text:
\section{Metric spaces}
\label{metric spaces}
\setcounter{equation}{0}

        A \emph{metric space} is a set $M$ equipped with a function
$d(x, y)$ defined for $x, y \in M$ such that $d(x, y)$ is a
nonnegative real number that is equal to $0$ exactly when $x = y$,
\begin{equation}
        d(y, x) = d(x, y)
\end{equation}
for every $x, y \in M$, and
\begin{equation}
        d(x, z) \le d(x, y) + d(y, z)
\end{equation}
for every $x, y, z \in M$, which is known as the \emph{triangle inequality}.

        Remember that the \emph{absolute value} of a real number $r$
is denoted $|r|$ and equal to $r$ when $r \ge 0$ and to $-r$ when $r
\le 0$.  It is easy to check that
\begin{equation}
        |r + t| \le |r| + |t|
\end{equation}
and
\begin{equation}
        |r \, t| = |r| \, |t|
\end{equation}
for any pair of real numbers $r$, $t$.  The standard metric on the
real line ${\bf R}$ is given by $|r - t|$, which is the first main
example of a metric space.

        If $(M, d(x, y))$ is a metric space, then $d(x, y)$ is called
the \emph{distance function} or \emph{metric} on $M$.  For each $x \in
M$ and $r > 0$, the \emph{open ball} in $M$ with center $x$ and radius
$r$ is
\begin{equation}
        B(x, r) = \{y \in M : d(x, y) < r\}.
\end{equation}
Similarly, the \emph{closed ball} with center $x$ and radius $r \ge 0$ is
\begin{equation}
        \overline{B}(x, r) = \{y \in M : d(x, y) \le r\}.
\end{equation}
Thus
\begin{equation}
        B(x, r) \subseteq \overline{B}(x, r) \subseteq B(x, t)
\end{equation}
when $r < t$.

        Let $a$, $b$ be real numbers with $a < b$.  The \emph{open
interval} $(a, b)$ in ${\bf R}$ is defined by
\begin{equation}
        (a, b) = \{r \in {\bf R} : a < r < b\},
\end{equation}
and the \emph{closed interval} $[a, b]$ is defined by
\begin{equation}
        [a, b] = \{r \in {\bf R} : a \le r \le b\}.
\end{equation}
One may also allow $a = b$ for the latter.  The \emph{length} of these
intervals is $b - a$.  Note that open and closed balls in the real
line with respect to the standard metric are open and closed
intervals.

\section{A little calculus}
\label{calculus}
\setcounter{equation}{0}

        Suppose that $a$, $b$ are real numbers with $a < b$, and that
$f(x)$ is a continuous real-valued function on the closed interval
$[a, b]$ in the real line.  The \emph{extreme value theorem} states
that there are elements $p$, $q$ of $[a, b]$ at which $f$ attains its
maximum and minimum, which is to say that
\begin{equation}
        f(q) \le f(x) \le f(p)
\end{equation}
for every $x \in [a, b]$.  This works as well for continuous
real-valued functions on compact subsets of metric spaces, or even
topological spaces.  If $p$ or $q$ is in the open interval $(a, b)$
and $f$ is differentiable there, then the derivative $f'(p)$ or
$f'(q)$ is equal to $0$.

        Suppose that $f(x)$ is differentiable at every point in $(a,
b)$.  If $f(a) = f(b) = 0$, then \emph{Rolle's theorem} states that
$f'(x) = 0$ for some $x \in (a, b)$.  This is because the maximum or
minimum of $f$ on $[a, b]$ is attained on $(a, b)$, or $f(x) = 0$ for
every $x \in [a, b]$.  No matter the values of $f(a)$, $f(b)$, the
\emph{mean value theorem} says that there is an $x \in (a, b)$ such that
\begin{equation}
        f'(x) = \frac{f(b) - f(a)}{b - a}.
\end{equation}
This follows from Rolle's theorem applied to $f - \phi$, where
$\phi(x) = \alpha \, x + \beta$ and $\alpha, \beta \in {\bf R}$ are
chosen so that $\phi(a) = f(a)$, $\phi(b) = f(b)$.

        Of course, the derivative of a constant function is $0$, and
the mean value theorem implies that a continuous function $f$ on $[a,
b]$ is constant if the derivative of $f$ exists and is equal to $0$ at
every point in $(a, b)$.  If $f$ is monotone increasing on $[a, b]$,
in the sense that $f(x) \le f(y)$ when $a \le x \le y \le b$, then
$f'(x) \ge 0$ for every $x \in (a, b)$ at which $f$ is differentiable.
Conversely, if $f$ is continuous on $[a, b]$, differentiable on $(a,
b)$, and $f'(x) \ge 0$ for each $x \in (a, b)$, then $f$ is monotone
increasing on $[a, b]$, by the mean value theorem.  If $f'(x) > 0$ for
every $x \in (a, b)$, then $f$ is strictly increasing on $[a, b]$, in
the sense that $f(w) < f(y)$ when $a \le w < y \le b$.  However, the
derivative of a strictly increasing function may be equal to $0$, as
when $f(x) = x^3$.

\section{Norms on ${\bf R}^n$}
\label{norms}
\setcounter{equation}{0}

        Let $n$ be a positive integer, and let ${\bf R}^n$ be the
space of $n$-tuples of real numbers.  This means that an element $x$
of ${\bf R}^n$ is of the form $x = (x_1, \ldots, x_n)$, where the
coordinates $x_1, \ldots, x_n$ of $x$ are real numbers.  Addition and
scalar multiplication on ${\bf R}^n$ are defined coordinatewise in the
usual way, so that ${\bf R}^n$ becomes a finite-dimensional vector
space over the real numbers.

        A \emph{norm} on ${\bf R}^n$ is a function $N(x)$ such that
$N(x)$ is a nonnegative real number for every $x \in {\bf R}^n$ which
is equal to $0$ exactly when $x = 0$,
\begin{equation}
\label{N(r x) = |r| N(x)}
        N(r \, x) = |r| \, N(x)
\end{equation}
for every $r \in {\bf R}$ and $x \in {\bf R}^n$, and
\begin{equation}
\label{N(x + y) le N(x) + N(y)}
        N(x + y) \le N(x) + N(y)
\end{equation}
for every $x, y \in {\bf R}^n$.  If $N$ is a norm on ${\bf R}^n$, then
\begin{equation}
        d_N(x, y) = N(x - y)
\end{equation}
is a metric on ${\bf R}^n$.

        For example, the absolute value function is a norm on ${\bf
R}$, for which the corresponding metric is the standard metric on the
real line.  The standard Euclidean norm on ${\bf R}^n$ is defined by
\begin{equation}
        |x| = \Big(\sum_{j = 1}^n x_j^2 \Big)^{1/2},
\end{equation}
and the corresponding metric is the standard Euclidean metric on ${\bf R}^n$.
It is not so obvious that this satisfies the triangle inequality, and hence
is a norm, and we shall discuss a proof of this fact in Section \ref{convex
sets}.

