufficiently large.  Our goal is to show that $g_{i,0}$ will converge to a Ricci-flat metric on $M$ in a weak sense. Let $g_{i}(t)$ be the Yamabe flow with initial metric $g_{i,0}$.

\begin{thm}\label{Sec3: Conv-1}
Let $M^n$ be a closed manifold with $\sigma(M)\leq 0$.
Suppose $g_{i,0}$ is a sequence of metrics on $M$ such that
\begin{enumerate}\setlength{\itemsep}{1mm}
\item[(a)] $g_{i,0}=u_i^\frac4{n-2}h_i$ for some metric $h_i$ on $M_i$ satisfying assumption {\bf (A)};
\item[(b)] $g_{i,0}$ satisfies assumption {\bf (B)} for $\delta=i^{-1}\rightarrow 0$ and $p_0>\frac{n}{2}+\frac{2n(n+4)}{(n-2)(n+2)}$.
\end{enumerate}
Then the Yamabe flow $g_{i}(t)$ with initial metric $g_{i,0}$ exists on $M\times [0,T_{0}]$, where $T_{0}(n,\Lambda,p_{0})$ is the constant in Proposition \ref{Sec2: YFC1-estimates}. Moreover, there is a sequence of diffeomorphisms $\Phi_i$ of $M$ and a Ricci-flat metric $g_\infty$ on $M$ such that after passing to subsequence, $\Phi_{i}^{*}g_{i}(t)$ converges to $g_\infty$ on $M$ in $C^{0}_{loc}(M\times (0,T_0])$.
\end{thm}

\begin{proof}
In the proof, for notational convenience all convergent sequence means convergent subsequence. By \cite{Anderson1990}, we can find a sequence of diffeomorphism $\Phi_i$ such that $\Phi_i^*h_i$ converges to some $C^{1,\beta}$ metric $h_\infty$ on $M$ in $C^{1,\gamma}$ for all $\gamma<\beta<1$ after passing to subsequence.
For notational convenience, we will pull-back all $g_{i,0}$ by $\Phi_i$ and omit $\Phi_i^*$.

Applying Proposition \ref{Sec2: YFC1-estimates} to each $g_{i}(t)$, we obtain a sequence of Yamabe flows $g_i(t)$ on $M^{n}\times[0,T_{0}]$ which is uniform bounded in $C^1$ on any $[a,T_{0}]\subset(0,T_{0}]$. Our goal is to show that $g_i(t)$ converges to a Ricci-flat metric on $M$ weakly. In case $h_i$ is uniformly regular in $C^\infty$, it is not difficult to see that $g_i(t)$ converges to a limiting solution of the Yamabe flow $g_\infty(t)$ for $t>0$ after passing to subsequence. Since $h_i$ is only mildly regular, it is not clear to us whether $g_\infty(t)$ exists smoothly although the flow is expected to be static. Instead, we will regularize it further using the Ricci flow.

\begin{claim}\label{Sec3: YF-limit1}
There is a Ricci-flat metric $g_{\infty}$ on $M$ such that $g_{i}(T_0)\rightarrow g_{\infty}$ in $C^0(M)$ as $i\rightarrow\infty$ after passing to subsequence and pulling back by a sequence of diffeomorphism.
\end{claim}
\begin{proof}[Proof of Claim~\ref{Sec3: YF-limit1}]
Since $g_i(T_0)$ is uniformly bounded in $C^1$ with respect to $h_i$, $g_i(T_0)$ is uniformly equivalent to $h_i$ and $h_i\rightarrow h_\infty$ in $C^{1,\gamma}$. By passing to subsequence, we may assume $g_i(T_0)\rightarrow g_\infty(T_0)$ in $C^{\gamma}$ for all $\gamma\in(0,1)$. In particular, for $\varepsilon>0$ sufficiently small, there is $N\in \mathbb{N}$ such that for all $i>N$,
\begin{equation}
(1-\varepsilon)\bar h\leq g_i(T_0)\leq (1+\varepsilon) \bar h,\quad |\nabla^{\bar h}g_i(T_0)|_{\bar h} \leq C(\bar{h},n,\Lambda,p_{0}),
\end{equation}
where $\bar h=g_N(T_0)$. Note that the constant $C$ depends on $\bar h$, but is independent of $i$. In the following, all constants may possibly depend on $\bar h$, but are independent of $i$.

Let $\tilde g_i(s)$ be Ricci flow starting from $g_{i}(T_{0})$. By \cite[Lemma 4.3]{ShiTam2018} with $\delta=0$ (see also \cite{Simon2002}), there is a constant $S_{0}(\bar{h},n,\Lambda,p_{0})>0$ such that $\tilde g_i(s)$ exists on $M^{n}\times[0,S_{0}]$ and satisfies
\begin{equation}\label{Sec3: Conv-1 eqn 1}
|\text{\rm Rm}(\tilde g_i(s))|\leq C(\bar{h},n,\Lambda,p_{0})s^{-1/2}
\end{equation}
provided that $\varepsilon$ is sufficiently small depending only on $n$. In particular, $\tilde g_i(s)$ is uniformly equivalent to $\tilde g_i(0)$ and hence $h_i$ for all $s\in [0,S_0]$ by integration on time. Here we have used the estimate of the Ricci-Deturck flow with reference metric $\bar h$ from \cite[Lemma 4.3]{ShiTam2018} and the fact that the Ricci-Deturck flow is diffeomorphic to the Ricci flow with the same initial data.

