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The Dividend Discount Model (DDM) is the key valuation technique for dividend stocks.
The most straightforward form of it is called the Gordon Growth Model. This guide explains how it works and the streamlined way to use it.
Determining the fair value of a company means using Discounted Cash Flow Analysis (DCFA). DCFA, put simply, states that the present value of a company is equal to the sum value of all future cash flows that the company produces. But each future cash flow must be discounted to translate it into today's dollars. This is logical: the purpose of a business is to produce cash flows, so the value of the business is equal to the sum value of all future discounted cash flows.
By discounted, what I mean is that due to the time value of money, a payment in the future is worth less than the same payment today. For example, if you can earn a 10% rate of return on your money over time, then a payment of $10,000 one year from now would only be worth $9,091 to you today, because if you had $9,091 today, you could invest it at a 10% rate of return and turn it into $10,000 a year from now. ($9,091 multiplied by 1.10 equals $10,000) So, the discounted version of, or the net present value of, $10,000 one year from now, is equal to $9,091.
Similarly, if you were to receive $10,000 in five years, then this sum would only be worth $6,209 to you today, because you could take $6,209 and compound it by 10% annually to get $10,000. ($6,209 multiplied by 1.10 five times in a row equals $10,000). So, the discounted version of, or the net present value of, $10,000 five years from now, is equal to $6,209.
To value a business, you would take the discounted values of all future annual expected cash flows, sum them together, and that's the fair value of the business. You're trading a present sum of money (the fair value), for a future series of expected cash flows, but each cash flow has to be translated into today's value to take into account the time value of money and your target rate of return on your current money.
The inputs you need are the current free cash flow figures, the projected growth rate of those cash flows, and your target rate of return to use as the discount rate.
Obviously there's a mix of art and science involved here. If appropriate inputs (expected cash flows) are used, the output (current intrinsic value) is objective. But since the inputs are future expected cash flows, there is uncertainty in those figures and it requires reasonably accurate estimates to be useful.
Enter the Dividend Discount Model
You can take that same approach, and tailor it specifically for analyzing a stock that pays good dividends, and this is the Dividend Discount Model. It's also called the Dividend Growth Model, and the most straightforward form is called the Gordon Growth Model.
The DDM is based on the exact same idea, except that the share of stock represents what we're valuing, and all future dividends represent all future cash flows of that share. The value of the stock is equal to the sum of the net present value of all future dividends.
For example, let's say you're analyzing a share of stock that pays $0.50 in dividends per quarter, or $2.00 per year. Furthermore, it's a dividend aristocrat that has raised the dividend consecutively every single year for 25 years or more. You look over its history, and find that it has increased the dividend by an average of 8% per year over the course of several decades, but that the growth is slowing down. So, you estimate that the dividend will continue to grow by an average of only 5% per year going forward.
If you desire an 11% rate of return on your money, which would represent pretty good returns, then you can use that as your discount rate. So for example, when the $2 in dividends goes up to $2.10 next year (because it grew by 5%), this $2.10 is only worth $1.89 to you today, because if you had $1.89 today, then you could turn it into $2.10 in a year if you could compound it by 11% during that period.
The following chart shows the estimated value of dividend payments over the next forty years.
The orange values are the actual dividends you expect to be paid if the dividend grows by 5% per year. The blue values are the discounted versions of those dividends; the dividends translated into today's value based on your discount rate of 11%. As can be seen, if this chart continues forever, the sum of all dividends would be infinite, but the sum of all discounted dividends is finite, because the discount rate is larger than the dividend growth rate.
To calculate the fair value of this stock, we need to sum up all of those discounted dividends. It can be done with fancy math, but after a number of mathematical cancellations, the accurate equation is extremely simple, and this is called the Gordon Growth Model:
In the formula, P is the fair price of the stock. D1 is next year's expected dividend, which would be $2.10 in this case. r is the discount rate, which is 1.11, and g is the dividend growth rate, which is 1.05.
When you plug everything into the equation, you get $35. That's the fair value of this hypothetical stock, assuming that the current dividend of $2.00 really grows at that 5% per year into the future, and assuming an 11% rate of return is desirable to you. The math shows that it would be fair to trade $35 in present value for the sum value of all future dividends, because when they are discounted by 11% per year, the sum of all of them is $35.
If your estimates are correct, you'll get a nice double-digit long-term rate of return.
The first flaw of the Gordon Growth Model is that it assumes a constant dividend growth rate; it's a constant growth model. This may be an acceptable estimate for a fairly high-yielding mature company, but for stocks with lower dividend yields and higher dividend growth, this may not be appropriate.
To fix this, you can move towards a more general two-stage or multi-stage Dividend Discount Model. The math gets a bit more tedious at this point, but a multi-stage DDM allows you to estimate that the dividend will grow at a certain rate for a number of years, and then slow down to another growth rate after that.
The second flaw of both the Gordon Growth Model and the whole Dividend Discount Model is that it's quite sensitive to the accuracy of the inputs. Because dividend growth rates tend to be fairly high (higher than core company net income growth usually, due to share buybacks), even minor variances between the estimated dividend growth and the real dividend growth. This is why you should always have a margin of safety in your estimates. This also means that the DDM tends to be better for high yielding dividend stocks with lower dividend growth, rather than lower yielding stocks with higher dividend growth rates.
Dividend Discount Model Spreadsheet
The one product I offer on this site is the Dividend Toolkit, which is a comprehensive stock guide that also comes with an easy-to-use valuation spreadsheet to calculate the fair price for dividend stocks.
Specifically, it's a spreadsheet tool that lets you put in a single set of inputs, and then it automatically calculates a range of fair values based on different outcomes.
See, all valuation methods rely on estimated inputs. The two main things you need are an estimate for the growth rate, and your discount rate. (Your discount rate should be your target rate of return.) But if those inputs are off even slightly, the whole valuation method will be off. With basic freebie valuation calculators, you need to put in more than one set of inputs if you want to see several output options, and you have no way of keeping track of the differences.
Suppose you want to calculate the fair value of a stock using the Dividend Discount Model (which is explained in significantly more detail in the book), and you estimate that the dividend will grow by 5% per year, and you're using 12% as your discount rate. First, you put the simple inputs into the Dividend Discount Model spreadsheet tool:
And the tool instantly updates the output chart to tell you the fair value of the stock:
This output chart will not only tell you the fair stock value based on those inputs, but will also tell you the fair stock value based on nearby inputs. In this example, in addition to calculating the results for 5% dividend growth and a 12% discount rate, it will automatically show what the fair value is if it turns out that the stock only grows its dividend by 4%, or if you use a discount rate of 11% instead.
This particular tool provides nine total output values, centered around the primary estimate. In this example, because I chose 5% for my estimated dividend growth, the output chart automatically adjusts to show the calculated fair values for 4%, 5%, and 6% growth. And because I chose 12% for my discount rate, the output chart automatically adjusts to show 11%, 12%, and 13% discount rates.
So there are nine total outputs. In this example, I was looking to estimate the value of this stock that paid $1.80 in dividends with estimates for 5% annual dividend growth and a 12% discount rate. The primary result is the $27.00 figure in the middle, which corresponds to 5% growth and 12% target rate of return. The result in the top left corner shows a related estimate, except that it shows the fair value for the same stock but with a 6% dividend growth rate and with a less aggressive 11% discount rate. The result in the bottom right corner shows another estimate, except that it shows the fair value if the stock only has a 4% dividend growth rate and if a 13% discount rate is used. The other cells show other combinations in between those two extremes.
Take a look at the Dividend Toolkit if you want that tool, or read the reviews. Hundreds of investors use it now. The guide covers how to make reasonable estimates for dividend growth as well.
Either way, make sure you're signed up for the monthly dividend newsletter, at no cost.
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redpajama_set_name: RedPajamaCommonCrawl


A duck was shot by a crossbow bolt prompting an RSPCA appeal for information.
The injured bird was found near Buckley Common in Flintshire and reported to the charity.
RSPCA inspector Rachael Davies, who rescued the duck, said: "Shockingly this duck appears to have been shot with the arrow going straight through his leg, which would have been by a crossbow bolt, possibly a pistol-crossbow bolt.
The duck has been taken to RSPCA Stapeley Grange Wildlife Centre for further treatment and it is hoped that he will be released back into the wild once fully recovered from his ordeal.
Inspector Davies added: "There is no question as to whether this was deliberate. Someone had deliberately shot an arrow at the duck with the intention of causing harm.
All wild birds are protected under the Wildlife and Countryside Act 1981 and it is illegal - except under licence - to take, injure or kill wild birds. Under the Wildlife and Countryside Act, the use of crossbows to kill or take birds is also prohibited. The maximum penalty, if found guilty, is six months in prison and/or a £5,000 fine.
For more information on what to do if you find an injured wild animal, visit the RSPCA website .


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text:
Tre navi della Royal Navy sono state battezzate HMS Hood in onore di membri della famiglia Hood, da cui provennero diversi famosi ufficiali di marina:
HMS Hood (1859) – una 91 cannoni di seconda classe, varata come Edgar. Nel 1860 venne ribattezzata in onore dell'ammiraglio Samuel Hood. Costruita di legno e con propulsione a vele, venne successivamente dotata di un motore a vapore. Venne radiata nel 1888.
HMS Hood (1891) – Una nave da battaglia basata sulla classe Royal Sovereign modificata, costruita a Chatham, Inghilterra e affondata come nave blocco nel 1914. Venne battezzata in onore del Primo Lord del Mare, ammiraglio sir Arthur Hood.
HMS Hood (1918) – incrociatore da battaglia classe Admiral battezzata in onore di Samuel Hood, costruita da John Brown and Company, Scozia. Venne affondata nel 1941 dalla nave da battaglia tedesca Bismarck e dall'incrociatore pesante Prinz Eugen nella Battaglia dello Stretto di Danimarca, solo tre membri dell'equipaggio su 1418 sopravvissero.