        One can check directly that
\begin{equation}
        \|x\|_1 = \sum_{j = 1}^n |x_j|
\end{equation}
and
\begin{equation}
        \|x\|_\infty = \max(|x_1|, \ldots, |x_n|)
\end{equation}
are norms on ${\bf R}^n$.  We shall see in Section \ref{convex sets} that
\begin{equation}
        \|x\|_p = \Big(\sum_{j = 1}^n |x_j|^p \Big)^{1/p}
\end{equation}
is a norm when $p \ge 1$, which includes the Euclidean norm as a
special case.

\section{Convex functions}
\label{convex functions}
\setcounter{equation}{0}

        A real-valued function $f(x)$ on the real line is said to be
\emph{convex} if
\begin{equation}
\label{convexity}
        f(t \, x + (1 - t) \, y) \le t \, f(x) + (1 - t) \, f(y)
\end{equation}
for every $x, y \in {\bf R}$ and $t \in [0, 1]$.  This is equivalent
to
\begin{equation}
\label{convexity, 2}
        \frac{f(w) - f(x)}{w - x} \le \frac{f(y) - f(w)}{y - w}
\end{equation}
for every $x, w, y \in {\bf R}$ such that $x < w < y$.  Applying this
condition twice, we get that
\begin{equation}
\label{convexity, 3}
        \frac{f(w) - f(x)}{w - x} \le \frac{f(z) - f(y)}{z - y}
\end{equation}
when $x < w < y < z$.  As another refinement of (\ref{convexity, 2}),
one can use (\ref{convexity}) to show that
\begin{equation}
\label{convexity, 4}
        \frac{f(w) - f(x)}{w - x} \le \frac{f(y) - f(x)}{y - x}
                                   \le \frac{f(y) - f(w)}{y - w}
\end{equation}
when $x < w < y$.

        If $f$ is differentiable and $f'$ is monotone increasing, then
the mean value theorem implies (\ref{convexity, 2}) and hence that $f$
is convex.  Conversely, (\ref{convexity, 3}) implies that the
derivative of $f$ is monotone increasing when $f$ is differentiable.
Actually, one can show that the right and left derivatives $f_+'(x)$,
$f_-'(x)$ exist for each $x \in {\bf R}$ when $f$ is convex, and
satisfy
\begin{equation}
        f_-'(x) \le f_+'(x)
\end{equation}
and
\begin{equation}
        f_+'(x) \le f_-'(y)
\end{equation}
when $x < y$.  One can also show that these conditions characterize
convexity, using analogues of Rolle's theorem and the mean value
theorem for functions with one-sided derivatives.

        A function $f : {\bf R} \to {\bf R}$ is \emph{strictly convex} if
\begin{equation}
        f(t \, x + (1 - t) \, y) < t \, f(x) + (1 - t) \, f(y)
\end{equation}
when $x \ne y$ and $0 < t < 1$.  This corresponds to strict inequality
in (\ref{convexity, 2}), (\ref{convexity, 3}), and (\ref{convexity,
4}) as well.  If $f$ is differentiable on ${\bf R}$, then $f$ is
strictly convex if and only if $f'$ is strictly increasing.
Otherwise, strict convexity can be characterized in terms of one-sided
derivatives by the requirement that
\begin{equation}
        f_+'(x) < f_-'(y)
\end{equation}
when $x < y$.  Alternatively, if a convex function $f$ on ${\bf R}$ is
not strictly convex, then $f$ is equal to an affine function on an
interval of positive length.

        For example, consider $f(r) = |r|^p$, $p > 0$.  If $p = 1$,
then $f(r) = |r|$ is convex but not strictly convex on ${\bf R}$.  If
$p = 2$, then $f(r) = r^2$ is twice-differentiable, $f''(r) = 2$, and
$f$ is strictly convex.  If $p > 2$, then $f$ is twice-differentiable
on ${\bf R}$, $f''(r) > 0$ when $r \ne 0$, $f''(0) = 0$, and $f$ is
strictly convex because $f'$ is strictly increasing.  If $1 < p < 2$,
then $f$ is differentiable on ${\bf R}$, twice-differentiable on ${\bf
R} \backslash \{0\}$, $f''(r) > 0$ when $r \ne 0$, and again $f$ is
strictly convex since $f'$ is strictly increasing.  If $0 < p < 1$,
then $f$ is twice-differentiable on ${\bf R} \backslash \{0\}$,
$f''(r) < 0$ when $r \ne 0$, and $f$ is not convex.

\section{Convex sets}
\label{convex sets}
\setcounter{equation}{0}

        A set $E \subseteq {\bf R}^n$ is said to be \emph{convex} if
\begin{equation}
        t \, x + (1 - t) \, y \in E
\end{equation}
for every $x, y \in E$ and $t \in (0, 1)$.  For example, open and
closed balls associated to metrics defined by norms on ${\bf R}^n$ are
convex.

        Conversely, suppose that $N(x)$ is a nonnegative real-valued
function on ${\bf R}^n$ such that $N(x) > 0$ when $x \ne 0$ and the
homogeneity condition (\ref{N(r x) = |r| N(x)}) holds for all $x \in
{\bf R}^n$ and $r \in {\bf R}$.  If the closed unit ball
\begin{equation}
\label{B_N}
        B_N = \{x \in {\bf R}^n : N(x) \le 1\}
\end{equation}
is convex, then $N$ satisfies the triangle inequality (\ref{N(x + y)
le N(x) + N(y)}) and hence is a norm.  Let $x, y \in {\bf R}^n$ be
given, and let us check (\ref{N(x + y) le N(x) + N(y)}).  We may
suppose that $x, y \ne 0$, since the inequality is trivial when $x =
0$ or $y = 0$.  Put
\begin{equation}
        x' = \frac{x}{N(x)}, \quad y' = \frac{y}{N(y)},
\end{equation}
so that $N(x') = N(y') = 1$.  By hypothesis,
\begin{equation}
        N(t \, x' + (1 - t) \, y') \le 1
\end{equation}
when $0 \le t \le 1$.  Applying this with
\begin{equation}
        t = \frac{N(x)}{N(x) + N(y)},
\end{equation}
we get (\ref{N(x + y) le N(x) + N(y)}), as desired.