Since $\mathrm{inj}(h_i)$ is bounded from below, we have a uniform lower bound on $\mathrm{inj}(\tilde g_i(s))$ thanks to the uniform equivalence of metrics. Together with Shi's estimates \cite{Shi1989} (see also \cite[Theorem 4.3]{Simon2002}) and Hamilton's compactness \cite{Hamilton1995}, we can pass $\tilde g_{i}(s)\rightarrow \tilde g_\infty(s)$ in $C^\infty$ Cheeger-Gromov sense on $M\times (0,S_{0}]$ after passing to subsequence. More precisely, there is a sequence of diffeomorphism $\Psi_i$ of $M$ such that $\Psi_i^*\tilde g_i(s)$ converges to $g_\infty(s)$ in $C^\infty_{loc}\left(M\times (0,S]\right)$. As usual, we will pull-back each metrics by  $\Psi_i$ and therefore will omit $\Psi_i^*$. We note that in this way, $\{\Psi_i^*h_i\}_{i=1}^\infty$ are not necessarily compact in $C^{1,\gamma}$ topology anymore due to the additional pull-back of diffeomorphism but it will remain uniformly equivalent to the Ricci flow $\tilde g_i(s)$ thanks to the curvature estimates in \eqref{Sec3: Conv-1 eqn 1}.

It is well-known that Ricci flow preserved the lower bound of the scalar curvature, we have
\begin{equation}\label{Sec3: Conv-1 eqn 2}
R_{\tilde g_i(s)} \geq R_{\tilde g_i(0)} = R_{g_i(T_{0})} \geq -i^{-1},
\end{equation}
where we used \eqref{Sec2: YFC1-estimates eqn 3} in the last inequality. Letting $i\rightarrow\infty$, we obtain $R_{\tilde g_\infty(s)}\geq 0$ for $s\in (0,S_{0}]$. By the assumption of $\sigma(M)\leq 0$ and uniqueness of the Ricci  flow, $\tilde g_\infty(s)\equiv g_{\infty}$ on $M$ for $s\in (0,S_{0}]$ for some Ricci-flat metric $g_\infty$. Combining \eqref{Sec3: Conv-1 eqn 1} and the equation of Ricci-flow
\begin{equation}
\frac{\partial \tilde{g}_{kl}}{\partial s} =-2\text{\rm Ric}(\tilde{g})_{kl}
\end{equation}
we obtain
\begin{equation}
\big|\tilde{g}_{i}(s)-g_{i}(T_{0})\big|_{\tilde g_i(S_0)}
= \big|\tilde{g}_{i}(s)-\tilde{g}_{i}(0)\big|_{\tilde g_i(S_0)} \leq C(\bar{h},n,\Lambda,p_{0})s^{\frac{1}{2}}.
\end{equation}
Combining this with $\tilde g_{i}(s)\rightarrow g_{\infty}$ in $C^{\infty}_{\mathrm{loc}}$ on $(0,S_{0}]$, we conclude $g_{i}(T_0)=\tilde g_i(0)\rightarrow g_{\infty}$ in $C^0(M)$ as $i\rightarrow\infty$.
\end{proof}

Next, we claim that $g_i(t)$ converges to the same Ricci-flat metric $g_{\infty}$.
\begin{claim}\label{Sec3: YF-limit}
For all $t\in (0,T_{0}]$, $g_{i}(t)\rightarrow g_{\infty}$ in $C^0$ as $i\rightarrow\infty$, where $g_{\infty}$ is the Ricci-flat metric obtain from Claim~\ref{Sec3: YF-limit1}.
\end{claim}
\begin{proof}
By Proposition \ref{Sec2: YFC1-estimates}, for $a\in (0,T_{0}]$, $g_i(t)$ is uniformly equivalent to $g_i(T_0)$ on $[a,T_{0}]$ and $-i^{-1}\leq R_{g_{i}(t)}\leq\lambda$. Using \eqref{evo-R}, we compute
\begin{equation}
\begin{split}
 \frac{\partial}{\partial t}\left(\int_{M} R_{g_{i}(t)}d\mu_{g_{i}(t)}\right) =& \left(1-\frac{n}{2}\right)\int_{M} R^2_{g_{i}(t)} d\mu_{g_i(t)}\\
\geq {} & -C(a,n,\Lambda,p_{0})\left(\int_{M} R_{g_{i}(t)} d\mu_{g_{i}(t)}+i^{-1}\right).
\end{split}
\end{equation}
Hence, for all $t\in [a,T_{0}]$,
\begin{equation}\label{Sec3: R-flat-YF}
\int_{M} R_{g_i(t)}d\mu_{g_i(t)}\leq C(a,n,\Lambda,p_{0})\left(\int_{M} R_{g_i(T_{0})}d\mu_{g_i(T_{0})}+i^{-1}\right).
\end{equation}
We now estimate the integral of scalar curvature on the right hand side. We will make use of the smooth convergence of the Ricci-flow.
The scalar curvature $R_{\tilde{g}_{i}(s)}$ satisfies (see e.g., \cite[(2.5.5)]{Topping2006})
\begin{equation}
\left(\frac{\partial}{\partial s}-\Delta_{\tilde{g}_{i}(s)}\right)R_{\tilde{g}_{i}(s)} = 2\left|\text{\rm Ric}(\tilde g_i(s))\right|_{\tilde{g}_{i}(s)}^2.
\end{equation}
By \eqref{Sec3: Conv-1 eqn 1} and \eqref{Sec3: Conv-1 eqn 2},
\begin{equation}
-i^{-1}\leq R_{\tilde g_i(s)}\leq C(\bar{h},n,\Lambda,p_{0})s^{-\frac{1}{2}}.
\end{equation}
Then along the Ricci flow, we have
\begin{equation}
\begin{split}
 \frac{\partial}{\partial s} \left( \int_{M} R_{\tilde g_i(s)}d\mu_{\tilde g_i(s)}\right)
= &\int_{M}\left(2\left|\text{\rm Ric}(\tilde g_i(s))\right|_{\tilde g_i(s)}^2-|R_{\tilde g_i(s)}|^2\right)d\mu_{\tilde g_i(s)}\\
\geq  & -C(a,\bar h,n,\Lambda,p_{0})s^{-\frac{1}{2}}\left(\int_{M} R_{\tilde g_i(s)}d\mu_{\tilde g_i(s)}+i^{-1}\right).
\end{split}
\end{equation}
This shows
\begin{equation}
\begin{split}
 \int_{M} R_{g_i(T_{0})}d\mu_{g_i(T_{0})} =& \int_{M} R_{\tilde g_i(0)}d\mu_{\tilde g_i(0)} \\
\leq {} & C(a,\bar h,n,\Lambda,p_{0})\left(\int_{M} R_{\tilde g_i(S_{0})}d\mu_{\tilde g_i(S_{0})}+i^{-1}\right).
\end{split}
\end{equation}
Using \eqref{Sec3: R-flat-YF}, for all $t\in [a,T_{0}]$,
\begin{equation}
\int_{M} R_{g_i(t)}d\mu_{g_i(t)}
\leq C(a,\bar{h},n,\Lambda,p_0) \left(\int_{M} R_{\tilde g_i(S_{0})}d\mu_{\tilde g_i(S_{0})}+i^{-1}\right).
\end{equation}
Combining this with $R_{g(t)}\geq-i^{-1}$ (see \eqref{Sec2: YFC1-estimates eqn 3}), we have
\begin{equation}
\int_{M}|R_{g_i(t)}|d\mu_{g_i(t)}
\leq C(a,\bar{h},n,\Lambda,p_0) \left(\int_{M} R_{\tilde g_i(S_{0})}d\mu_{\tilde g_i(S_{0})}+i^{-1}\right).
\end{equation}
By the similar computation of \eqref{L1} and the fact that $g_i(t)$ is uniformly equivalent to $h_i$ and hence $\tilde g_i(S_0)$, we have
\begin{equation}
\begin{split}
\int_{M} \big|g_i(t)-g_i(T_{0})\big|_{g_i(T_0)}d\mu_{g_i(T_0)} \leq & C\int_{a}^{T_{0}} \int_M |R_{g_i(s)}| d\mu_{g_i(T_0)}ds \\
\leq {} & C(a,\bar{h},n,\Lambda,p_0) \left(\int_{M} R_{\tilde g_i(S_{0})}d\mu_{\tilde g_i(S_{0})}+i^{-1}\right).
\end{split}
\end{equation}
Using Claim \ref{Sec3: YF-limit1}, the smooth convergence of $\tilde g_{i}(S_{0})$ to $g_{\infty}$ and the fact that $\text{\rm Ric}(g_\infty)=0$, we see that $g_{i}(t)\rightarrow g_{\infty}$ in $L^1(M,g_\infty)$ as $i\rightarrow\infty$ for all $t\in [a,T_{0}]$. By the uniform $C^1$ estimates from Proposition \ref{Sec2: YFC1-estimates}, we can improve the convergence to $C^0$. This completes the proof since $a\in(0,T_{0}]$ is arbitrary.
\end{proof}
\end{proof}