Hood

redpajama_set_name: RedPajamaWikipedia


We investigate leukocyte migration processes in zebrafish embryos during inflammation, which as part of the innate immune system form the first line of response to infection and trauma. Fluorescence microscopy has been used to generate time-lapse images that follow migrating leukocytes during inflammation. Image processing, including object detection and tracking, allows statistical analysis of the leucocyte trajectories as a function of time and space. With this information we can link intracellular signalling dynamics to tissue-level processes. Immune response mechanisms span both scales and automated image analysis will allow us to elucidate this intrinsically multi-scale problem. Zebrafish embryos, because of their transparency, provide a powerful model organism to study the spatio-temporal aspects of the innate immune response in vivo; in particular for the inference of mathematical multi-scale models it has probably unique advantages. We will illustrate the power of the model system in describing multi-scale dynamics by coupling image processing with explicit mathematical models for leukocyte chemotaxis.


Immediately, I thought absolutely true! You choose to read books that appeal to your interests, or ones that you hope will teach you a piece of valuable information. Even books you read "for entertainment" only make you laugh if you identify with the situations the author presents.
DNA For Dummies, 101 Homeopathic Headache Remedies, The US Penal System for the Proactive, Birth Order Backlash, Belittled Women, DSM-5 Diagnostic and Statistical Manual of Mental Disorders, and one I bought just because I was fooled by its cover.
Uhhhh……alrighty then. I think that makes it pretty clear about who I am. What does your book collection say about you?
This entry was posted in Uncategorized and tagged #books, #frightening, #hapercollins by thebufferzoneday. Bookmark the permalink.


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•What is the delinquency rate on HOA fees by the existing owners?
The No Bind Box is defined by precise latitude and longitudes and actually covers a considerable area. It encompasses Haiti, almost the entire Gulf of Mexico and as far north as Memphis.The "Buyer Beware" caveat to be conscious of here is that not all insurance providers will stop binding even if a "named storm" enters the box. They do however reserve that right.


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Parliament criticises UK for opposing Juncker as next Commission president
The European Parliament told EU council president Herman Van Rompuy on Thursday to nominate EPP Jean Claude Juncker as the next commission president. Parliament leaders warned that the Chamber is ready to veto any other candidates.
Despite having the support of German Chancellor Angela Merkel, Juncker is facing fierce opposition from some countries including the UK, the Netherlands and Sweden.
"Unfortunately we observe a rather dirty campaign against Mr Juncker and I think that this isn't the way that we should proceed as democrats." said EPP chairman Manfred Weber.
British Prime Minister David Cameron has repeatedly emphasised his opposition to Juncker; he views the former Luxembourg Prime Minister as an old-style federalist who might obstruct his push for a reformed EU.
But EPP chairman Manfred Weber underlined that the UK has no right to veto and cannot dictate the EU's political agenda.
"The UK is one state out of 28, there is no right of veto and it is not right for the UK to dictate the agenda across Europe." said Weber.
EU leaders are expected to choose their candidate for the presidency at a summit at the end of this month.
Commission presidency


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'This must be what heaven feels like' - storm over Doncaster star's church heroin taking scene on TV drama
A Doncaster actor has caused outrage after his character on Coronation Street took heroin in a church and described it as 'heaven'.
Saturday, 03 March, 2018, 13:13
Richard Crehan, who plays Lee Mayhew, centre.
Richard Crehan, aged 28, who grew up in Scawthorpe, has caused quite a stir in his time playing heroin addict Lee Mayhew on the ITV soap opera.
Doncaster actor feels at home on Corrie's cobbles
During this moment, Lee said: "This must be what heaven feels like."
But the scenes left a number of people unhappy, with 82 complaints being made to TV regulator Ofcom.
In addition, a number of viewers took to Twitter to voice their concerns over the scene.
Katie Taylor tweeted: "Advertising heroin as what heaven feels like, nice one Corrie. A bit much before 9pm, jacking up in a church!"
Zoe Lamb added: "A heroin addict on Corrie saying, while high, 'this must be what heaven feels like', isn't exactly the right message to be sending out."
Raj Lyall posted: "I was bit disturbed watching Corrie last night and where Billy was taking heroin in the church.
"It really wasn't appropriate to film this in a place of worship.
"What kind of message does this send out to the younger audience and worse before the 9pm watershed."
An Ofcom spokesperson said: "We are assessing these complaints under our broadcasting rules, before deciding whether or not to investigate."
An ITV spokesperson made it clear that the show will portray the "very negative side" of their decision to take drugs.
In a statement, they added: "Coronation Street has a strong tradition of tackling serious contemporary issues in a responsible way.
"In upcoming episodes we will be showing the very negative side of drug taking as Billy's life starts to fall apart."


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text:
\section{Introduction}

In this paper we introduce a relative entropy $\cH(A,B)$ of two bounded operators $A$ and $B$ on a Hilbert space and discuss its elementary properties. Our relative entropy $\cH(A,B)$ could be useful in many practical situations including quantum information theory. In a separate work~\cite{LewSab-13a}, we will use it to discuss the orbital stability of certain stationary states for the Hartree equation of an interacting gas containing infinitely many quantum particles.

For simplicity we work on the interval $I=[0,1]$ but all our results can be generalized to any finite interval. We consider two $N\times N$ hermitian matrices $A,B$ whose spectrum is a subset of $I$. For $\phi\in C^0([0,1],\R)\cap C^1((0,1),\R)$ a \emph{convex} function, we introduce the following relative entropy\footnote{This expression always makes sense if $0<B<1$. If the spectrum of $B$ contains $0$ or $1$ and if $\phi$ is not differentiable at these points, then we let $\cH(A,B)=+\ii$ except when $A=B$ on $\ker B$, $\ker(1-B)$, or $\ker B\oplus\ker(1-B)$. In this case the trace is by definition taken on the orthogonal to these subspaces.}
\begin{equation}
\boxed{\cH(A,B)=\tr\Big(\phi(A)-\phi(B)-\phi'(B)(A-B)\Big).}
\label{eq:def_relative_entropy}
\end{equation}

We remind the reader of Klein's lemma.

\begin{lemma}[Klein~{\cite[Prop. 3.16]{OhyPet-93},~\cite[Thm 2.5.2]{Ruelle}}]\label{lem:Klein}
If $f_k,g_k:[a,b]\to\R$ are $K$ measurable functions such that $\sum_{k=1}^K f_k(x)g_k(y)\geq0$ for all $x,y\in [a,b]$, then we have
$\sum_{k=1}^K \tr\,\big( f_k(A)g_k(B)\big)\geq0$ for all hermitian matrices $a\leq A,B\leq b$.
\end{lemma}

Because of our assumption that $\phi$ is convex, we obtain from the lemma that the trace in~\eqref{eq:def_relative_entropy} is always non-negative. If $\phi$ is strictly convex, it can be proved that $\cH(A,B)=0$ only when $A=B$. Therefore, $\cH(A,B)$ is an appropriate object to measure how $A$ is close to $B$.

The formula~\eqref{eq:def_relative_entropy} is motivated by several physical situations. First, the usual von Neumann entropy
$$\cH_{\text{vN}}(A,B)=\tr A\big(\log A-\log B\big)$$
corresponds to taking
$$\phi_{\rm vN}(x)=x\log(x),\qquad x\in [0,1].$$
This is not the only function of interest, however. If we take two quasi-free states $\alpha$ and $\beta$ on the CAR (resp. CCR) algebra of a (finite-dimensional) Hilbert space $\gH$, then the corresponding von Neumann entropy can be expressed in terms of the one-particle density matrices $A$ and $B$ of $\alpha$ and $\beta$, leading to the expression~\eqref{eq:def_relative_entropy} with, this time,
$$\phi_{\text{CAR}}(x)=x\log(x)+(1-x)\log(1-x),\qquad x\in [0,1]$$
and
$$\phi_{\text{CCR}}(x)=x\log(x)-(1+x)\log(1+x),\qquad x\in [0,1],$$
respectively~\cite{BacLieSol-94}. It seems therefore natural to study the properties of $\cH$ for a general convex function $\phi$, and this is the main purpose of this paper. We will particularly study the possibility to define $\cH$ for two bounded operators in an \emph{infinite} dimensional Hilbert space $\gH$. \emph{Monotonicity} then plays a crucial role and we discuss it in the next section.

\section{Monotonicity}

Our goal is to define $\cH(A,B)$ for $A$ and $B$ two bounded operators on an infinite dimensional Hilbert space. The usual technique employed in statistical mechanics is to first project the system into a finite dimensional space and then to take the limit when the dimension goes to $+\ii$. This means that we will define
\begin{equation}
\cH(A,B):=\lim_{k\to\ii}\cH(P_kAP_k,P_kBP_k)
\label{eq:def_limit}
\end{equation}
where $P_k$ is an increasing sequence of finite-rank orthogonal projectors on $\gH$. For a general (convex) function $\phi$, the limit is not guaranteed to exist. Furthermore, the limit could be different for two sequences $P_k$ and $P'_k$. One natural way to ensure a good behavior is to add the constraint that $\cH$ is \emph{monotone}, that is, satisfies the inequality
\begin{equation}
\boxed{\cH(PAP,PBP)\leq \cH(A,B) }
\label{eq:monotonicity}
\end{equation}
for every finite self-adjoint matrices $0\leq A,B\leq 1$ and every orthogonal projector $P$. The limit on the right of~\eqref{eq:def_limit} is then monotone and $\cH(A,B)$ is well defined in $\R^+\cup\{+\ii\}$. Our purpose in this section is to characterize the convex functions $\phi$ for which $\cH$ is monotone in the sense of~\eqref{eq:monotonicity}.