        For example, suppose that $N(x) = \|x\|_p$, $1 < p < \infty$.
Let $x, y \in {\bf R}^n$ with $\|x\|_p, \|y\|_p \le 1$ be given, so that
\begin{equation}
        \sum_{j = 1}^n |x_j|^p, \ \sum_{j = 1}^n |y_j|^p \le 1.
\end{equation}
We would like to show that
\begin{equation}
        \|t \, x + (1 - t) \, y\|_p \le 1
\end{equation}
when $0 \le t \le 1$, which is the same as
\begin{equation}
        \sum_{j = 1}^n |t \, x_j + (1 - t) \, y_j|^p \le 1.
\end{equation}
The convexity of $|r|^p$ on ${\bf R}$ implies that
\begin{equation}
        |t \, x_j + (1 - t) \, y_j|^p \le t |x_j|^p + (1 - t) \, |y_j|^p
\end{equation}
for each $j$, and the desired inequality follows by summing this over
$j$.

        A norm $N$ on ${\bf R}^n$ is said to be \emph{strictly convex}
if the unit ball $B_N$ is strictly convex in the sense that
\begin{equation}
        N(t \, x + (1 - t) \, y) < 1
\end{equation}
when $x, y \in {\bf R}^n$, $N(x) = N(y) = 1$, $x \ne y$, and $0 < t <
1$.  It is easy to see that the absolute value function is strictly
convex as a norm on ${\bf R}$, if not as a general function as in the
previous section.  One can also check that $\|x\|_p$ is a strictly
convex norm on ${\bf R}^n$ when $p > 1$, using the strict convexity of
$|r|^p$ on ${\bf R}$ and computations as in the preceding paragraph.
However, $\|x\|_1$ and $\|x\|_\infty$ are not strictly convex norms on
${\bf R}^n$ when $n \ge 2$.

\section{A little more calculus}
\label{a little more calculus}
\setcounter{equation}{0}

        Let $f$ be a continuous real-valued function on a closed
interval $[a, b]$, $a < b$.  The integral
\begin{equation}
        \int_a^b f(t) \, dt
\end{equation}
can be defined in the usual way as a limit of finite sums.  The
convergence of the finite sums to the integral uses the fact that
continuous functions on $[a, b]$ are actually uniformly continuous.
It is well known that continuous functions on compact subsets of any
metric space are uniformly continuous.

        Consider the indefinite integral
\begin{equation}
        F(x) = \int_a^x f(t) \, dt.
\end{equation}
This defines a continuous function on $[a, b]$ which is differentiable
on $(a, b)$ and satisfies $F'(x) = f(x)$.  Similarly, $F$ has
one-sided derivatives at the endpoints $a$, $b$ that satisfy the same
condition.  If another differentiable function on $(a, b)$ has
derivative $f$, then the difference of $F$ and this function is
constant, by the mean value theorem.

        Clearly $F$ is monotone increasing on $[a, b]$ if $f \ge 0$ on
the whole interval.  If $f > 0$ on $[a, b]$, then $F$ is strictly
increasing.  The same conclusion holds if $f \ge 0$ on $[a, b]$ and $f
> 0$ at some point in any nontrivial subinterval of $[a, b]$.
Equivalently, if $f \ge 0$ on $[a, b]$, and if $F$ is not strictly
increasing on $[a, b]$, then $f = 0$ at every point in a nontrivial
subinterval.

        Suppose that $f' \ge 0$ on $(a, b)$, or simply that $f$ is
monotone increasing on $[a, b]$.  This implies that
\begin{equation}
        f(x) \le \frac{F(y) - F(x)}{y - x} \le f(y)
\end{equation}
when $a \le x < y \le b$.  In particular,
\begin{equation}
        \frac{F(w) - F(x)}{w - x} \le \frac{F(y) - F(w)}{y - w}
\end{equation}
when $a \le x < w < y \le b$.  If $f$ is strictly increasing, then
these inequalities are strict as well.

\section{Supremum and infimum}
\label{supremum and infimum}
\setcounter{equation}{0}

        A real number $b$ is said to be an \emph{upper bound} for a
set $A \subseteq {\bf R}$ if $a \le b$ for every $a \in A$.  We say
that $b_1 \in {\bf R}$ is the \emph{least upper bound} or
\emph{supremum} of $A$ if $b_1$ is an upper bound for $A$ and $b_1 \le
b$ for every upper bound $b$ of $A$.  If $b_2 \in {\bf R}$ also
satisfies these two conditions, then $b_1 \le b_2$ and $b_2 \le b_1$,
and hence $b_1 = b_2$.  Thus the supremum of $A$ is unique when it
exists, in which case it is denoted $\sup A$.  The \emph{completeness
property} of the real line states that every nonempty set with an
upper bound has a least upper bound.

        More precisely, this is completeness with respect to the
ordering on the real line, which can be defined for other ordered
sets.  There is also completeness for metric spaces, which means that
every Cauchy sequence converges.  Both forms of completeness hold on
the real line, and are basically equivalent to each other in this
particular situation.  However, the two notions are distinct, because
they can be applied in different circumstances.  There are
completeness conditions concerning the existence of solutions of
ordinary differential equations as well, which may be related to
completeness for an associated metric space.

       Similarly, a real number $c$ is said to be a \emph{lower bound}
for $A \subseteq {\bf R}$ if $c \le a$ for every $a \in A$, and $c_1
\in {\bf R}$ is the \emph{greatest lower bound} or \emph{infimum} of
$A$ if $c_1$ is a lower bound for $A$ and $c \le c_1$ for every lower
bound $c$ of $A$.  This is unique when it exists for the same reasons
as before, and is denoted $\inf A$.  It follows from completeness that
a nonempty set $A \subseteq {\bf R}$ with a lower bound has a greatest
lower bound, which can be characterized as the supremum of the set of
lower bounds of $A$.  Alternatively, the infimum of $A$ is equal to
the negative of the supremum of $-A = \{-a : a \in A\}$.

\section{Bounded sets}
\label{bounded sets}
\setcounter{equation}{0}

        Let $(M, d(x, y))$ be a metric space.  A set $E \subseteq M$
is said to be \emph{bounded} if there is a $p \in M$ and an $r \ge 0$
such that
\begin{equation}
        d(p, x) \le r
\end{equation}
for every $x \in E$.  This implies that for every $q \in M$ there is a
$t \ge 0$ such that $d(q, x) \le t$ for every $x \in E$, by taking $t
= r + d(p, q)$.

        Equivalently, $E \subseteq M$ is bounded if the set of
distances $d(x, y)$ for $x, y \in E$ has an upper bound in ${\bf R}$.
If $E$ is nonempty and bounded, then the \emph{diameter} of $E$ is defined by
\begin{equation}
        \mathop{\rm diam} E = \sup \{d(x, y) : x, y \in E\}.
\end{equation}
The diameter of the empty set may be interpreted as $0$.

        If $E_1 \subseteq E_2 \subseteq M$ and $E_2$ is bounded, then
$E_1$ is bounded, and
\begin{equation}
        \mathop{\rm diam} E_1 \le \mathop{\rm diam} E_2.
\end{equation}
The union of two bounded subsets of $M$ is also bounded, but the diameter of
the union may be much larger than the sum of the diameters of the two subsets.