We now prove the stability Theorem for manifolds with $\sigma(M)\leq 0$. This will imply Theorem \ref{Sec1: maintheorem} and \ref{Sec1: maintheorem2} when $M=\mathbb{T}^n$ by Remark~\ref{torus-Yama} as Ricci-flat metric on torus is flat from the splitting Theorem of Cheeger-Gromov \cite{CheegerGromov1971}.
\begin{cor}
Under the assumption in Theorem \ref{Sec3: Conv-1}, if $\mathrm{diam}(M,g_{i,0})$ are uniformly bounded from above, then after passing to subsequence, $g_{i,0}$ converges to a Ricci-flat metric $g_{\infty}$ in the volume preserving intrinsic flat sense.
\end{cor}
\begin{proof}
By \eqref{Sec2: YFC1-estimates eqn 3}, the Yamabe flow $g_i(t)$ in Theorem \ref{Sec3: Conv-1} satisfies
\begin{equation}
\frac{\partial g_{i}}{\partial t} = -R_{g_{i}}g_{i} \leq i^{-1}g_{i},
\end{equation}
which implies
\begin{equation}
g_i(t) \leq e^{-i^{-1}t}g_{i}(0) = e^{-i^{-1}t}g_{i,0}.
\end{equation}
In particular,
\begin{equation}
g_i(T_{0}) \leq e^{-i^{-1}T_{0}}g_{i,0}.
\end{equation}
By Theorem \ref{Sec3: Conv-1}, $F_i^*g_i(T_0)$ converges to a Ricci-flat metric $g_{\infty}$ in $C^0(M)$ where $F_i=\Phi_i\circ \Psi_i$. By relabelling the index, we may assume without loss of generality that
\begin{equation}\label{proof of maintheorem eqn 1}
\left(1-\frac1i\right)g_{\infty}\leq F_i^*g_{i,0}.
\end{equation}
The intrinsic flat convergence follows from \eqref{proof of maintheorem eqn 1}, volume convergence in Proposition \ref{Prop: Lp} and \cite[Theorem 2.1]{AllenPeralesSormani2020}.
\end{proof}

\begin{cor}[Theorem \ref{Sec1: maintheorem2}]
Under the assumption in Theorem~\ref{Sec3: Conv-1}, if $u_i$ are uniformly bounded from above, then there is a sequence of diffeomorphisms $F_i$ of $M$ such that after passing to subsequence, $F_i^*g_{i,0}$ converges to a Ricci-flat metric $g_{\infty}$ in $L^p(M,g_{\infty})$ for all $p>0$. Moreover, $g_{i,0}$ converges to $g_\infty$ in the measured Gromov-Hausdorff sense.
\end{cor}
\begin{proof}
The $L^p$ convergence follows from Proposition \ref{Prop: Lp} and Theorem \ref{Sec3: Conv-1}. As shown in proof of Theorem \ref{Sec3: Conv-1}, $F_i^*h_{i}\leq C(n,\Lambda)\cdot F_i^*g_{i}(T_0)$ and $F_i^*g_{i}(T_0)$ converges to $g_\infty$ in $C^0(M)$. Combining this with \eqref{proof of maintheorem eqn 1}, we have
\begin{equation}\label{proof of maintheorem2 eqn 1}
\left(1-\frac1i\right)g_{\infty} \leq F_i^*g_{i,0}\leq C(n,\Lambda)g_{\infty}.
\end{equation}
for $i$ sufficiently large.
The measured Gromov-Hausdorff convergence follows from \eqref{proof of maintheorem2 eqn 1}, the $L^p$ convergence and \cite[Theorem 1.2]{AllenSormani2020}.
\end{proof}