Let us emphasize that another notion of monotonicity is often used in the literature. Assume that our (finite-dimensional) Hilbert space is the tensor product $\gH=\gH_1\otimes\gH_2$. Define $\Phi(A)=\tr_{\gH_2}(A)$, the partial trace with respect to the second Hilbert space. Then the von Neumann relative entropy (corresponding to $\phi=\phi_{\rm vN}$) satisfies that
\begin{equation}
\cH_{\rm vN}(\Phi(A),\Phi(B))\leq \cH_{\rm vN}(A,B)
\label{eq:strong_monotonicity}
\end{equation}
for every $0\leq A,B\leq 1$ with $\tr(A)=\tr(B)$. Indeed, the property holds for any trace-preserving completely positive map $\Phi$ between two Hilbert spaces~\cite{OhyPet-93}. Since $A\mapsto PAP+P^\perp AP^\perp$ is completely positive and trace-preserving, it is easy to deduce that the von Neumann entropy also satisfies~\eqref{eq:monotonicity}. Furthermore, it can be shown (see~\cite[Chap. 2]{OhyPet-93}) that $\phi(x)=ax\log(x)+bx+c$ with $a\geq0$ are the only functions for which~\eqref{eq:strong_monotonicity} is satisfied.

The weaker monotonicity property~\eqref{eq:monotonicity} is associated with a decomposition of the Hilbert space $\gH$ into a direct sum $\gH=\gH_1\oplus\gH_2$, instead of a tensor product. Indeed, if we have a quasi-free state on the CCR (resp. CAR) algebra of a Hilbert space $\gH$, the restriction of this state to a sub-algebra of observables in $\gH_1\subset\gH$ is obtained by taking a \emph{partial trace for the state} and by \emph{replacing the one-particle density matrix $A$ by $P_1AP_1$}. The tensor structure at the level of states is transformed into a direct sum at the level of density matrices. From this discussion we conclude that the relative entropy associated with $\phi_{\rm CCR/CAR}$ (defined above) must also satisfy the monotonicity property~\eqref{eq:monotonicity}, even if they do not satisfy the stronger monotonicity~\eqref{eq:strong_monotonicity}.

In this section we want to characterize the convex functions $\phi$ for which $\cH$ is monotone in the sense of~\eqref{eq:monotonicity}. In a recent paper~\cite{AudHiaPet-10}, Audenaert, Hiai and Petz have studied the strong sub-additivity of the generalized entropy $A\mapsto -\tr\;\phi(A)$ when $\phi'$ is operator monotone. Petz then conjectured that the operator monotonicity of $\phi'$ is necessary. Here we express everything in terms of the relative entropy~\eqref{eq:def_relative_entropy} and we prove Petz's conjecture in this setting.

\begin{theorem}[Monotonicity]\label{thm:monotonicity}
Let $\phi\in C^0([0,1],\R)\cap C^1((0,1),\R)$ be a convex function. Then the following are equivalent

\begin{enumerate}
\item $\phi'$ is operator monotone on $(0,1)$;

\smallskip

\item For any linear map $X:\gH_1\to\gH_2$ on finite-dimensional spaces $\gH_1$ and $\gH_2$ with $X^*X\leq1$, and for any $0\leq A,B\leq 1$ on $\gH_1$, we have
\begin{equation}
\cH(XA X^*,XBX^*)\leq \cH(A,B),
\label{eq:monotonicity_2}
\end{equation}
with $\cH(A,B)$ defined in~\eqref{eq:def_relative_entropy}.
\end{enumerate}
\end{theorem}

Typical functions $\phi$ satisfying the conditions in Theorem~\ref{thm:monotonicity} are provided in~\cite{AudHiaPet-10} and include
$$\phi(x)=\begin{cases}
(t+x)\log(t+x),&t\geq0\\
-\log(t+x),&t>0,\\
-x^m,&0<m\leq 1,\\
x^m,&1\leq m\leq 2.\\
\end{cases}$$
By using L\"owner's characterization of operator-monotone functions, it is possible to write an integral representation formula for the allowed functions $\phi$. This is done later in the proof, see~\eqref{eq:Lowner2}.

\begin{proof}
As a preliminary we remark that $x\mapsto x\phi'(x)$ possesses a limit when $x\to0^+$. Indeed, since $\phi$ is convex by assumption, then $x\mapsto x\phi'(x)-\phi(x)$ must be increasing on $(0,1)$. Also, $x\phi'(x)-\phi(x)\geq -\phi(0)$ and we conclude that $x\mapsto x\phi'(x)-\phi(x)$ possesses a limit when $x\to0^+$. Since $\phi$ is continuous by assumption, then $x\phi'(x)$ must have a limit as well, and this limit is $\geq0$. The same argument shows that $(1-x)\phi'(x)$ has a limit when $x\to 1^-$.

Next we prove that (2) implies (1).
Let $A$ be any matrix such that $0<\eta<A<1-\eta$, and $P$ be any orthogonal projector. We use the shorthand notation $A_{11}=PAP$, $A_{22}=P^\perp AP^\perp$ and so on. Let $X$ be any self-adjoint matrix such that $P^\perp X=XP^\perp=P^\perp$ and $0< X\leq1$. Then we have by assumption
\begin{align*}
\cH\Big(XA X,XA_{11}X+A_{22}\Big)&=\cH\Big(XA X,X(A_{11}+A_{22})X\Big)\\
&\leq \cH\big(A,A_{11}+A_{22}\big).
\end{align*}
Note that
\begin{multline*}
\tr_\gH\; \phi'\big(A_{11}+A_{22}\big)\big(A-A_{11}-A_{22}\big)\\
=\tr_{P\gH\oplus P^\perp \gH}
\begin{pmatrix}
\phi'\big(A_{11}\big) & 0\\
0 & \phi'\big(A_{22}\big)
\end{pmatrix}
\begin{pmatrix}
0 & A_{12}\\
A_{21} & 0
\end{pmatrix}=0.
\end{multline*}
A similar argument for the term involving $X$ gives
\begin{align*}
 0&\leq \cH\big(A,A_{11}+A_{22}\big) - \cH\Big(XA X,XA_{11}X+A_{22}\Big)\\
&=\tr_\gH\Big(\phi(A)-\phi(A_{11}+A_{22})-\phi(XAX)+\phi(XA_{11}X+A_{22}) \Big)\\
&=\tr_\gH\Big(\phi(A)-\phi(XAX)\Big)-\tr_{P\gH}\Big(\phi(A_{11})-\phi((XAX)_{11}) \Big).
\end{align*}
Now we choose $X:=\exp(-\epsilon C)$ for some $C>0$ living in the range of $P$ and $\epsilon>0$. We remark that $XAX=A-\epsilon(CA+AC)+O(\epsilon^2)$. Using the differentiability of $A\mapsto \tr\phi(A)$, we obtain
$$\tr_\gH\Big(\phi(A)-\phi(e^{-\epsilon C}Ae^{-\epsilon C})\Big)=2\epsilon \tr_\gH \big(CA\phi'(A)\big)+o(\epsilon).$$
The same argument for the other term and the fact that the inequality is valid for all $C>0$ on $P\gH$ and all $\epsilon>0$ gives
$$\big(A\phi'(A)\big)_{11}\geq A_{11}\phi'(A_{11}).$$
This inequality is valid for all $0<A<1$. We have already seen that $x\mapsto x\phi'(x)$ can be extended by continuity at $x=0$. So the property stays true for all $0\leq A< 1$. By~\cite[Thm V.2.3 \& Thm V.2.9]{Bhatia}, this exactly proves that the function $\phi'$ is operator monotone on $I$. These results give us the additional information that $\lim_{x\to0^+}x\phi'(x)\leq0$, and since we already had the other inequality we deduce that $\lim_{x\to0^+}x\phi'(x)=0$.

Next we turn to the proof that (1) implies (2), for which we follow arguments from~\cite{AndPet-09,AudHiaPet-10}. By L\"owner's theorem (see~\cite[Corollary V.4.5]{Bhatia}), we can write $\phi'$ as
\begin{equation}
\phi'(x)=a+b\int_{-1}^1 \frac{2x-1}{1-\lambda(2x-1)}\,d\nu(\lambda)
\label{eq:Lowner}
\end{equation}
with $b\geq0$ and $\nu$ a Borel probability measure on $[-1,1]$.
Integrating this formula, we find
$$\phi(x)=ax+c-\frac{b}2\int_{-1}^1\left(\frac{2x-1}{\lambda}+\frac{\log\big(1+\lambda (1-2x)\big)}{\lambda^2}\right)\,d\nu(\lambda).$$
The fact that $\phi$ possesses a limit at $0$ and $1$ implies that $\nu(\{-1\})=\nu(\{1\})=0$. From this we can actually prove that
\begin{equation}
\lim_{x\to0^+}x\phi'(x)=\lim_{x\to1^-}(1-x)\phi'(x)=0.
\label{eq:limits_phi}
\end{equation}
After letting $t=-(1+\lambda)/(2\lambda)$ for $\lambda\in(-1,0)$ and $t=(1-\lambda)/(2\lambda)$ for $\lambda\in(0,1)$, $\phi$ can also be written as
\begin{multline}
\phi(x)=a'x+c'-\int_{0}^\ii\left(\log(t+x)-\log(t+1/2)-\frac{2x-1}{2t+1}\right)\,d\nu_1(t)\\
-\int_{0}^\ii\left(\log(t+1-x)-\log(t+1/2)+\frac{2x-1}{2t+1}\right)\,d\nu_2(t)
\label{eq:Lowner2}
\end{multline}
with $\nu_1$ and $\nu_2$ two positive Borel measures satisfying
$$\int_{0}^\ii \frac{\big(d\nu_1+d\nu_2\big)(t)}{(2t+1)^2}<\ii.$$
The constraint that $\phi$ admits limits at $0$ and $1$ now means that
\begin{equation}
-\int_0^1\log(t)\big(d\nu_1+d\nu_2\big)(t)<\ii
\label{eq:measure_finite_log}
\end{equation}
and it implies $\nu_1(\{0\})=\nu_2(\{1\})=0$, as we have already said before. The parameters are given in the following table for the usual distributions.