        Suppose that $M$ is ${\bf R}^n$ equipped with a norm $N$ and
its associated metric $d_N(x, y)$.  The \emph{convex hull}
$\widehat{E}$ of a set $E \subseteq {\bf R}^n$ consists of all convex
combinations of elements of $E$.  More precisely, $\widehat{E}$ is the
set of all finite sums of the form
\begin{equation}
        \sum_{i = 1}^k r_i \, x(i),
\end{equation}
where $k$ is a positive integer, $r_1, \ldots, r_k$ are
nonnegative real numbers such that
\begin{equation}
        \sum_{i = 1}^k r_k = 1,
\end{equation}
and $x(1), \ldots, x(k)$ are elements of $E$.  It is well known that
one can take $k = n + 1$ here, but we shall not need this fact.  By
construction, $\widehat{E}$ is a convex set in ${\bf R}^n$ that
contains $E$.  Moreover, $\widehat{E}$ is the smallest such set, in
the sense that $\widehat{E}$ is contained in any convex set in ${\bf
R}^n$ that contains.  If $E$ is bounded, so that $E$ is contained in a
ball, then $\widehat{E}$ is contained in the same ball, and hence
$\widehat{E}$ is bounded.  Let us check that
\begin{equation}
        {\mathop{\rm diam}}_N \widehat{E} \le {\mathop{\rm diam}}_N E,
\end{equation}
where the subscript $N$ indicates that the diameter uses the norm $N$.
Let
\begin{equation}
 \xi = \sum_{i = 1}^k r_i \, x(i), \quad \eta = \sum_{j = 1}^l t_j \, y(j)
\end{equation}
be arbitrary elements of $\widehat{E}$, as before.  Thus
\begin{equation}
        \xi - \eta = \sum_{i = 1}^k \sum_{j = 1}^l r_i \, t_j \, (x(i) - y(j)),
\end{equation}
and therefore
\begin{equation}
 N(\xi - \eta) \le \sum_{i = 1}^k \sum_{j = 1}^l r_i \, t_j \, N(x(i) - y(j)),
\end{equation}
by the properties of norms.  This implies that
\begin{equation}
 N(\xi - \eta) \le \max \{N(x(i) - y(j)) : 1 \le i \le k, \, 1 \le j \le l\},
\end{equation}
and consequently $N(\xi - \eta) \le {\mathop{\rm diam}}_N E$, as desired.

\section{Lipschitz mappings}
\label{lipschitz mappings}
\setcounter{equation}{0}

        Let $(M_1, d_1(x, y))$ and $(M_2, d_2(u, v))$ be metric
spaces.  A mapping $f : M_1 \to M_2$ is said to be \emph{Lipschitz} if
\begin{equation}
        d_2(f(x), f(y)) \le C \, d_1(x, y)
\end{equation}
for some $C \ge 0$ and all $x, y \in M$.  More precisely, this means
that $f$ is Lipschitz of order $1$, and we shall discuss other
Lipschitz conditions later.  One can also say that $f$ is
$C$-Lipschitz or $C$-Lipschitz of order $1$ to mention the constant
$C$ explicitly.

         Thus $f$ is $C$-Lipschitz with $C = 0$ if and only if $f$ is
constant.  Note that Lipschitz mappings are uniformly continuous.
Suppose that $(M_3, d_3(w, z))$ is another metric space, and that $f_1
: M_1 \to M_2$ and $f_2 : M_2 \to M_3$ are Lipschitz mappings with
constants $C_1$, $C_2$, respectively.  The composition $f_2 \circ f_1$
is the mapping from $M_1$ to $M_2$ defined by
\begin{equation}
         (f_2 \circ f_1)(x) = f_2(f_1(x)),
\end{equation}
and it is easy to check that this is Lipschitz with constant equal to
the product of $C_1$ and $C_2$.

         If $f : M_1 \to M_2$ is $C$-Lipschitz and $E \subseteq M_1$
is bounded, then
\begin{equation}
         f(E) = \{f(x) : x \in E\}
\end{equation}
is bounded in $M_2$, and
\begin{equation}
         {\mathop{\rm diam}}_2 f(E) \le C \, {\mathop{\rm diam}}_1 E.
\end{equation}
Here the subscripts indicate in which metric space the diameter is
taken.  This is easy to verify, directly from the definitions, and
suggests another way to look at the composition of Lipschitz mappings,
as in the previous paragraph.

\section{Real-valued functions}
\label{real functions}
\setcounter{equation}{0}

        Let $f$ be a real-valued function on an open interval $(a, b)$
in the real line.  If $f$ is $C$-Lipschitz with respect to the
standard metric on the domain and range, then
\begin{equation}
        |f'(x)| \le C
\end{equation}
at every point $x \in (a, b)$ at which $f$ is differentiable, by
definition of the derivative.  Conversely, if $f$ is differentiable
and satisfies this condition everywhere on $(a, b)$, then $f$ is
$C$-Lipschitz, by the mean value theorem.

        Now let $(M, d(x, y))$ be a metric space.  A function $f : M
\to {\bf R}$ is $C$-Lipschitz with respect to the standard metric on
${\bf R}$ if and only if
\begin{equation}
        f(x) \le f(y) + C \, d(x, y)
\end{equation}
for every $x, y \in M$.  This follows easily from the definitions.  In
particular, $f_p(x) = d(p, x)$ is $1$-Lipschitz for every $p \in M$.

        If $A \subseteq M$, $A \ne \emptyset$, and $x \in M$, then put
\begin{equation}
        \mathop{\rm dist}(x, A) = \inf \{d(x, a) : a \in A\}.
\end{equation}
For each $x, y \in M$ and $a \in A$,
\begin{equation}
        \mathop{\rm dist}(x, A) \le d(x, a) \le d(x, y) + d(y, a),
\end{equation}
and therefore
\begin{equation}
        \mathop{\rm dist}(x, A) \le \mathop{\rm dist}(y, A) + d(x, y).
\end{equation}
This shows that $\mathop{\rm dist}(x, A)$ is $1$-Lipschitz on $M$.

         Suppose that $f_1, f_2 : M \to {\bf R}$ are Lipschitz with
constants $C_1$, $C_2$, respectively.  For any $r_1, r_2 \in {\bf R}$,
$r_1 \, f_1 + r_2 \, f_2$ is Lipschitz with constant $|r_1| \, C_1 +
|r_2| \, C_2$.  Suppose also that $f_1$, $f_2$ are bounded on $M$, with
\begin{equation}
        |f_1(x)| \le k_1, \quad |f_2(x)| \le k_2
\end{equation}
for some $k_1, k_2 \ge 0$ and every $x \in M$.  Because
\begin{eqnarray}
\lefteqn{f_1(x) \, f_2(x) - f_1(y) \, f_2(y)}   \\
 & & = (f_1(x) - f_1(y)) \, f_2(x) + f_1(y) \, (f_2(x) - f_2(y)) \nonumber
\end{eqnarray}
for every $x, y \in M$, $f_1 \, f_2$ is Lipschitz on $M$ with constant
$k_2 \, C_1 + k_1 \, C_2$.