redpajama_set_name: RedPajamaArXiv


In this article, I will be looking at how to compare the power curves of two engines that have different rev ranges.
To make a comparison, I have made two power curves that look exactly the same; they both have a rev range that is 2000RPM across. Do these engines have the same average power as each other?
Figure 1: Two visually identical power curves, one from 3000RPM to 5000RPM, the other from 6000RPM to 8000RPM.
To compare the capabilities of the two engines, they need to be given the same work to do. Let's say we want to accelerate a rotating shaft from 1500RPM to 2500RPM. This would be about the same as accelerating a car from 110mph to 190mph if it had 27" diameter tires.
Engine 2 is starting at 6000RPM, so it will need a 4:1 gearbox to turn the shaft at 1500RPM. Below is a graph showing the two engines at work.
Figure 2: Graphs showing the engines passing through their rev ranges. Note that the 4:1 gearbox given to Engine 2 only gets the shaft to 2000RPM, not 2500.
No single gear ratio satisfies the situation.
Gearboxes act linearly, that is, if the input speed is doubled, the gearbox will exactly double the output speed, no matter what the ratio is. Also, if the gearbox is turned at 0RPM, the gearbox will turn the output shaft at 0RPM. Engine 2 can not obey these rules while doing the work we wanted.
This means that the two engines do not have the same power as each other when called upon to do the same task. Engine 1 is the better engine.
Figure 3: Power at the output shaft when driven by the two engines.
At this point, it may look like Engine 2 is not capable of accelerating the shaft from 1500RPM to 2500RPM at all. It is already known that there is no single gear ratio that will work, and that Engine 2 is not capable of revving up to 10000RPM. Changing the gear ratio "on the fly" can be done instead. This is the equivalent to shifting gears on a car, and is one of the reasons why cars have gearboxes with multiple gears.
Figure 4: Engine 2 fitted with a two-speed gearbox, with a 4:1 ratio and a 3:1 ratio. Engine 1 retains its single-speed gearbox, with a 2:1 ratio.
Engine 2 only changes its revs by 1.33x (33%) as it passes from 6000RPM to 8000RPM. This means that Engine 1 has a wider rev range, even though they both have a rev change of 2000RPM.
In general, the scale of the width of the rev ranges between two engines can be determined using the formula seen below.
Figure 5: Rev range width formula.
In the case where M=1, the power curves of two engines can be superimposed by scaling one engine's maximum RPM by the other's, as seen in the equation below.
Figure 6: Rev range scale formula.
In the case where M≠1, the power curves of two engines can still be superimposed by scaling one engine's maximum RPM by the other's, but one of the engines will "fall short" when doing the work. This can be seen in Figure 3.
The power curves of two engines are not comparable directly if their rev ranges differ. The rev ranges must be corrected to show the average power when doing the same work.


Deep in the bowels of the Marriott Hotel, at 7 AM this morning, over two hundred Local 2 hotel workers have gathered, in the first of seven mass meetings scheduled today to give the hotel bosses in San Francisco a message.
We will be going to work soon. But, now, here, early this morning, we are voting to give our union's leadership the authority to call a strike if necessary. If that happens, it will mean long, hard days on the picket lines. But we are used to long, hard days -- cleaning hotel rooms, catering to businessmen and tourists in hotel lobbies, toiling in hotel kitchens and restaurants. Long, hard days are nothing new for us.
First up is Earl Streeter, a bellman at the Fairmont Hotel, and a member of union's negotiating committee. He tells us that the contract fight has been a long, hard struggle. We know this, but we are glad to hear it acknowledged. It started two years ago, when four hotels went out on what was supposed to be a two-week strike, but ended up being a nearly two-month long lockout for over 4,000 workers at 14 hotels. And, for two years now, all of us have been working without a contract.
But, says Earl, we are ready. Whatever it takes. He outlines the issues on the bargaining table: decent wages, comprehensive and affordable health care, liveable pensions, a workload that doesn't destroy our bodies or our minds, job security, the right to organize. These are basic demands, things that the american dream says we all deserve, though few of us really have.
We are a multi-national union, ethnically and racially diverse. There is hardly a white face in the room. There are workers from every corner of the globe, from China, from Mexico and Latin America, from the Philippines. Some are here legally, some are not. Some are descended from slaves.
We are facing off against giant multi-national corporations and conglomerates. The union is bargaining now with the Multi-Employer Group (MEG), a cabal of thirteen of the City's biggest hotels -- including Hilton, Hyatt, Starwood (The Palace, Westin St. Francis), and Intercontinental (Mark Hopkins). The investors who make their money in the hotel business make it off our sweat and strain. They know it, and we know it.
Next up is Mike Casey, long-time President of Local 2. He tells us that the hotel bosses are making record profits. He tells us that we have to stand strong, not just for ourselves, but for hotel workers and our families in the next generation -- just like previous generations of hotel workers fought and struggled and sacrificed so we could enjoy the rights and benefits we have today.
Mike pledges that the leadership will never accept proposals like the MEG's plan for a two-tier health care system -- where new employees pay more to get less. He calls such proposals "inexcusable." Here, Mike is rejecting what many other workers in other unions have been forced to accept in recent years. Local 2, he says, will hold the line.
The current battle with the multi-national hotel companies is not limited to San Francisco. Mike reminds us that on Monday and Tuesday, hotel workers in Chicago voted in overwhelming numbers to authorize a strike in that windy city. On Wednesday, hotel workers in Hawaii and in Monterey, California took strike votes. Hotel workers in Toronto have also authorized a strike. Negotiations continue on hotel contracts in a number of other cities nationwide. It is, Mike says, an "historic week."
Finally, Mike tells us that the union will be negotiating every day next week, and that the leadership will do everything possible to avoid a strike -- if it can be avoided without giving in to the hotel bosses. But, just in case, the union has scheduled a "City-Wide Action."
Date: Next Thursday, August 31.
Place: Local 2 Plaza (Market Street between 3rd and 4th Streets).
The flyer is very simple: "No Justice, No Peace!"
As the corporate media stories have been telling us all week, one year ago, in the summer of 2005, Hurricane Katrina battered the people of New Orleans and the Gulf. Katrina exposed, for all the world to see the poverty, injustice and racism that is so much a part of this country.
Two years ago, in the summer of 2004, the hotel bosses in San Francisco forced thousands of workers out onto the streets, exposing for all the world to see the poverty, injustice and racism that is so much a part of San Francisco, despite what so many of us want to believe about our "progressive" city.
San Francisco, like New Orleans, has suffered tremendous natural disasters. We can survive hurricanes and floods, earthquakes and fires. But the intolerable disasters are the ones that are brought on only by power elites that don't care one bit about the lives of working class people and our families, especially when our faces are black, brown or yellow.