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
 & $d\nu_1/dt$&$d\nu_2/dt$&$a'$&$c'$\\
\hline
von Neumann & 1 & 0 & $1-\log2$ &$-1/2$ \\
Fermi-Dirac & 1 & 1 & 0 &$-\log2$\\
Bose-Einstein & $\1(0\le t\le1)$ & 0 & $-\log3$ & $\log2-\log3$ \\
\hline
\end{tabular}
\end{center}

It suffices to prove the monotonicity for $\phi(x)=-\log(t+x)$ and for $\phi(x)=-\log(t+1-x)$ with $t\geq0$. We start by considering the case $t=0$ and we let $\phi(x)=-(1/2)\log(x)$. We follow the method of~\cite{AndPet-09}. For $A$ a positive $n\times n$ matrix, we introduce the corresponding normalized Gaussian, defined by
$$f_A(x):=\frac{\sqrt{\det(A)}}{(2\pi)^{n/2}}\exp\left(-\frac{x^TAx}{2}\right).$$
A simple calculation shows that the \emph{classical} entropy of $f_A$ is
\begin{align*}
-\int_{\R^n}f_A(x)\log f_A(x)\,dx&=-\frac{\log\det(A)}{2}+ \frac{n\log(2\pi e)}{2}\\
&=\frac{\tr\log (A^{-1})}{2}+ \frac{n\log(2\pi e)}{2},
\end{align*}
and that the \emph{classical} relative entropy of two Gaussians $f_A$ and $f_B$ is
\begin{equation*}
\cH_{\rm class}(f_A,f_B)=\int_{\R^n}f_A(x)\log \frac{f_A(x)}{f_B(x)}\,dx=\cH(A^{-1},B^{-1}),
\end{equation*}
see~\cite[Eq. (21)--(22)]{AndPet-09}. Now, let $P$ be any orthogonal projection on $\gH$. Up to a change of variables we may assume for simplicity that $x=(x_1,x_2)$ with $Px=(x_1,0)$ for all $x\in\R^n$. The marginal of a Gaussian is computed in~\cite[Lemma 4]{AndPet-09} and it is equal to
$$\int f_A(x_1,x_2)\,dx_2=f_{A_{11}-A_{12}A_{22}^{-1}A_{21}}(x_1)$$
where we have used the decomposition
$$A=\begin{pmatrix}
A_{11}&A_{12}\\
A_{21}&A_{22}
\end{pmatrix}.$$
Noticing that $(A_{11}-A_{12}A_{22}^{-1}A_{21})^{-1}=(A^{-1})_{11}$ by the Schur complement formula, we deduce that
$$\cH_{\rm class}\left(\int f_{A^{-1}} dx_2,\int f_{B^{-1}}dx_2\right)=\cH\big(A_{11},B_{11}\big)=\cH(PA P,PB P).$$
The inequality $\cH(PA P,PB P)\leq \cH(A,B)$ now follows from the well-known monotonicity of the classical relative entropy in terms of marginals. The argument is exactly the same if $\log(x)$ is replaced by $\log(t+x)$ and $\log(t+1-x)$, since $(t+A)_{11}=t+A_{11}$ and $(t+1-A)_{11}=t+1-A_{11}$.

We have proved the monotonicity for a projection $P$. If $U:\gH_1\to\gH_2$ is a partial isometry between two spaces such that $U^*U=1_{\gH_1}$ and $UU^*\leq 1_{\gH_2}$, it is obvious that $\cH(UAU^*,UBU^*)=\cH(A,B)$ since $U\gH_1$ is then isometric to a subspace of $\gH_2$. On the other hand, if we have $UU^*=1_{\gH_2}$ and $U^*U\leq 1_{\gH_1}$, then $\cH(UAU^*,UBU^*)\leq \cH(A,B)$ by the previous result on projections, since $U^*\gH_2$ is now isometric to a subspace of $\gH_1$.

In order to deduce the result for an arbitrary operator $X:\gH\to\gH$ with $X^*X\leq1$, we double the dimension of the space, that is, we introduce the partial isometry
$$\begin{array}{rccrcl}
U:&\gH&\to&\gH&\oplus&\gH\\
&f&\mapsto&Xf&\oplus &\sqrt{1-X^*X}f
\end{array}$$
and we infer $\cH(UA U^*,UB U^*)= \cH(A,B)$, since $U^*U=1_{\gH}$.
Note that
$$UAU^*=\begin{pmatrix}
XA X & XA\sqrt{1-X^*X}\\
\sqrt{1-X^*X}A X & \sqrt{1-X^*X}A\sqrt{1-X^*X}\\
\end{pmatrix}.$$
If we project onto the first Hilbert space of $\gH\oplus\gH$, this decreases the relative entropy and we obtain the result. The case of $X:\gH_1\to\gH_2$ with different Hilbert spaces and $X^*X\leq1$ follows from the polar decomposition $X=UY$ with $U$ a partial isometry and $0\leq Y=\sqrt{X^*X}\leq1$. This concludes our proof that (1) implies (2).
\end{proof}

After having characterized the functions $\phi$ for which $\cH$ is monotone, we quickly mention the consequences on the entropy $S(A)=-\tr\;\phi(A)$, for completeness.

One of the fundamental properties of the von Neumann entropy (defined with $\phi_{\rm vN}(x)=x\log(x)$) is its strong subadditivity (SSA), which was proved by Lieb and Ruskai in \cite{LieRus-73a,LieRus-73b}. The fact that SSA is very important for large quantum systems was first remarked by Robinson and Ruelle in \cite{RobRue-67}, see also \cite{Wehrl-78} and~\cite{HaiLewSol_2-09}. It is well-known that SSA is equivalent to the monotonicity of the relative von Neumann entropy $\cH$ under completely positive trace-preserving maps~\cite{OhyPet-93,Carlen-10,Wehrl-78}, as in~\eqref{eq:strong_monotonicity}.

The usual definition of SSA is expressed using a decomposition of the ambient Hilbert space $\gH$ into a tensor product of the form $\gH=\gH_1\otimes\gH_2\otimes\gH_3$, and we do not recall it here. In our setting there is another concept of SSA, which is associated with a decomposition of $\gH$ into a direct sum $\gH=\gH_1\oplus\gH_2\oplus\gH_3$ of three spaces (instead of a tensor product), and which was considered before in~\cite{AudHiaPet-10}. For $\phi=\phi_{\rm CCR/CAR}$, this is just the usual SSA for quasi-free states, expressed in terms of the corresponding one-particle density matrices.
We denote by $P_j$ the corresponding orthogonal projections, as well as $P_{12}=P_1+P_2$ and $P_{123}=1$.

\begin{corollary}[SSA for the entropy]
We assume that $\phi\in C^0([0,1],\R)$ and that $\phi'$ is operator monotone on $(0,1)$.
Let $S(A)=-\tr_\gH\,\phi(A)$, with $\dim(\gH)<\ii$. Then we have
$$S(P_{123}AP_{123})+S(P_2AP_2)\leq S(P_{12}AP_{12})+S(P_{23}AP_{23}),$$
for all self-adjoint matrices $0\leq A\leq 1$ on $\gH$.
\end{corollary}

\begin{proof}
By using the monotonicity of the relative entropy, we deduce that
$\cH(P_{123}AP_{123},P_{23}AP_{23})\geq\cH(P_{12}AP_{12},P_{2}AP_{2})$.
It is straightforward to verify that $\cH(P_{123}AP_{123},P_{23}AP_{23})=S(P_{123}AP_{123})-S(P_{23}AP_{23})$ and that $\cH(P_{12}AP_{12},P_{2}AP_{2})=S(P_{12}AP_{12})-S(P_{2}AP_{2})$ and the result follows.
\end{proof}

\begin{remark}
Since an operator monotone function is always $C^\ii$, there is no need to assume anymore that $\phi\in C^1$.
\end{remark}

\section{Definition in infinite-dimensional spaces}

We now turn to the definition of $\cH(A,B)$ in an infinite-dimensional space $\gH$. For any two self-adjoint operators $0\leq A,B\leq 1$, we let
\begin{equation}
\cH(A,B):=\lim_{k\to\ii}\cH(P_kA P_k,P_kB P_k)
\label{eq:def_relative_entropy_abstract}
\end{equation}
where $P_k$ is any increasing sequence of finite-dimensional projections on $\gH$, such that $P_k\to 1$ strongly.
The following result says that the limit exists in $[0,\ii]$ and provides important properties of $\cH$.

\begin{theorem}[Generalized relative entropy in infinite dimension]\label{thm:relative_entropy}
We assume that $\phi\in C^0([0,1],\R)$ and that $\phi'$ is operator monotone on $(0,1)$.

\smallskip

\noindent$\bullet$ \emph{($\cH$ is well-defined).} For any increasing sequence $P_k$ of finite-dimensional projections on $\gH$ such that $P_k\to 1$ strongly, the sequence
$\cH(P_kA P_k,P_kB P_k)$ is monotone and possesses a limit in $\R^+\cup\{+\ii\}$. This limit does not depend on the chosen sequence $P_k$ and hence $\cH(A,B)$ is well-defined in $\R^+\cup\{+\ii\}$.