\section{${\bf R}^n$-valued functions}
\label{R^n functions}
\setcounter{equation}{0}

        Let $N$ be a norm on ${\bf R}^n$.  Thus $N$ is $1$-Lipschitz
as a real-valued function on ${\bf R}^n$ with the metric $d_N(x, y)$
associated to $N$, as in the previous section.  One can also show that
$N$ is bounded by a constant multiple of the standard Euclidean norm
on ${\bf R}^n$.  This uses the finite-dimensionality of ${\bf R}^n$ in
an essential way, and it implies that $N$ is Lipschitz with respect to
the standard metric on ${\bf R}^n$.

        Suppose that $f$ is a continuous ${\bf R}^n$-valued function
on a closed interval $[a, b]$ in the real line.  As an extension of
the triangle inequality for $N$,
\begin{equation}
        N\Big(\int_a^b f(t) \, dt\Big) \le \int_a^b N(f(t)) \, dt.
\end{equation}
Indeed, the analogous statement for the finite sums follows from the
triangle inequality for $N$.  The integral of $f$ can be approximated
by finite sums, and continuity of $N$ as in the preceding paragraph
can be employed to pass to the limit.  Alternatively, one can use
duality, as follows.  For any linear functional $\phi : {\bf R}^n \to
{\bf R}$,
\begin{equation}
        \phi\Big(\int_a^b f(t) \, dt\Big) = \int_a^b \phi(f(t)) \, dt.
\end{equation}
If $|\phi(w)| \le N(w)$ for every $w \in {\bf R}^n$, then we get that
\begin{equation}
        \biggl|\int_a^b \phi(f(t)) \, dt\biggr| \le \int_a^b N(f(t)) \, dt.
\end{equation}
A famous theorem states that for each $v \in {\bf R}^n$ there is such
a $\phi$ with $\phi(v) = N(v)$, which permits one to estimate the norm
of the integral.  We shall not discuss the proof of this here, but one
can take $\phi(w)$ to be the standard inner product of $w$ with $v /
|v|$ when $v \ne 0$ and $N$ is the Euclidean norm on ${\bf R}^n$, and
there are also explicit expressions for $\phi$ when $N(w) = \|w\|_p$,
$1 \le p \le \infty$.

        Suppose now that $F : [a, b] \to {\bf R}^n$ is $C$-Lipschitz
with respect to the standard metric on ${\bf R}$ and the metric $d_N$
on ${\bf R}^n$.  If $F$ is differentiable at a point $x \in (a, b)$,
then $N(F'(x)) \le C$.  This follows from the definition of the
derivative, as in the real-valued case.  Conversely, if $F$ is
continuously differentiable on $[a, b]$ and $N(F') \le C$, then one
can use the fundamental theorem of calculus and the integral form of
the triangle inequality to show that that $F$ is $C$-Lipschitz with
respect to $N$.  One can use duality to get the same conclusion when
$F$ is continuous on $[a, b]$ and differentiable on $(a, b)$ with
$N(F') \le C$, by applying the mean value theorem to $\phi \circ F$
for linear functionals $\phi : {\bf R}^n \to {\bf R}$.

        Let $(M, d(x, y))$ be a metric space, and let $F = (F_1,
\ldots, F_n)$ be a mapping from $M$ into ${\bf R}^n$.  If ${\bf R}^n$
is equipped with the norm $\|w\|_\infty$, then it is easy to see that
$F$ is $C$-Lipschitz if and only if $F_1, \ldots, F_n$ are
$C$-Lipschitz as real-valued functions on $M$.  Of course, one can
estimate Lipschitz conditions for $F$ in terms of Lipschitz conditions
for $F_1, \ldots, F_n$ for other norms on ${\bf R}^n$, and vice-versa,
but the relationship between the constants is normally not quite as
simple as for the norm $\|w\|_\infty$.

\section{Bounded variation}
\label{bounded variation}
\setcounter{equation}{0}

        Let $f$ be a real-valued function on a closed interval $[a,
b]$.  A \emph{partition} of $[a, b]$ is a finite sequence $\{t_j\}_{j
= 0}^n$ of real numbers such that
\begin{equation}
        a = t_0 < t_1 < \cdots < t_n = b.
\end{equation}
For each partition $\mathcal{P} = \{t_j\}_{j = 0}^n$ of $[a, b]$, consider
\begin{equation}
        V_\mathcal{P}(f) = \sum_{j = 1}^n |f(t_j) - f(t_{j - 1})|.
\end{equation}
This measures the variation of $f$ on the partition $\mathcal{P}$.  We
say that $f$ has \emph{bounded variation} on $[a, b]$ if there is an
upper bound for $V_\mathcal{P}(f)$ over all partitions $\mathcal{P}$
of $[a, b]$.  In this case, the \emph{total variation} $V_a^b(f)$ of
$f$ on $[a, b]$ is defined by
\begin{equation}
\label{V_a^b(f)}
        V_a^b(f) = \sup \{ V_\mathcal{P} : \mathcal{P}
                                    \hbox{ is a partition of } [a, b]\}.
\end{equation}
Thus $V_a^b(f) = 0$ if and only if $f$ is constant on $[a, b]$.

        Using the partition that consists of only $a$, $b$, we get that
\begin{equation}
        |f(b) - f(a)| \le V_a^b(f).
\end{equation}
If $f$ is monotone increasing on $[a, b]$, then
\begin{equation}
        V_\mathcal{P}(f) = f(b) - f(a)
\end{equation}
for every partition $\mathcal{P}$ of $[a, b]$.  Hence $f$ has bounded
variation on $[a, b]$, and
\begin{equation}
        V_a^b(f) = f(b) - f(a).
\end{equation}
Conversely, if $f$ has bounded variation on $[a, b]$ and
\begin{equation}
        V_a^b(f) = |f(b) - f(a)|,
\end{equation}
then $f$ is either monotone increasing or decreasing on $[a, b]$.