text:
Articles by: Regan Fletcher
In this week's issue of our Content Marketing All-Stars series, we caught up with Matt Anchin, SVP of Global Communications for Monster Worldwide. Monster has been a leading online platform for job seekers since 1999, and content has always been a key part of that equation. With thousands of searches for job-hunting tips happening every day, capturing the attention of their audience through compelling and useful content is critical. Our interview with Matt highlights the way that Monster has built the premiere web destination for all things related to your job hunt.
While some may just be waking up to the idea of content marketing, Joe Pulizzi has been preaching the benefits since all the way back in 2001. As the recognized authority on all things content, Joe has made a name for himself as the founder of Content Marketing Institute and the author of three books on the subject. The opportunity to catch up with Joe and chat about the industry from his perspective was a truly unique experience for us, and in our interview he offered some insights that all marketers can learn from.
Content Marketing All Stars: Q&A with Arati Randolph of Wells Fargo
As Wells Fargo's Director of Internal Communications Arati Randolph is charged with being the brand's official storyteller. Whether it is sharing stories of community involvement or feel good anecdotes about a teller making a child's day with a lollipop, Wells Fargo Stories strives to deliver content that customers and readers can make an emotional connection with. We caught up with Arati to hear more about how Wells Fargo is looking at their content strategy and what the future looks like for their branded content channels.
Content Marketing All-Stars: Q&A with Ken Kaplan of iQ by Intel
Ken Kaplan lives and breathes cool content as managing editor of iQ by Intel. The tech-culture hub covers the likes of sports, health, music and art from Intel's own employees and the company's partners. We chatted with Kaplan to get his takes on branded storytelling, content promotion and why a combination of journalism and marketing can create the perfect balance. Read on for more insight from Kaplan.
Content Marketing All-Stars: Q&A with Helen Loh of Charles Schwab
This week we are continuing our "Content Marketing All-Stars" series with Helen Loh, Senior Vice President of Owned Media, Content & Digital Marketing at Charles Schwab. Schwab has been a leader in the financial services industry for the ways they have leveraged their blog content to spread awareness around their brand and corporate values. Helen sat down with us to discuss how Schwab is thinking about new channels for marketing, and her opinion of what the future looks like for content marketing.
Content Marketing All-Stars: Q&A with Michael Brenner of SAP
The rise in content marketing efforts by brands has been dramatic. CMOs and marketing leaders have recognized that strategies beyond traditional advertising and media are critical to building customer awareness, and are utilizing content programs as a new way to reach potential customers. It's not just consumer brands that have been bit by the content bug. Enterprise organizations have created resources and web properties to engage their customers with more than just traditional sales-focused rhetoric.
Happy Valentine's Day To All Our Users, but Especially This One…
Sometimes with the Hallmark haze in the air, it's easy to forget the true meaning of Valentine's Day– love. So when Kevin Lombardo contacted us to have his wife's newly launched food blog added to News360 as a Valentine's Day surprise, we couldn't resist. Call us a sucker for Cupid. Usually it takes three to four business days to go through the process of adding a new feed, but our content team caught the love bug and turned the request around within two days. We were thrilled to help Kevin surprise his wife today and think you'll enjoy the delicious […]
Native Ads and the New York Times: End of Publishing as We Know It?
Last week, the New York Times rolled out a sleek new website design, which included the publication's first ever native ad. The debut made headlines throughout the industry and many were abuzz about what this means for the publishing industry, both good and bad. Native advertising allows brands to engage with consumers by delivering content where they are actively looking and engaged. From the chatter surrounding the New York Time's roll out of native ads, you'd think this was a brand new phenomenon. The idea of native advertising has been around for some time, but the fact that one of […]