\smallskip

\noindent$\bullet$ \emph{(Monotonicity).} The so-defined relative entropy $\cH(A,B)$ is \emph{monotone}: for any linear map $X:\gH_1\to\gH_2$ such that $X^*X\leq1$ and any $0\leq A,B\leq 1$ on $\gH_1$, we have
\begin{equation}
\cH(A,B)\geq \cH(XA X^*,XBX^*).
\label{eq:monotonicity_3}
\end{equation}

\smallskip

\noindent$\bullet$ \emph{(Approximation).} If $X_k:\gH_1\to\gH_k$ is a sequence such that $X_k^*X_k\leq1$ and $X_k^*X_k\to 1$ strongly in $\gH_1$, then
\begin{equation}
\cH(A,B)=\lim_{k\to\ii}\cH(X_kA X_k^*,X_kBX_k^*).
\label{eq:strong_CV}
\end{equation}

\smallskip

\noindent$\bullet$ \emph{(Weak lower semi-continuity).} The relative entropy is \emph{weakly lower semi-continuous}: if $0\leq A_n,B_n\leq 1$ are two sequences such that $A_n\wto A$ and $B_n\wto B$ weakly-$\ast$ in $\cB(\gH)$, then
\begin{equation}
\cH(A,B)\leq\liminf_{n\to\ii} \cH(A_n,B_n).
\label{eq:wlsc}
\end{equation}
\end{theorem}

Most of the statements of the theorem easily follow from the monotonicity in finite dimension proved in Theorem~\ref{thm:monotonicity}, showing the importance of this concept.

\begin{proof}
 We split the proof into several steps.

\subsubsection*{Step 1. Lower semi-continuity in finite dimension}

We want to prove that $\cH(A,B)\leq\liminf_{n\to\ii}\cH(A_n,B_n)$ if $A_n\to A$ and $B_n\to B$ are convergent sequences of hermitian matrices. All the terms in the definition of $\cH(A,B)$ are continuous, except possibly for $(A,B)\mapsto \tr\, \phi'(B) (A-B)$. We write this term as
\begin{multline*}
\tr\, \phi'(B) (B-A)\\= -\tr\, \phi'(B)\1(\phi'(B)\leq0)A+\tr\, \phi'(B)\1(\phi'(B)\geq0)(1-A)\\
+\tr\, B\phi'(B)\1(\phi'(B)\leq0)
-\tr\, (1-B)\phi'(B)\1(\phi'(B)\geq0).
\end{multline*}
Using that the first two terms involve non-negative matrices and that the next ones are continuous by~\eqref{eq:limits_phi}, we obtain the result.

\subsubsection*{Step 2. The definition in infinite dimension does not depend on $(P_k)$}
Let us consider two sequences $(P_k)$ and $(P'_\ell)$ of increasing orthogonal projections, such that $P_k\to1$ and $P'_\ell\to1$ strongly. By the monotonicity in finite dimension, we have
$\cH(P_kP'_\ell A P'_\ell P_k,P_kP'_\ell B P'_\ell P_k)\leq \cH(P'_\ell AP'_\ell,P'_\ell BP'_\ell)$.
Taking first $\ell\to\ii$ using that $P_kP'_\ell A P'_\ell P_k\to P_kP'_\ell B P'_\ell P_k$ and the lower semi-continuity of $\cH$ in the finite dimensional space ${\rm Ran}(P_k)$, we get
$$\cH(P_kA P_k,P_kB P_k)\leq \lim_{\ell\to\ii} \cH(P'_\ell A P'_\ell,P'_\ell B P'_\ell).$$
Taking then $k\to\ii$ shows that
$$\lim_{k\to\ii}\cH(P_kA P_k,P_kB P_k)\leq \lim_{\ell\to\ii} \cH(P'_\ell A P'_\ell,P'_\ell B P'_\ell).$$
Exchanging $P_k$ and $P'_\ell$ gives the other inequality.

\subsubsection*{Step 3. Monotonicity in infinite dimension}
Let $X:\gH_1\to\gH_2$ be such that $X^*X\leq1_{\gH_1}$. For fixed sequences $(P_k)$ and $(P'_\ell)$ of finite-rank projections on $\gH_1$ and $\gH_2$ respectively, we have
\begin{align*}
\cH(P'_\ell XP_k A P_kX^*P'_\ell,P'_\ell XP_kB P_kX^*P'_\ell)&\leq \cH(XP_k A P_kX^*,XP_kB P_kX^*)\\
&\leq \cH(P_k A P_k,P_kB P_k).
\end{align*}
Taking first $k\to\ii$ using that $P'_\ell XP_k A P_kX^*P'_\ell\to P'_\ell X A X^*P'_\ell$ with the finite rank projection $P'_\ell$ fixed, and only then $\ell\to\ii$, gives the monotonicity.

\subsubsection*{Step 4. Weak lower semi-continuity}
If $ A_n\wto A$ and $B_n\wto B$ weakly-$\ast$ in $\cB(\gH)$, then
$$\cH(P_k A P_k,P_kB P_k)\leq\liminf_{n\to\ii}\cH(P_k A_n P_k,P_kB_n P_k)\leq \liminf_{n\to\ii}\cH( A_n,B_n)$$
for any fixed sequence $(P_k)$ as before. Here we have used that $P_k A_n P_k\to P_k A P_k$ for a finite-rank projection and the lower semi-continuity in finite dimension. Taking $k\to\ii$ gives the weak lower semi-continuity.

\subsubsection*{Step 5. Convergence for any sequence $X_k\to1$}
Let $X_k:\gH\to\gH$ be any sequence of self-adjoint operators, such that $X_k^*X_k\leq1$ and $X_k^*X_k\to1$ strongly. We want to prove that $\cH(X_k A X_k^*,X_kB X_k^*)\to \cH(A,B)$ but, by the polar decomposition $X_k=U_kY_k$, we can always assume that $X_k$ is self-adjoint, hence that $X_k\to1$ strongly. We have $\cH(X_k A X_k,X_kB X_k)\leq \cH(A,B)$. On the other hand, $X_k A X_k\to  A$ strongly and thus the weak lower semi-continuity implies the reverse inequality. If $X_k:\gH_1\to\gH_2$, the argument is similar.
\end{proof}

\section{Klein inequalities and consequences}

In this last section, we derive some useful bounds on $A-B$ when the relative entropy $\cH(A,B)$ is finite. For instance, we will prove that $A-B$ must be a Hilbert-Schmidt (hence compact) operator. Our main result is inspired of~\cite[Thm 1]{HaiLewSei-08} and of~\cite[Lemma 1]{FraHaiSeiSol-12} and it is mainly based on Klein's Lemma~\ref{lem:Klein}, hence the name ``Klein inequalities''.

\begin{theorem}[Klein inequalities]\label{thm:Klein}
We assume that $\phi\in C^0([0,1],\R)$, and that $\phi'$ is operator monotone on $(0,1)$ and not constant.

\smallskip

\noindent$\bullet$ \emph{(Lower bound).} There exists $C>0$ (depending only on $\phi$) such that
\begin{equation}
\cH(A,B)\geq C\,\tr\,\big(1+|\phi'(B)|\big)(A-B)^2,
\label{eq:Klein}
\end{equation}
for all $0\leq A,B\leq 1$. Hence, $A-B$ is Hilbert-Schmidt when $\cH(A,B)<\ii$.

\smallskip

\noindent$\bullet$ \emph{(Upper bound).} Similarly, we have
\begin{equation}
\cH(A,B)\leq C\,\tr\,\left(\frac{1}{B^{2}}+\frac{1}{(1-B)^2}\right)(A-B)^2,
\label{eq:Klein_upper}
\end{equation}
for all $0\leq A,B\leq1$.

\smallskip

\noindent$\bullet$ \emph{(Regularity).} If $0\leq A\leq1$ and $0<\epsilon\leq A',B\leq 1-\epsilon$ are such that $A-A'$ and $A'-B$ are Hilbert-Schmidt operators, then
\begin{equation}
\big|\cH(A,B)-\cH(A',B)\big|\leq C_\epsilon\Big(\norm{A-A'}_{\gS^2}^2+\norm{A'-B}_{\gS^2}\norm{A-A'}_{\gS^2}\Big)
\label{eq:Klein_Lipschitz}
\end{equation}
where $C_\epsilon$ depends on $\phi$ and on $\epsilon$.
\end{theorem}

We remark that the inequality~\eqref{eq:Klein_upper} is far from optimal if $\phi(x)$ is very smooth at $0$ or $1$. Similarly,~\eqref{eq:Klein_Lipschitz} holds without the assumptions that the density matrices have their spectrum in $[\epsilon,1-\epsilon]$, if $\phi''$ is bounded on $[0,1]$.

\begin{proof}[Proof of Theorem \ref{thm:Klein}]
We have
\begin{equation*}
-\frac{\log(t+x)-\log(t+y)-\frac{x-y}{t+y}}{(x-y)^2}=\frac{1}{(t+y)^{2}}\,\frac{h-\log(1+h)}{h^2},
\end{equation*}
with $h=(x-y)/(t+y)$. Using that $h\mapsto h^{-2}(h-\log(1+h)$ is decreasing, we deduce that
$$\min_{0\leq x\leq 1}\left(-\frac{\log(t+x)-\log(t+y)-\frac{x-y}{t+y}}{(x-y)^2}\right)=\frac{\frac{1-y}{t+y}-\log\left(1+\frac{1-y}{t+y}\right)}{(1-y)^2}.$$
Using now that
$$\frac{h-\log\left(1+h\right)}{h^2}\geq \frac{1}{2(h+1)}$$
for all $h\geq0$, we conclude that
\begin{equation}
-\left(\log(t+x)-\log(t+y)-\frac{x-y}{t+y}\right)\\ \geq \frac{(x-y)^2}{2(1+t)(t+y)}.
\label{eq:Klein_x_y_1}
\end{equation}
By replacing $x$ by $1-x$ and $y$ by $1-y$, we get the same estimate with $t+1-y$ in place of $t+y$ in the denominator on the right side. Inserting in~\eqref{eq:Lowner2} we find that, for all $0<x,y<1$,
\begin{align*}
&\phi(x)-\phi(y)-\phi'(y)(x-y)\\
&\qquad \geq \frac{b}2\left(\int_0^\ii \frac{d\nu_1(t)}{(1+t)(t+y)}+\int_0^\ii \frac{d\nu_2(t)}{(1+t)(t+1-y)}\right)(x-y)^2\\
&\qquad \geq C\left(\int_1^\ii \frac{d\nu_1(t)+d\nu_2(t)}{(2t+1)^2}+\int_0^1 \frac{d\nu_1(t)}{t+y}+\int_0^1 \frac{d\nu_2(t)}{t+1-y}\right)(x-y)^2.
\end{align*}
From the integral representation of $\phi$ in~\eqref{eq:Lowner2}, we see that
$$\phi'(x)\underset{x\to0^+}{\sim}-b\int_0^1\frac{d\nu_1(t)}{(t+x)}+C,\qquad \phi'(x)\underset{x\to1^-}{\sim}b\int_0^1\frac{d\nu_2(t)}{(t+1-x)}+C.$$
Therefore we have proved that
\begin{equation*}
\forall 0\leq x,y\leq1,\quad  \phi(x)-\phi(y)-\phi'(y)(x-y)\geq C(1+|\phi'(y)|)(x-y)^2.
\end{equation*}
The result follows for finite matrices by Klein's Lemma~\ref{lem:Klein}, and for general operators by the definition of $\cH$ using projections on finite-dimensional spaces.