        If $f$ is $C$-Lipschitz on $[a, b]$, then
\begin{equation}
        V_\mathcal{P}(f) \le C \, (b - a)
\end{equation}
for every partition $\mathcal{P}$ of $[a, b]$.  Hence $f$ has bounded
variation on $[a, b]$, and
\begin{equation}
        V_a^b(f) \le C \, (b - a).
\end{equation}
If $\phi : {\bf R} \to {\bf R}$ is $C$-Lipschitz, then
\begin{equation}
        V_\mathcal{P}(\phi \circ f) \le C \, V_\mathcal{P}(f)
\end{equation}
for every $f : [a, b] \to {\bf R}$ and partition $\mathcal{P}$ of $[a, b]$.
If $f$ has bounded variation on $[a, b]$, then it follows that $\phi
\circ f$ has bounded variation on $[a, b]$, and
\begin{equation}
        V_a^b(\phi \circ f) \le C \, V_a^b(f).
\end{equation}

        Let $f_1, f_2 : [a, b] \to {\bf R}$ and $r_1, r_2 \in {\bf R}$
be given.  For any partition $\mathcal{P}$ of $[a, b]$,
\begin{equation}
        V_\mathcal{P}(r_1 \, f_1 + r_2 \, f_2)
           \le |r_1| \, V_\mathcal{P}(f_1) + |r_2| \, V_\mathcal{P}(f_2).
\end{equation}
If $f_1$, $f_2$ have bounded variation on $[a, b]$, then it follows
that $r_1 \, f_1 + r_2 \, f_2$ also has bounded variation, with
\begin{equation}
 V_a^b(r_1 \, f_1 + r_2 \, f_2) \le |r_1| \, V_a^b(f_1) + |r_2| \, V_a^b(f_2).
\end{equation}

        Suppose that $f_1$, $f_2$ are bounded on $[a, b]$, so that
\begin{equation}
        |f_1(x)| \le k_1, \quad |f_2(x)| \le k_2
\end{equation}
for some $k_1, k_2 \ge 0$ and every $x \in [a, b]$.  It is easy to
check that
\begin{equation}
        V_\mathcal{P}(f_1 \, f_2) \le k_2 \, V_\mathcal{P}(f_1)
                                     + k_1 \, V_\mathcal{P}(f_2)
\end{equation}
for every partition $\mathcal{P}$ of $[a, b]$.  If $f_1$, $f_2$ have
bounded variation on $[a, b]$, then $f_1 \, f_2$ has bounded variation, and
\begin{equation}
        V_a^b(f_1 \, f_2) \le k_2 \, V_a^b(f_1) + k_1 \, V_a^b(f_2).
\end{equation}
This is analogous to the earlier estimate for the Lipschitz constant
of the product of bounded Lipschitz functions, and to the Leibniz rule
for differentiating the product of two functions.

        Suppose that $a_1$, $b_1$ are real numbers such that $a \le
a_1 \le b_1 \le b$.  If $f$ has bounded variation on $[a, b]$, then
$f$ has bounded variation on $[a_1, b_1]$, and
\begin{equation}
        V_{a_1}^{b_1}(f) \le V_a^b(f).
\end{equation}
This is because every partition of $[a_1, b_1]$ can be extended to a
partition of $[a, b]$.  In particular, $f$ is bounded on $[a, b]$ when
it has bounded variation.

        A partition $\mathcal{P}'$ of $[a, b]$ is said to be a
\emph{refinement} of a partition $\mathcal{P}$ of $[a, b]$ if
$\mathcal{P}'$ contains all of the terms in $\mathcal{P}$.
In this case, one can check that
\begin{equation}
        V_\mathcal{P}(f) \le V_{\mathcal{P}'}(f)
\end{equation}
for every $f : [a, b] \to {\bf R}$, using the triangle inequality.
Also, any finite collection of partitions of $[a, b]$ has a common
refinement.

        Suppose that $f$ has bounded variation on $[a, b]$, and that
$x \in (a, b)$.  Thus the restrictions of $f$ to $[a, x]$ and to $[x,
b]$ have bounded variation, and moreover
\begin{equation}
        V_a^x(f) + V_x^b(f) = V_a^b(f).
\end{equation}
Indeed, any partitions $\mathcal{P}_1$, $\mathcal{P}_2$ of $[a, x]$,
$[x, b]$, respectively, can be combined to get a partition
$\mathcal{P}_3$ of $[a, b]$ for which
\begin{equation}
        V_{\mathcal{P}_1}(f) + V_{\mathcal{P}_2}(f) = V_{\mathcal{P}_3}(f),
\end{equation}
which implies that $V_a^x(f) + V_x^b(f) \le V_a^b(f)$.  To get the
opposite inequality, note that every partition of $[a, b]$ can be
refined if necessary to contain $x$, and hence to be a combination of
partitions of $[a, x]$ and $[x, b]$.  The same argument shows that $f$
has bounded variation on $[a, b]$ if it has bounded variation on $[a,
x]$ and on $[x, b]$.

        Suppose that $f$ is continuously differentiable on $[a, b]$.
If $a \le r \le t \le b$, then
\begin{equation}
        |f(t) - f(r)| = \biggl|\int_r^t f'(\xi) \, d\xi\biggr|
                      \le \int_r^t |f'(\xi)| \, d\xi.
\end{equation}
This implies that
\begin{equation}
        V_\mathcal{P}(f) \le \int_a^b |f'(\xi)| \, d\xi
\end{equation}
for every partition $\mathcal{P}$ of $[a, b]$.  One can show that
\begin{equation}
        V_a^b(f) = \int_a^b |f'(\xi)| \, d\xi,
\end{equation}
using very fine partitions $\mathcal{P}$ of $[a, b]$.

        For each $r \in {\bf R}$, put $r_+ = r$ when $r \ge 0$ and
$r_+ = 0$ when $r \le 0$, and $r_- = -r$ when $r \le 0$ and $r-_ = 0$
when $r \ge 0$, so that
\begin{equation}
        r_+ - r_- = r, \quad r_+ + r_- = |r|.
\end{equation}
Given $f : [a, b] \to {\bf R}$ and a partition $\mathcal{P} =
\{t_j\}_{j = 0}^n$ of $[a, b]$, put
\begin{equation}
        P_\mathcal{P}(f) = \sum_{j = 1}^n (f(t_j) - f(t_{j - 1}))_+
\end{equation}
and
\begin{equation}
        N_\mathcal{P}(f) = \sum_{j = 1}^n (f(t_j) - f(t_{j - 1}))_-.
\end{equation}
Thus
\begin{equation}
        P_\mathcal{P}(f) + N_\mathcal{P}(f) = V_\mathcal{P}(f)
\end{equation}
and
\begin{equation}
        P_\mathcal{P}(f) - N_\mathcal{P}(f) = f(b) - f(a).
\end{equation}
Suppose that $f$ has bounded variation on $[a, b]$, and put
\begin{equation}
        P_a^b(f) = \sup \{P_\mathcal{P}(f) : \mathcal{P}
                                       \hbox{ is a partition of } [a, b]\}
\end{equation}
and
\begin{equation}
        N_a^b(f) = \sup \{N_\mathcal{P}(f) : \mathcal{P}
                                       \hbox{ is a partition of } [a, b]\}.
\end{equation}
One can check that
\begin{equation}
        P_a^b(f) + N_a^b(f) = V_a^b(f)
\end{equation}
and
\begin{equation}
        P_a^b(f) - N_a^b(f) = f(b) - f(a).
\end{equation}
Similarly,
\begin{equation}
        P_a^x(f) - N_a^x(f) = f(x) - f(a)
\end{equation}
when $a \le x \le b$.  This implies that $f$ can be expressed as the
difference of two monotone increasing functions on $[a, b]$, since
$P_a^x(f)$, $N_a^x(f)$ are monotone increasing in $x$.