redpajama_set_name: RedPajamaCommonCrawl


text:
MorphoSys appoints Tim Demuth as new Chief Research and Development Officer, following the retirement of Malte Peters
Tim Demuth, M.D., Ph.D., has more than 20 years of broad leadership experience in drug development, with a focus in oncology
Posted by MorphoSys US
MorphoSys AG (FSE: MOR; NASDAQ: MOR) announced today that Malte Peters, M.D., the company's Chief Research and Development Officer and Management Board member, has decided to retire at the end of 2022. His successor will be Tim Demuth, M.D., Ph.D., current Chief Medical Officer of Pieris Pharmaceuticals, Inc., a clinical-stage biotechnology company. Tim will start his new role on October 1. He will report to MorphoSys' CEO, Jean-Paul Kress, M.D., and be a member of the company's Executive Committee. At this time, Malte will step down from his operational responsibilities.
"On behalf of the entire Supervisory Board, I want to thank Malte for the remarkable contributions he has made to science, modern medicine and human health over the course of his distinguished career," said Marc Cluzel, M.D., Ph.D., Chairman of the Supervisory Board of MorphoSys. "We are grateful for the strong pipeline and team Malte has built at MorphoSys. We wish him all the best as he begins this new chapter, and we look forward to working with Tim."
Malte joined MorphoSys in 2017. He was instrumental in building an accomplished Clinical Development organization that successfully developed and obtained accelerated approval of Monjuvi® (tafasitamab-cxix) in the U.S. and European Union, as well as other countries worldwide. Most recently, he spearheaded the successful integration of Constellation Pharmaceutical's Research and Development organization and the optimization of the pivotal, Phase 3 MANIFEST-2 study of pelabresib in first-line myelofibrosis.
"I am very thankful for everything Malte has done during his tenure at MorphoSys. Malte pushed us to think differently in how we discover and develop new medicines, driven by an unshakeable determination to put these breakthroughs into the hands of patients as quickly as possible," said Jean-Paul Kress, M.D., Chief Executive Officer of MorphoSys. "We are very excited that Tim is joining MorphoSys as our new Chief Research and Development Officer. His well-rounded clinical development experience and scientific expertise in oncology make him an asset to MorphoSys and the ideal leader to drive our potential best-in-class pipeline forward."
"It has been a privilege and an honor to lead MorphoSys' Research and Development organization. The strategic and innovative decisions we have taken to focus the company over the last five years have strengthened our pipeline and put us on a strong path towards the future," said Malte Peters, M.D., Chief Research and Development Officer of MorphoSys. "I am looking forward to this new chapter in my life. But I will miss working with my courageous, dedicated and inspiring MorphoSys colleagues who come to work every day to make a difference in patients' lives. Tim will be a great leader for the Research and Development organization, and I wish him and the team continued success."
Tim Demuth is a seasoned biopharma executive with broad leadership experience in drug development, with a focus in oncology. Prior to being named the Chief Medical Officer of Pieris, he was Vice President and Head of Global Clinical Development Oncology at Merck KGaA, where he was responsible for the early- and late-stage oncology pipeline of small molecules, antibodies and antibody-drug conjugates, as well as for the end-to-end development strategy for oncology. Previously, he led the Clinical Development Unit of Hematology Biopharma Development at Sandoz AG, a division of Novartis AG. Dr. Demuth held various other leadership roles at Novartis AG, Italfarmaco S.p.A and Merck & Co. He received M.D. and Ph.D. degrees from Johannes Gutenberg University, Mainz, Germany and was trained in Mainz and at Ludwig Maximilian University, Munich, Germany. He has also served as a Post-Doctoral Fellow at the Translational Genomics Research Institute in Phoenix, Arizona.
"I am thrilled to join MorphoSys at this pivotal and exciting time," said Tim Demuth, M.D., Ph.D. "Clinical data to date highlight the enormous potential of the company's pipeline. I look forward to overseeing the efforts to advance MorphoSys' pivotal Phase 3 trials, which are investigating pelabresib in first-line myelofibrosis and tafasitamab in first-line diffuse large B-cell lymphoma. These two medicines, if approved, may have the potential to transform the standard of care in aggressive and difficult-to-treat blood cancers, and I am excited to be part of that journey to make these medicines available to patients."
About Pelabresib
Pelabresib (CPI-0610) is an investigational selective small molecule designed to promote anti-tumor activity by inhibiting the function of bromodomain and extra-terminal domain (BET) proteins to decrease the expression of abnormally expressed genes in cancer. Pelabresib is being investigated as a treatment for myelofibrosis and has not yet been evaluated or approved by any regulatory authorities.
About Tafasitamab
Tafasitamab is a humanized Fc-modified CD19 targeting immunotherapy. In 2010, MorphoSys licensed exclusive worldwide rights to develop and commercialize tafasitamab from Xencor, Inc. Tafasitamab incorporates an XmAb® engineered Fc domain, which mediates B-cell lysis through apoptosis and immune effector mechanism including Antibody-Dependent Cell-Mediated Cytotoxicity (ADCC) and Antibody-Dependent Cellular Phagocytosis (ADCP).
In the United States, Monjuvi® (tafasitamab-cxix) is approved by the U.S. Food and Drug Administration in combination with lenalidomide for the treatment of adult patients with relapsed or refractory DLBCL not otherwise specified, including DLBCL arising from low grade lymphoma, and who are not eligible for autologous stem cell transplant (ASCT). This indication is approved under accelerated approval based on overall response rate. Continued approval for this indication may be contingent upon verification and description of clinical benefit in a confirmatory trial(s).
In Europe, Minjuvi® (tafasitamab) received conditional marketing authorization in combination with lenalidomide, followed by Minjuvi monotherapy, for the treatment of adult patients with relapsed or refractory diffuse large B-cell lymphoma (DLBCL) who are not eligible for autologous stem cell transplant (ASCT).
Tafasitamab is being clinically investigated as a therapeutic option in B-cell malignancies in several ongoing combination trials.
Monjuvi® and Minjuvi® are registered trademarks of MorphoSys AG. Tafasitamab is co-marketed by Incyte and MorphoSys under the brand name Monjuvi® in the U.S., and marketed by Incyte under the brand name Minjuvi® in Europe, the UK and Canada.
XmAb® is a registered trademark of Xencor, Inc.
About MorphoSys:
At MorphoSys, we are driven by our mission: More life for people with cancer. As a global commercial-stage biopharmaceutical company, we use groundbreaking science and technologies to discover, develop, and deliver innovative cancer medicines to patients. MorphoSys is headquartered in Planegg, Germany, and has its U.S. operations anchored in Boston, Massachusetts. To learn more, visit us at www.morphosys.com and follow us on Twitter and LinkedIn.
This communication contains certain forward-looking statements concerning the MorphoSys group of companies. The forward-looking statements contained herein represent the judgment of MorphoSys as of the date of this release and involve known and unknown risks and uncertainties, which might cause the actual results, financial condition and liquidity, performance or achievements of MorphoSys, or industry results, to be materially different from any historic or future results, financial conditions and liquidity, performance or achievements expressed or implied by such forward-looking statements. In addition, even if MorphoSys' results, performance, financial condition and liquidity, and the development of the industry in which it operates are consistent with such forward-looking statements, they may not be predictive of results or developments in future periods. Among the factors that may result in differences are that MorphoSys' expectations may be incorrect, the inherent uncertainties associated with competitive developments, clinical trial and product development activities and regulatory approval requirements, MorphoSys' reliance on collaborations with third parties, estimating the commercial potential of its development programs and other risks indicated in the risk factors included in MorphoSys' Annual Report on Form 20-F and other filings with the U.S. Securities and Exchange Commission. Given these uncertainties, the reader is advised not to place any undue reliance on such forward-looking statements. These forward-looking statements speak only as of the date of publication of this document. MorphoSys expressly disclaims any obligation to update any such forward-looking statements in this document to reflect any change in its expectations with regard thereto or any change in events, conditions or circumstances on which any such statement is based or that may affect the likelihood that actual results will differ from those set forth in the forward-looking statements, unless specifically required by law or regulation.
Thomas Biegi
Tel.: +49 (0)89 / 89927 26079
thomas.biegi@morphosys.com
Investor Contacts:
Dr. Julia Neugebauer
Tel: +49 (0)89 / 899 27 179
julia.neugebauer@morphosys.com
Eamonn Nolan
eamonn.nolan@morphosys.com