Maximizing instead of minimizing in~\eqref{eq:Klein_x_y_1}, we find
\begin{align*}
-\left(\log(t+x)-\log(t+y)-\frac{x-y}{t+y}\right) &\leq (x-y)^2\frac{-\frac{y}{t+y}-\log\left(1-\frac{y}{t+y}\right)}{y^2}\\
&\leq (x-y)^2\frac{1-\log\left(\frac{t}{t+y}\right)}{(t+y)^2}
\label{eq:Klein_x_y_2}
\end{align*}
and a similar estimate for $1-x$ and $1-y$. After integrating with respect to $t$, we see that the integral converges for large $t$, uniformly in $y$. The only possibly diverging terms are
$$\int_0^1\frac{\log\left(\frac{t}{t+y}\right)}{(t+y)^2}d\nu_1(t)+\int_0^1\frac{\log\left(\frac{t}{t+1-y}\right)}{(t+1-y)^2}d\nu_2(t)$$
and they can always be estimated by a constant times $y^{-2}+(1-y)^{-2}$, by~\eqref{eq:measure_finite_log}. (This is far from optimal if $\nu_1$ or $\nu_2$ vanish sufficiently fast at $0$).

Now we turn to the proof of~\eqref{eq:Klein_Lipschitz}, which only has to be done in finite dimension, by the definition of $\cH$. We remark that
$$\cH(A,B)-\cH(A',B)=\cH(A,A')+\tr \big(\phi'(A)-\phi'(B')\big)(A-A').$$
Since $\phi''$ is bounded on $[\epsilon,1-\epsilon]$, we have
$$\tr \big(\phi'(A)-\phi'(B')\big)^2\leq C\tr \big(A'-B\big)^2,$$
by Klein's Lemma~\ref{lem:Klein}.
By H\"older's inequality this gives
$$\left|\tr \big(\phi'(A)-\phi'(B')\big)(A-A')\right|\leq C\norm{A'-B}_{\gS^2}\norm{A-A'}_{\gS^2}.$$
Using~\eqref{eq:Klein_upper} and the assumption that $\epsilon\leq A'\leq1-\epsilon$ to estimate $\cH(A,A')$ gives~\eqref{eq:Klein_Lipschitz}.
\end{proof}

Using the previous Klein inequalities, we can get some information on the set of one-particle density matrices $A$ which have a finite relative entropy with respect to a given $B$. Of particular interest is fact that any operator $A$ such that $\cH(A,B)<\ii$ can be approximated in an appropriate sense by a sequence of finite rank perturbations of $B$.

\begin{corollary}[Density of finite rank perturbations]\label{cor:density}
We assume that $\phi\in C^0([0,1],\R)$ and that $\phi'$ is operator monotone on $(0,1)$ and that $0\leq A,B\leq1$ have a finite relative entropy, $\cH(A,B)<\ii$. Then for every $\epsilon>0$, there exists $0\leq A'\leq1$ such that
\begin{itemize}
\item $A'-A$ has a finite rank and its eigenvectors belong to $\ker(B)$, to $\ker(1-B)$, or to the domain of $B^{-2}+(1-B)^{-2}$ in the orthogonal of these two subspaces;
\item $\tr (1+|\phi'(B)|)(A-A')^2\leq \epsilon$;
\item $\big|\cH(A,B)-\cH(A',B)|\leq \epsilon$.
\end{itemize}
\end{corollary}

\begin{proof}
We have several cases to look at, depending whether $\phi'$ diverges at the end points $0$ and $1$ and depending whether $B$ has the end points as eigenvalues or not. First we note that if $\phi'$ diverges at $0$ and $\ker B\neq\{0\}$, then $A=B$ on $\ker B$ and therefore there is nothing to do on this subspace and we can argue on the orthogonal of $\ker B$.
So let us now assume that $\ker B=\ker(1-B)=\{0\}$. We introduce the spectral projector $\Pi_\eta:=\1(\eta\leq B\leq 1-\eta)$. By the spectral theorem, we have $\Pi_\eta\to1$ strongly when $\eta\to0$. By Theorem~\ref{thm:relative_entropy}, we deduce that
$$\lim_{\eta\to0}\cH(\Pi_\eta A\Pi_\eta,\Pi_\eta B\Pi_\eta)=\lim_{\eta\to0}\cH(B+\Pi_\eta(A-B)\Pi_\eta,B)=\cH(A,B).$$
In the first equality we have used that
\begin{align*}
\cH(B+\Pi_\eta(A-B)\Pi_\eta,B)&=
\cH(\Pi_\eta^\perp B\Pi_\eta^\perp+\Pi_\eta A\Pi_\eta,\Pi_\eta^\perp B\Pi_\eta^\perp+\Pi_\eta B\Pi_\eta)\\
&=\cH(\Pi_\eta A\Pi_\eta,\Pi_\eta B\Pi_\eta)+\cH(\Pi_\eta^\perp B\Pi_\eta^\perp,\Pi_\eta^\perp B\Pi_\eta^\perp)\\
&=\cH(\Pi_\eta A\Pi_\eta,\Pi_\eta B\Pi_\eta),
\end{align*}
since $\Pi_\eta$ commutes with $B$. By Klein's inequality~\eqref{eq:Klein}, we know that $(1+|\phi'(B)|)^{1/2}(A-B)$ is a Hilbert-Schmidt operator,  and it is then clear that
$$\lim_{\eta\to0}\tr (1+|\phi'(B)|)\Big(\Pi_\eta(A-B)\Pi_\eta-(A-B)\Big)^2=0.$$
The density matrix $A_\eta:=B+\Pi_\eta(A-B)\Pi_\eta$ satisfies all the desired properties, except for the finite rank one, and we need to approximate it further.

In order to approximate $A_\eta$, we will work in the space $\gH_\eta:=\Pi_\eta\gH$. We first remark that, by Klein's inequality, $\Pi_\eta A\Pi_\eta$ is a Hilbert-Schmidt perturbation of $\Pi_\eta B$ and, therefore, the essential spectrum of $\Pi_\eta A\Pi_\eta$ (restricted to $\gH_\eta$) is included in the interval $[\eta,1-\eta]$. The end points $0$ and $1$ can only be isolated eigenvalues of finite multiplicity. Let us denote by $P_0$ and $P_1$ the associated projections (if they exist). Then $A'_\eta:=\Pi_\eta A\Pi_\eta+\epsilon P_0-\epsilon P_1$ is a finite rank perturbation of $A_\eta$ and a simple calculation shows that
\begin{multline*}
\cH(\Pi_\eta A\Pi_\eta+\epsilon P_0-\epsilon P_1,\Pi_\eta B)-\cH(\Pi_\eta A\Pi_\eta,\Pi_\eta B)=\big(\phi(\epsilon)-\phi(0)\big)\tr P_0\\
+ \big(\phi(1-\epsilon)-\phi(1)\big)\tr P_1+\epsilon\tr\big(\Pi_\eta \phi'(B) (P_1-P_0)\big)
\end{multline*}
(the computation must be done in a finite sequence of dimensional spaces $V_k\nearrow \gH_\eta$ and it is simpler to assume that the eigenvectors of $\Pi_\eta A\Pi_\eta$ corresponding to the end points belong to these spaces).
This goes to $0$ as $\epsilon\to0$ with $\eta$ fixed. At this step we have approximated $\Pi_\eta A\Pi_\eta$ by an operator $A'_\eta$ which has its spectrum away from $0$ and $1$, and which is a Hilbert-Schmidt perturbation of $\Pi_\eta B$. Now we can take any sequence $Q_k$ of smooth finite rank operators which converges to $A'_\eta-\Pi_\eta B$ in the Hilbert-Schmidt norm. For large $k$ the operator $A_k:=\Pi_\eta B+Q_k$ satisfies $0\leq A_k\leq1$ and its relative entropy converges to that of $A'_\eta$, by~\eqref{eq:Klein_Lipschitz} in Theorem~\ref{thm:Klein}. This concludes the proof in the case where $\ker B=\ker(1-B)=\{0\}$.