        Functions of bounded variation do not have to be continuous,
but they can only have jump discontinuities.  More precisely, if $f$
has bounded variation on $[a, b]$, then $f$ has one-sided limits from
both sides at every point in $(a, b)$, and from the right and left
sides at $a$, $b$, respectively.  This follows from the analogous
statement for monotone functions and the fact that a function of
bounded variation can be expressed in terms of monotone functions, and
it can also be shown more directly.

\section{Lengths of paths}
\label{lengths of paths}
\setcounter{equation}{0}

        Let $(M, d(x, y))$ be a metric space, let $a$, $b$ be real
numbers with $a \le b$, and let $f$ be a function on $[a, b]$ with
values in $M$.  For each partition $\mathcal{P} = \{t_j\}_{j = 0}^n$
of $M$, consider
\begin{equation}
        \Lambda_\mathcal{P}(f) = \sum_{j = 1}^n d(f(t_j), f(t_{j - 1})).
\end{equation}
This is the same as the variation $V_\mathcal{P}(f)$ of $f$ on
$\mathcal{P}$ when $M$ is the real line with the standard metric.
If there is an upper bound for $\Lambda_\mathcal{P}$ over all
partitions $\mathcal{P}$ of $[a, b]$, then we say that the path $f :
[a, b] \to M$ has finite length, and the length of the path is defined by
\begin{equation}
         \Lambda_a^b(f) = \sup \{\Lambda_\mathcal{P} : \mathcal{P}
                                          \hbox{ is a partition of } [a, b]\}.
\end{equation}
This is the same as the total variation $V_a^b(f)$ of $f$ when $M =
{\bf R}$.  As in the previous case, $\Lambda_a^b(f) = 0$ if and only
if $f$ is constant.  If $a \le r \le t \le b$, then
\begin{equation}
        d(f(r), f(t)) \le \Lambda_\mathcal{P}(f)
\end{equation}
for any partition $\mathcal{P}$ of $[a, b]$ that contains $r$, $t$.
Hence $f([a, b])$ is a bounded set in $M$ when $f : [a, b] \to M$ has
finite length, with
\begin{equation}
        \mathop{\rm diam} f([a, b]) \le \Lambda_a^b(f).
\end{equation}
Of course, $f([a, b])$ is a compact set in $M$ when $f$ is continuous,
and therefore bounded.  If $f$ is continuous, then $f([a, b])$ is also
a connected set in $M$.

        If $f : [a, b] \to M$ is $C$-Lipschitz, then $f$ has finite
length, and
\begin{equation}
        \Lambda_a^b(f) \le C \, (b - a).
\end{equation}
Let $(\widetilde{M}, \widetilde{d}(u, v))$ be another metric space,
and suppose that $\phi : M \to \widetilde{M}$ is $C$-Lipschitz.  For
any $f : [a, b] \to M$ and partition $\mathcal{P}$ of $[a, b]$,
\begin{equation}
 \widetilde{\Lambda}_\mathcal{P}(\phi \circ f) \le C \, \Lambda_\mathcal{P}(f),
\end{equation}
where $\widetilde{\Lambda}$ is the analogous quantity for $\widetilde{M}$.
If $f : [a, b] \to M$ has finite length, then $\phi \circ f : [a, b]
\to \widetilde{M}$ does too, and
\begin{equation}
        \widetilde{\Lambda}_a^b(\phi \circ f) \le C \, \Lambda_a^b(f).
\end{equation}
In particular, if $f$ has finite length and $\phi : M \to {\bf R}$ is
Lipschitz, then $\phi \circ f$ has bounded variation.

        If $f : [a, b] \to M$ has finite length and $a \le a_1 \le b_1
\le b$, then the restriction of $f$ to $[a_1, b_1]$ has finite length, and
\begin{equation}
        \Lambda_{a_1}^{b_1}(f) \le \Lambda_a^b(f).
\end{equation}
If $\mathcal{P}$, $\mathcal{P}'$ are partitions of $[a, b]$ and
$\mathcal{P}'$ is a refinement of $\mathcal{P}$, then
\begin{equation}
        \Lambda_\mathcal{P}(f) \le \Lambda_{\mathcal{P}'}(f)
\end{equation}
for any $f : [a, b] \to M$, as in the case of real-valued functions in
the previous section.  As before, one can use this to show that
\begin{equation}
        \Lambda_a^x(f) + \Lambda_x^b(f) = \Lambda_a^b(f)
\end{equation}
for every $x \in (a, b)$ when $f$ has finite length.  If $a \le x <
b$, then
\begin{equation}
        \lim_{y \to x+} \Lambda_a^y(f)
\end{equation}
exists, because $\Lambda_a^x(f)$ is monotone increasing in $x$, and hence
\begin{equation}
        \lim_{y \to x+} \sup \{\Lambda_w^y(f) : x < w \le y\} = 0.
\end{equation}
This implies that
\begin{equation}
        \lim_{y \to x+} \mathop{\rm diam} f((x, y]) = 0,
\end{equation}
since
\begin{eqnarray}
        \mathop{\rm diam} f((x, y]) & = & \sup \{\mathop{\rm diam} f([w, y]) : x < w \le y\}  \\
             & \le & \sup \{\Lambda_w^y(f) : x < w \le y\}. \nonumber
\end{eqnarray}
If $M$ is complete, then it follows that $f$ has a limit from the
right at $x$, and similarly there is a limit from the left when $a < x
\le b$.