text:
Q: How to capture image from camera and upload to Retrofit multipart? I am doing the following to capture image from camera and upload it to web service through retrofit multipart image upload, but i am facing problem in converting bitmap to image file to be able to fill the image file multipart parameter of the retrofit web service call. I have used URI to convert bitmap from camera to file and i am getting null pointer exception in that function.
Intent cameraIntent = new Intent(android.provider.MediaStore.ACTION_IMAGE_CAPTURE);
startActivityForResult(cameraIntent, CAMERA_REQUEST);

@Override
public void onActivityResult(int requestCode, int resultCode, Intent data) {
    super.onActivityResult(requestCode, resultCode, data);
    if (requestCode == CAMERA_REQUEST && resultCode == Activity.RESULT_OK)
    {
        final Bitmap photo = (Bitmap) data.getExtras().get("data");

        // CALL THIS METHOD TO GET THE URI FROM THE BITMAP
        Uri tempUri = getImageUri(getActivity(), photo);

        // CALL THIS METHOD TO GET THE ACTUAL PATH
        File file = new File(getRealPathFromURI(tempUri));

        MyWebService webService =
                MyWebService.retrofit.create(MyWebService.class);

        MultipartBody.Part filePart = MultipartBody.Part.createFormData("file", file.getName(), RequestBody.create(MediaType.parse("image/*"), file));

        Call<com.syamantak.footsteps.models.response.Response> call = webService.set_profile_photo(filePart, "Bearer " + MainActivity.api_token);
        MainActivity.setProgressBarVisible();
        call.enqueue(new Callback<com.syamantak.footsteps.models.response.Response>() {
            @Override
            public void onResponse(Call<com.syamantak.footsteps.models.response.Response> call, Response<com.syamantak.footsteps.models.response.Response> response) {
                MainActivity.setProgressBarInVisible();
                com.syamantak.footsteps.models.response.Response jsonResponse = response.body();
                selfieImage.setImageBitmap(photo);
                Toast.makeText(getActivity(), jsonResponse.getMessage(), Toast.LENGTH_SHORT).show();
            }

            @Override
            public void onFailure(Call<com.syamantak.footsteps.models.response.Response> call, Throwable t) {
                t.getLocalizedMessage();
            }
        });
    }
}

public Uri getImageUri(Context inContext, Bitmap inImage) {
    ByteArrayOutputStream bytes = new ByteArrayOutputStream();
    inImage.compress(Bitmap.CompressFormat.JPEG, 100, bytes);
    String path = MediaStore.Images.Media.insertImage(inContext.getContentResolver(), inImage, "Title", null);
    return Uri.parse(path);
}

public String getRealPathFromURI(Uri uri) {
    String path = "";
    if (getActivity().getContentResolver() != null) {
        Cursor cursor = getActivity().getContentResolver().query(uri, null, null, null, null);
        if (cursor != null) {
            cursor.moveToFirst();
            int idx = cursor.getColumnIndex(MediaStore.Images.ImageColumns.DATA);
            path = cursor.getString(idx);
            cursor.close();
        }
    }
    return path;
}

Following is my manifest file
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.syamantak.footsteps">

<uses-feature android:name="android.hardware.sensor.accelerometer" android:required="true" />
<uses-feature android:name="android.hardware.camera"/>
<uses-permission android:name="android.permission.READ_EXTERNAL_STORAGE" />
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />
<uses-permission android:name="android.permission.INTERNET"/>

I am getting following error
E/MediaStore: Failed to insert image
java.lang.SecurityException: Permission Denial: writing com.android.providers.media.MediaProvider uri content://media/external/images/media from pid=28108, uid=10237 requires android.permission.WRITE_EXTERNAL_STORAGE, or grantUriPermission()

and
java.lang.RuntimeException: Failure delivering result ResultInfo{who=null, request=67424, result=-1, data=Intent { act=inline-data (has extras) }} to activity {com.syamantak.footsteps/com.syamantak.footsteps.activities.MainActivity}: java.lang.NullPointerException: uriString

What am I doing wrong?

A: I arrived at following solution
        final Bitmap photo = (Bitmap) data.getExtras().get("data");
        File file = savebitmap(photo);

        MyWebService webService =
                MyWebService.retrofit.create(MyWebService.class);

        MultipartBody.Part filePart = MultipartBody.Part.createFormData("photo", file.getName(), RequestBody.create(MediaType.parse("image/*"), file));

        Call<com.syamantak.footsteps.models.response.Response> call = webService.set_profile_photo(filePart, "Bearer " + MainActivity.api_token);
        MainActivity.setProgressBarVisible();
        call.enqueue(new Callback<com.syamantak.footsteps.models.response.Response>() {
            @Override
            public void onResponse(Call<com.syamantak.footsteps.models.response.Response> call, Response<com.syamantak.footsteps.models.response.Response> response) {
                MainActivity.setProgressBarInVisible();
                com.syamantak.footsteps.models.response.Response jsonResponse = response.body();
                selfieImage.setImageBitmap(photo);
                Toast.makeText(getActivity(), jsonResponse.getMessage(), Toast.LENGTH_SHORT).show();
            }

            @Override
            public void onFailure(Call<com.syamantak.footsteps.models.response.Response> call, Throwable t) {
                t.getLocalizedMessage();
            }
        });

and following function to convert bitmap to file
private File savebitmap(Bitmap bmp) {
    String extStorageDirectory = Environment.getExternalStorageDirectory().toString();
    OutputStream outStream = null;
    // String temp = null;
    File file = new File(extStorageDirectory, "temp.png");
    if (file.exists()) {
        file.delete();
        file = new File(extStorageDirectory, "temp.png");

    }

    try {
        outStream = new FileOutputStream(file);
        bmp.compress(Bitmap.CompressFormat.PNG, 100, outStream);
        outStream.flush();
        outStream.close();

    } catch (Exception e) {
        e.printStackTrace();
        return null;
    }
    return file;
}

It is working well.