When $0$ or $1$ is an eigenvalue of $B$ and $\phi'$ is bounded at these points, the argument is exactly the same and we leave it to the reader. The projection $P_\eta$ must include all (if the multiplicity if finite) or a sequence of eigenvectors (if the multiplicity is infinite) at the end points.
\end{proof}

\bigskip

\noindent\textbf{Acknowledgements.}
We acknowledge financial support from the European Research Council under the European Community's Seventh Framework Programme (FP7/2007-2013 Grant Agreement MNIQS 258023), and from the French ministry of research (ANR-10-BLAN-0101).

redpajama_set_name: RedPajamaArXiv


Century 21 Beacon Realty, Ada, will hold an open house from 2 to 4 p.m., Sunday, June 10, at a residence listing at 6764 County Road 45, Alger.
Contact Darla Crownhart at 567-674-0765 or e-mail: c21.darla@yahoo.com for more information.
Open the attachment at the bottom of this story for complete details on the property.


SBI clerk main examination 2018 will be conducted on 5th August 2018. We bring for you SBI Clerk mains 2018 financial and computer knowledge pdf capsule. This capsule will help you in quick revision of these two major sections of the examination.
SBI Clerk Financial Awareness: Topic Wise Preparation Tips.
Business News: Mergers Acquisitions and Buyouts.
History of Banking in India: Which was the first bank to be set up in Madras province etc. Nationalization of Banks.
Reserve Bank of India's role in Indian banking system.
Foreign Exchange reserves of India.
Non Banking Financial companies: Definition, Difference from bank etc.
Financial Inclusion and various initiatives for it.
Banking Abbreviations, terms and concepts.
SBI Clerk Financial Awareness Study Plan.
Keep an eye on all the current happenings of the world.
At least give 60 minuted daily for Financial Awareness preparation.
Read current affairs magazine like CST, Chronicle or Wizard.
Please comment below if you have any suggestions or query related to SBI CLerk 2018 examination.


The ability to turn images in any direction - say image is horizontal and you fancy turning it 45 degrees.
Resizing without losing image original place and having to resize the whole image again.
A description of how to use two or more instances of EasyGif to great extent in the help folder.
The ability to amend text placed onto an image that needs editing or correcting later.
I'm not sure what you mean by "Rotational and resizing placeholders", could you elaborate?
You can pull it either side to make it larger or smaller either side up or down.
With rotation the placeholders are placed angular where the rectangular edges should be allowing one to move the rectangle around in either direction without having to resize.
I can give you a picture later if you like but I recommend you look at sitespinner's Webdwarf.
Link below. Free version in case you don't fancy having shareware on your pc.
Last edited by Limewire on Thu Jun 30, 2011 8:50 am, edited 1 time in total.
With regular placeholders. The one with the arrow on the outer edges will move the image in any direction as further below will show.
Now you see the whole image has rotated in a different angle allowing for my slightly hazy hand when teaking the snapshot as I wanted you to see the degree figures.
Thank you for the screenshots. That makes sense.


This beautifully shot and edited third production in the series explains how to show your German Shepherd Dog successfully. This title begins by detailing the judging routine inside the ring from the judges perspective. Only from understanding what the judge is trying to see in the ring can the handler efficiently show the dog. Because the culture of the german ring is unique, interview footage with professional german handlers Klaus Pasch and Walter K�tters is interspersed.
The Basic Training section guides the viewer through the mindset of the competitive show dog owner discussing housing, interaction, early socialization and the various stages of preparation of the show dog from neutralizing the dog to the physical examination and gunfire to successfully teaching the dog to stack and stand.
5. General Procedure for the "Stack"
Copyright Canine Training Systems�, 85 minutes, DVD.


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text:
Arcade Fire: British Summer Time, Hyde Park – 03.07.2014
04/07/2014 Adrienne Law Live Reviews 0
If ever one performance could cement a band's importance at any moment in time, then Arcade Fire's sublime Glastonbury headline slot might well have done the job. Regaling 70,000 people in one field is one thing, but to hold rapt millions of television viewers as they did so perfectly is truly astonishing, especially given that they are a band who, by Win Butler's own appraisal, haven't had a hit single. What they have had, however, is four of the best albums to be produced in the last decade. While they may be kicking off the British Summer Time season, tonight also marks the final date of their phenomenal Reflektor tour, and indeed what may be their last live show in the UK for a very long time.
Perhaps because of a curfew imposed by the show's Hyde Park location, the band take to the stage at 8:30pm, the hottest day of the year still far from over. It means that a set list adjustment is in order. With the Reflektor tour so heavily invested in the (literal) smoke and mirrors that make their shows such a stunning visual experience, there are a few songs whose impact can't really be felt until the sun starts to set. It means that usual opener 'Reflektor' is bumped down the list, the band instead choosing to start with the swinging rock'n'roll influenced 'Normal Person'. It doesn't pack quite the same punch as usual, perhaps because the sound is simply too quiet at this point (something Butler later addresses), but it's a brilliant choice to get things started.
As they power through a set that draws on material from each of their four albums, it becomes clear not only how much they are adored by their fans, but how much they adore performing for them. Their show is filled with theatrical drama, as in the brilliantly staged 'It's Never Over (Oh Orpheus)', (which sees Régine and a troupe of dancers face off against Win's main stage performance on a smaller platform in amongst the crowd), but the moments in between are filled with gratitude and love for the fans who are there. It's hard to believe that the band are unaware of the £2.50 ticket debacle, but whatever the price they paid, every fan gets their money's worth.
New tracks like 'Afterlife' and 'Joan of Arc' slot in perfectly alongside anthems like 'The Suburbs' and 'Neighbourhood #1 (Tunnels)'. 'No Cars Go' gets a fantastic reception, as does the riotous 'Ready To Start'. As they bring their main set to a close with the ever perfect 'Sprawl II (Mountains Beyond Mountains)', the lighting technicians outdo themselves, dropping the crowd into the gathering darkness perfectly on beat as Régine playfully begs, "I need the darkness, someone please cut the lights".
While they do technically play an encore, one of the joys of the Reflektor tour has been seeing the weird and wonderful 'interval' entertainment. Tonight Hyde Park is treated to the sight of the Pope dancing around to 'Sympathy for the Devil', undoubtedly a very deliberate song choice. Once Arcade Fire return to the stage, they play through brilliant renditions of 'Here Comes the Night Time' (timed perfectly as the sun finally sets), and 'Neighbourhood #3 (Power Out)', each one conjuring up the kind of ecstasy that can only exist in a crowd that knows the end is nigh.
Closing the set, as always, is 'Wake Up', the most anthemic song in their back catalogue, and probably one of the best live songs it is possible to experience. There's nothing quite like a chorus the size of a small town to emphasise how truly euphoric this evening has been. "It's easy to fall in love in London on a night like this" Win Butler remarks at one point, and it's hard not to believe he's talking about the band, and their sublime final performance.
Adrienne Law
Neon Bible
The Suburbs
New Acts Announced For Bestival
The Libertines Announce Two London Shows for September


Black Panther Book of the Month!
As promised early this week, here's a full complete story, a silent tale in the tradition of LULLUBY and my recent FURY and JONAH HEX short tales.
Some of you will remember this story: originally aired here on Pulp Sunday in December 2007, in a strip format, I am re-airing today in re-mastered first-time-in-color version, and in a comic format.
Click on the pages for a bigger and better viewing/reading experience.
As part of this Pulpy Xmas Week - and partially inspired by my working on another Sherlock Holmes cover today and having the heater out of service at the same time - here's my cover for Dynamite's SHERLOCK HOLMES YEAR ONE #3, solicited in January Previews.
A little late but I finally got around doing my Xmas card for this year.
Pulpy Xmas Week continues with the cover for KATO ORIGINS #11 (coming in January Previews).
I can finally spill the beans on another cool cover gig I have been working on recently: a set of four variant covers for Mike Mignola and Dark Horse on Hellboy related book that will be out in 2011.
The first cover has been recently revealed so it's safe now to post it in here too, since the look and feel I have been given these is pulp/vintage paperback.
I am very happy to see Black Panther (#513) finally out because I know all of you pulp fans will enjoy it. One of the reasons I was chosen to draw this is actually Black Beetle, and I was asked to keep the visuals in that vein (as you can see from the attached images).
Not only the pages will look very pulp, but writer (and award winning) David Liss has been cooking a very noirish gritty tale for this.
Here's my cover for Dynamite's SHERLOCK HOLMES YEAR ONE #2, currently in December Previews.
On December 5, 1933, the ratification of the Twenty-first Amendment signed the end fo the Prohibition.
while I am still happy for all the praise that my run on Detective (with those talented Snyder and Jock fellas) is getting, I found out today that in the new Wizard Magazine (now on the shelves) my other BIG-TWO debut book with writer David Liss, Black Panther Man Without Fear #513, has been picked as "BOOK OF THE MONTH" in the PULL LIST section of the magazine by the Wizard staff.
The book will be in stores on December 15th - mark your calendars!
Will share some preview when we get closer to the release.


Fruits are nature's fascinating medicines loaded with all sorts of vitamins, minerals, anti-oxidants and many phyto-nutrients; And a good source for creative minds to create fruit arts!
Here is our collection of the most amazing and creative fruit arts we found in the web. For the health benefits of different fruits click here.