A: If you want to send image with more different parameters in Multipart/formData then use this below code. It will solve your Problem.
    ` **1) Put this line above onCreate() method.**

    File selectedFile = null;

    **2) In onActivityResult() method of Camera Intent put this below code -**

    final Bitmap photo = (Bitmap) data.getExtras().get("data");
                imageViewPhoto.setImageBitmap(photo);
                selectedFile = new File(this.getFilesDir(), "image" + ".jpg");
                FileOutputStream outputStream = null;
                try {
                    outputStream = new FileOutputStream(selectedFile);
                    photo.compress(Bitmap.CompressFormat.PNG, 100, outputStream);
                    outputStream.flush();
                    outputStream.close();
                } catch (FileNotFoundException e) {
                    e.printStackTrace();
                } catch (IOException e) {
                    e.printStackTrace();
                }

**3) In API Interface Class -**

 @Multipart
    @POST("photouploadWithdata.php")
    Call<YourResponseModal> startDuty(@PartMap HashMap<String, RequestBody> map, @Part MultipartBody.Part image);

**4) In Activity when you call API-**

 RequestBody tokens = RequestBody.create(MediaType.parse("multipart/form-data"), token);
        RequestBody dates = RequestBody.create(MediaType.parse("multipart/form-data"), date);
        RequestBody times = RequestBody.create(MediaType.parse("multipart/form-data"), time);
        RequestBody latitudes = RequestBody.create(MediaType.parse("multipart/form-data"), latitude);
        RequestBody longitudes = RequestBody.create(MediaType.parse("multipart/form-data"), longitude);

        MultipartBody.Part image = MultipartBody.Part.createFormData("avatar", selectedFile.getName(),
                RequestBody.create(MediaType.parse(URLConnection.guessContentTypeFromName(selectedFile.getName())), selectedFile));

        HashMap<String, RequestBody> map = new HashMap<>();
        map.put("token", tokens);
        map.put("date", dates);
        map.put("time_in", times);
        map.put("latitude", latitudes);
        map.put("longitude", longitudes);

        RetrofitAPI retrofitAPI = APIClient.getRetrofitInstance().create(RetrofitAPI.class);
        Call<StartDutyResponseModal> call = retrofitAPI.startDuty(map, image);
        call.enqueue(new Callback<StartDutyResponseModal>() {
            @Override
            public void onResponse(Call<StartDutyResponseModal> call, Response<StartDutyResponseModal> response) {
                if (response.body().getStatus() == true){
                    progressBar.setVisibility(View.GONE);
                    Intent intent = new Intent(StartDutyActivity.this, StoreListActivity.class);
                    startActivity(intent);
                    finish();
                }

                else {
                    progressBar.setVisibility(View.GONE);
                    AppUtils.showToast(response.body().getMessage(), StartDutyActivity.this);
                }
            }

            @Override
            public void onFailure(Call<StartDutyResponseModal> call, Throwable t) {
                progressBar.setVisibility(View.GONE);
                AppUtils.showToast(t.getMessage(),StartDutyActivity.this);
            }
        });
    }

`

redpajama_set_name: RedPajamaStackExchange


text:
Meet February's Employee of the Month
We're pleased to announce that security officer Shaun Young has been named Gough & Kelly's first Employee of the Month for February.
Working on site at a high-profile school in Leeds, Shaun was most recently commended for quickly resolving a potential intrusion of what could have been hostile reconnaissance, something the client was extremely grateful for.
Ian Crawford, group operations director, commented: "Having started with us in April last year, Shaun quickly made himself an indispensable member of the company and is a great example of what makes Gough & Kelly such an integral part of the day to day operations of the clients we work with."
As recognition of his hard work, Shaun was presented with a certificate by Gough & Kelly's Sean Morgan, and will also receive an extra day of annual leave.
Return to the News page
Gough and Kelly can help you resume 'business as usual'
Gough and Kelly to the rescue at new coronavirus testing centre
We are here to help in uncertain times
Gough and Kelly wear red to raise funds for Children's Heart Surgery charity.
Sixty Second Security – Lets Defeat Terrorism
Dash to raise funds for Cancer Research UK


Art on this site was created and copyrighted by the artist, with all rights reserved. The contents of this site may not be sold, published or duplicated in either written or electronic form. Permission from the artist is required if you want to use any image for any purpose. Thank you for your support of the arts.


Alexandra is a freelance photographer as well as the founder and creative director of UNCONDITIONAL magazine. She graduated with a BFA in Photography & Imaging from the Art Center College of Design in Pasadena. She was born and raised between Paris and Los angeles, and is currently based in New York.


text:
sudo apt-get remove docker docker-engine docker.io

# Update the apt package index
sudo apt-get update

# Install packages to allow apt to use a repository over HTTPS
sudo apt-get install \
    apt-transport-https \
    ca-certificates \
    curl \
    gnupg2 \
    software-properties-common

# Add Docker's official GPG key:
curl -fsSL https://download.docker.com/linux/debian/gpg | sudo apt-key add -

# Set up the stable repository
sudo add-apt-repository \
    "deb [arch=amd64] https://download.docker.com/linux/debian \
    $(lsb_release -cs) \
    stable"

# Update the apt package index again
sudo apt-get update

# Install the latest version of Docker CE
sudo apt-get install docker-ce

# Test installation by running the hello-world-container
sudo docker run hello-world