Concealed hinges are great for when door's and panel's surfaces require a clean look.
Concealed hinges have (2) obround holes for mounting.
Hinges are .118" thick with a length of 2.36" and a .94" offset.
Concealed hinge is made from steel and has a mill / zinc plated finish.
Quality forming and casting. Works perfectly for our detachable doors.


text:
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Home / Section: Comic strips
CSotD: Re-inventing the Internet
Posted by Mike Peterson
Not to single out Benjamin Slyngstad in particular. He's just one of a whole lot of cartoonists and late night hosts who have decided that Joe Biden should not be president because he said "Malarkey."
And, however he means it, what he says here is probably true.
We may be reaching the "Al Gore is a Liar" stage of the Biden campaign.
Al Gore was branded a liar for claiming to have invented the Internet, discovered the pollution of Love Canal and served as a character model for Erich Segal's best-selling "Love Story."
Whatever policies he had in mind took a back seat to hilarious gags about what a crazy, ridiculous liar he was.
As you may recall — though a whole lot of people don't — he hadn't claimed to have "invented the Internet." He said he had helped bring about the legislation that allowed it to happen.
Which was true.
And he didn't claim to have discovered Love Canal, though he did say he was among the early champions of cleaning it up.
Which was also true.
And Erich Segal said that, while Oliver Barrett was mostly based on Gore's roommate, Tommy Lee Jones, Gore himself was part of the character, too.
So that was more of a mix-up than a lie.
However, Jay Leno couldn't get boffo laughs from jokes about mix-ups and press errors and misunderstandings, so Al Gore became a punchline for jokes about liars.
How many votes did those jokes cost Gore? Who knows?
But enough that we ended up counting hanging chads in Florida.
And dead bodies in Iraq.
Anyway, as Paul Fell suggests, some people use a less polite word for nonsense, and we expect a candidate's opponents to throw that stuff around.
The real malarkey in all this is the notion that voters see past the jokes and cheap shots and analyze a politician's policy proposals.
You can't see what isn't being covered, and, besides, we all like a good laugh.
And then cartoonists and late-night comedians got a whole second set of jokes over that hilarious "counting chads" thingie.
Win-win.
Which brings us to this related
Juxtaposition of the Day
(Mike Luckovich)
(Pat Bagley)
When motion pictures were invented, President Wilson eagerly proclaimed that they would be a great medium for teaching history.
The movie that so thrilled him was "Birth of a Nation."
And when Al Gore invented the Internet, it was seen as a means for people to be able to gather information and discover facts for themselves.
And then Facebook happened.
At least Woodrow Wilson wanted a racist view of history to be taught.
Zuckerberg doesn't seem to care what sorts of distortions and malarkey/bullshit (your choice) are being spread on his money machine, particularly if the spreaders pay him for the privilege.
His pose as a neutral figure in all this would be more convincing if he hadn't had a super secret dinner at the White House which he refuses to discuss.
CBS had good coverage, to the extent that Zuckerberg would admit he was even in the room:
(Gayle) King pointed out some people would say "the optics weren't good," adding, "Did he try to lobby you in any way?"
"No. I mean, I don't think that that's– that– I think some of the stuff that people talk about or think gets discussed and these discussions are not really how– how that works… I also want to respect that it was a private dinner and… private discussion," Zuckerberg said.
He genuinely doesn't get it.
You can either be head of Facebook and pal around with politicians or you can have private discussions.
You can either eat dinner with the president or you can have private dinners.
I once interviewed Hal Kennedy, the long-time anchor of KKTV's local news, and he said he mostly enjoyed being well-known and having people come up to him in public, but it had its drawbacks.
He told me he'd recently been swapping out his old water heater and, halfway through the project, realized he needed a part.
It meant, he said, that he had to wash up and change his clothes, because, if he went to the hardware store the way he was, someone would say, "I saw Hal Kennedy the other day, and he'd been drinking."
That's how celebrity works, Zuck.
If you want to be a private citizen, give it all away and go live in a cabin on a mountain somewhere.
Granted, you're a geek. But you're a very rich geek who made his pile by inventing a platform on which people could spread malarkey.
You can afford to buy a clue.
Or perhaps you'll understand our response to your "private dinner" with the president better if we put it in meme form:
Special Holiday Juxtaposition
(Dave Granlund)
(Between Friends)
So, apparently, there's a shortage of Christmas trees, which has driven up prices.
It's Jay Leno's fault.
If he hadn't turned Al Gore into a laughingstock, Dick Cheney would never have become Acting President and crashed the economy such that, back in 2008, tree farmers couldn't afford to plant trees and a decade later, we have this shortage.
If I lived in the city, I'd have an artificial tree, though not with music, thanks.
Half the joy of a tree is in cutting it yourself. The other half is in putting it in your livingroom before it dries out, drops needles and becomes a fire hazard like those net-wrapped, long-haul city trees.
Then, when the holidays end, you drag it into the woods behind the house and it becomes a hiding place for birds until it just becomes part of the forest floor.
As opposed to leaving it out on the curb for the city to deal with.
Though maybe I'll write a book about a magic pelletizing plant that springs up every January and turns all the Christmas trees in the city into wood pellets, then disappears again like Brigadoon.
Yeah, I know.
But these would be magical wood pellets.
This is the only holiday pollution I worry about.
#1 Brad Walker
December/4/2019
Speaking of Malarkey, who else remembers Walt Kelly's Simple J. Malarkey, his sendup of Joseph McCarthy? His best performance was when Kelly cast him as the King of Hearts in the trial from Alice in Wonderland.
#2 Mike Peterson
It occurs to me that, before the Internet fragmented us all into tiny groups, we were much more culturally literate, not only knowing phrases like (Simple J) Malarkey, but learning, from Mad Magazine and the standup comedians on Ed Sullivan, a lot of Yiddish.
Aside from whether or not we respected — much less "revered" — our elders, we were just a whole lot more universal in our worldview back when we gathered at the video hearth in the evening.
You can't turn back the clock, but you don't have to be so damn proud of your provincialism and ignorance, consarn it all.
#3 Kip Williams
I used to get an eyeful of the latest fake tree technology whenever I walked through Hested's Department Store. Some years they were green. Other years, white and silver predominated, with a spotlight that shone through colored gels to put a lightly queasy tone over it. One year, they were self-snowing fake trees, which blew little styro pellets into the air over themselves so they would fall and tumble through the branches, and then the majority would land in a large cone at the bottom, as if the vet was trying to keep the tree from chewing on its stand.
The corridor by the self-snowing trees was littered with the little pellets, which must have eventually all escaped by being part of the small percentage that missed the cone each time around.
I have two fake trees. After a couple of years of stringing lights on the first one, I went ahead and splurged on one with its own lights wired to the fake branches. I got both of them cheap by working at the place that sold them. Last year I didn't even bring the tree down from the attic. This year, I'm using my energy to practice holiday tunes on the piano, in anticipation of a paying gig next week.
#4 Mark B
FYI, more modern interpretations of "Birth of a Nation" note that the Wilson fandom story is as much myth as reality. The whole "he loved it" story was created by the film's author, not by the eyewitnesses, who said he left without saying a word.
– It was not the first film shown at the White House and it certainly was not the first film Wilson ever saw!
– It wasn't shown at the White House to honor the film, but because the family was in mourning and could not go to the theater.
– The major film endorsements came from the showing to Congress and the Supreme Court afterwards at the Press Club
– and while the film quoted Wilson in the title cards, they did not quote his(mild) denunciation of the Klan that was in the same book.
Of course Wilson was racist. But he was well within the mainstream of the time. Teddy Roosevelt's comments on race were as bad and often worse. The awful thing is that Wilson's racism was within the mainstream because the mainstream was really, really awful.
We didn't have any choice, there was only one set. Programs had to appeal to a wide spectrum, unlike today's narrowcasting. Variety shows like Ed Sullivan and Red Skelton booked rock performers despite the ill fit.
© 2005 - 2021 The Daily Cartoonist.


This blog post is adapted from comments Jennifer Zinth gave at the White House release event for the strategic plan.
Build strong foundations for STEM literacy.
Increase diversity, equity and inclusion in STEM.
Prepare the STEM workforce for the future.
At a time when relatively few states have a strategic plan for K-12 STEM education — yet representatives of education and business/industry are both calling for greater student exposure to high-quality STEM experiences — this strategic plan may provide a helpful starting point for fruitful state discussions. In the absence of a state strategic plan, the STEM policy landscape is a jumble of well-intentioned but random, unconnected efforts, with the potential to create inefficiencies or duplication of effort, or gaps in student access to high-quality programs and instruction.
However, the new federal strategic plan indicates it "is intended to serve as a 'North Star' for the broader community to help achieve the goals, pathways, and objectives" it sets forth. The federal plan also provides a common language for states should they choose to pursue interstate collaboration on any specific objectives in the plan. For example, many states have computer science initiatives underway. The objectives under the Build Computational Literacy pathway provide a framework for state collaboration on any number of fronts related to computational literacy — saving time, money and effort.
Mention throughout of early exposure to STEM: This is particularly encouraging considering the strategic plan goals of equity, inclusion and preparing the future workforce. High school is too late to start if we're talking about high-quality STEM experiences for every student.
The commitment to operating with transparency and accountability: So often – and rightly so – policymakers want to know which state programs have the greatest impact, particularly among students underrepresented in STEM. Reporting STEM participation among underrepresented populations, using common metrics to measure progress, and ensuring program performance and outcomes are publicly available will make it possible for policymakers to replicate the most effective programs and make course corrections on less impactful approaches.
Priority-setting: The new federal plan sets forth nine topical objectives related to the three educational pathways — strategic partnerships, the convergence of disciplines, and computational literacy — plus five additional objectives related to the Operate with Transparency and Accountability pathway. If states use the federal plan as a template, how will they determine which objectives they'll tackle in the short term versus the next one to two years — and which, if any, might be too great a stretch on vision and resources to address in the foreseeable future? How will states determine whose are the right voices around the table to be making those decisions?
Consensus-building: Beyond the consensus-building necessary to identify short- and long-term priorities for a state-level strategic plan for STEM, what consensus-building is necessary among various government, education, nonprofit and private partners for faithful implementation of the state plan at the state and local levels?
Coordination, funding and evaluation: As pointed out in A State Policymaker's STEM Playbook, a successful state-level STEM program requires statewide coordination, adequate and reliable funding and quality assurance/program evaluation. Without just one of these three elements, the most well-thought out state STEM plan is unlikely to achieve its hoped-for results. Are states that are missing one or more of these three components ready to make the necessary financial and human resources commitments to give their plans the greatest chances of success?
Here's to a 2019 full of productive state STEM-planning based on the foundation the new federal strategic plan provides!