        Suppose now that $M$ is ${\bf R}^n$, equipped with a norm $N$,
and thus the metric $d_N(x, y)$ associated to $N$ too.  If $f_1, f_2 :
[a, b] \to {\bf R}^n$ have finite length and $r_1, r_2 \in {\bf R}$,
then $r_1 \, f_1 + r_2 \, f_2$ has finite length, and
\begin{equation}
        \Lambda_a^b(r_1 \, f_1 + r_2 \, f_2)
                   \le |r_1| \, \Lambda_a^b(f_1) + |r_2| \, \Lambda_a^b(f_2).
\end{equation}
This is similar to the case of real-valued functions, and one can also
treat the product of a real-valued function and an ${\bf R}^n$-valued
function on $[a, b]$ in the same way as before.  If $f : [a, b] \to
{\bf R}^n$ is continuously differentiable, then one can show that $f$
has finite length and that
\begin{equation}
         \Lambda_a^b(f) = \int_a^b N(f'(\xi)) \, d\xi,
\end{equation}
in practically the same way as before.  It can be interesting to
consider integral norms
\begin{equation}
         \Big(\int_a^b N(f'(\xi))^p \, d\xi\Big)^{1/p}
\end{equation}
as well, $1 \le p < \infty$.  The $p = \infty$ case corresponds to the
maximum of $N(f')$ on $[a, b]$.  This integral norm is especially
interesting when $p = 2$ and $N$ is the standard Euclidean norm on
${\bf R}^n$.  For other $p$, there is some simplification when $N(v) =
\|v\|_p$.  If $N(v) = \|v\|_1$, then the length of any path of finite
length in ${\bf R}^n$ is equal to the sum of the total variations of
the coordinates of the path.  This uses the fact that any finite
collection of partitions of $[a, b]$ has a common refinement, so that
independent partitions for the coordinate functions are equivalent to
using the same partition for the whole path.

\section{Snowflake metrics}
\label{snowflakes}
\setcounter{equation}{0}

        Let $\alpha$ be a positive real number, with $\alpha < 1$.
For any pair of nonnegative real numbers $u$, $v$,
\begin{equation}
\label{u^alpha, v^alpha}
        (u + v)^\alpha \le u^\alpha + v^\alpha.
\end{equation}
To see this, observe that
\begin{equation}
        \max(u, v) \le (u^\alpha + v^\alpha)^{1/\alpha},
\end{equation}
and hence
\begin{equation}
        u + v \le \max(u, v)^{1 - \alpha} \, (u^\alpha + v^\alpha)
               \le (u^\alpha + v^\alpha)^{1/\alpha}.
\end{equation}
Note that the inequality is strict in (\ref{u^alpha, v^alpha}) when $u, v > 0$.

        If $(M, d(x, y))$ is a metric space, then it follows from
(\ref{u^alpha, v^alpha}) that $d(x, y)^\alpha$ is also a metric on
$M$.  This does not change the topology of $M$, but it does change the
geometry.  Many familiar examples of snowflake curves in the plane
have approximately this type of geometry, for instance.

        Suppose that $f : [a, b] \to M$ is a continuous path with
finite length with respect to $d(x, y)^\alpha$.  This means that
\begin{equation}
        \sum_{j = 1}^n d(f(t_j), f(t_{j - 1}))^\alpha \le A
\end{equation}
for some $A \ge 0$ and every partition $\{t_j\}_{j = 0}^n$ of $[a, b]$.
Let $\epsilon > 0$ be given.  By continuity and compactness, $f$
is uniformly continuous, and so there is a $\delta > 0$ such that
\begin{equation}
        d(f(r), f(w)) < \epsilon
\end{equation}
for every $r, w \in [a, b]$ such that $|r - w| < \delta$.  Hence
\begin{equation}
        \sum_{j = 1}^n d(f(t_j), f(t_{j - 1})) \le \epsilon^{1 - \alpha} \, A
\end{equation}
when $t_j - t_{j - 1} < \delta$ for each $j = 1, \ldots, n$.  Every
partition of $[a, b]$ has a refinement with this property, which
implies that the length $\Lambda_a^b(f)$ of $f$ with respect to $d(x,
y)$ satisfies
\begin{equation}
        \Lambda_a^b(f) \le \epsilon^{1 - \alpha} \, A.
\end{equation}
Thus $\Lambda_a^b(f) = 0$, since $\epsilon > 0$ is arbitrary, and $f$
must be constant.

\section{H\"older continuity}
\label{holder}
\setcounter{equation}{0}

        Let $(M_1, d_1(x, y))$ and $(M_2, d_2(w, z))$ be metric
spaces.  A mapping $f : M_1 \to M_2$ is said to be \emph{H\"older
continuous} of order $\alpha$, $0 < \alpha < 1$, if
\begin{equation}
        d_2(f(x), f(y)) \le C \, d_1(x, y)^\alpha
\end{equation}
for some $C \ge 0$ and every $x, y \in M_1$.  One might also say that
$f$ is Lipschitz of order $\alpha$ in this case, but it will be
convenient to refer to this as a Lipschitz condition when $\alpha = 1$
and H\"older continuity when $0 < \alpha < 1$.  Similar names are
sometimes used for other related conditions as well.

        As in the previous section, $d_1(x, y)^\alpha$ is a metric on
$M_1$, and therefore $f$ is H\"older continuous of order $\alpha$ with
respect to $d_1(x, y)$ if and only if $f$ is Lipschitz with respect to
$d_1(x, y)^\alpha$.  Thus many basic properties of H\"older continuous
mappings follow from the corresponding statements for Lipschitz
mappings.  In particular,
\begin{equation}
        f_p(x) = d_1(p, x)^\alpha
\end{equation}
is a real-valued H\"older continuous function of order $\alpha$ on
$M_1$ with $C = 1$ for each $p \in M_1$.

        Let $(M, d(x, y))$ be a metric space, and consider the case of
a continuous path $f : [a, b] \to M$.  If $f$ is $C$-Lipschitz, then
$f$ is H\"older continuous of order $\alpha$ for each $\alpha \in (0,
1)$ with constant $C \, (b - a)^{1 - \alpha}$.  Of course, $f$ also
has finite length $\le C \, (b - a)$ when $f$ is $C$-Lipschitz.
However, continuous paths of finite length need not be H\"older
continuous of any positive order, and there are couterexamples already
for monotone increasing real-valued functions.  Similarly, H\"older
continuous paths may not have finite length.

        Suppose that $f : [a, b] \to M$ is H\"older continuous of
order $\alpha$ with constant $C > 0$.  For each $\rho > 0$, $f([a,
b])$ is contained in the union of $O(\rho^{-1/\alpha})$ subsets of $M$
with diameter $\le \rho$, because $[a, b]$ is the union of
$O(\rho^{-1/\alpha})$ subintervals of length $\le
(\rho/C)^{1/\alpha}$.  This implies that the Minkowski dimension of
$f([a, b])$ is $\le 1/\alpha$, and hence the Hausdorff dimension of
$f([a, b])$ is $\le 1/\alpha$ too.  This is an analogue for H\"older
continuous paths of the fact that Lipschitz paths have finite length.

\section{Coverings}
\label{coverings}
\setcounter{equation}{0}

        If $[a, b], [a_1, b_1], \ldots, [a_n, b_n]$ are closed
intervals in the real line such that
\begin{equation}
        [a, b] \subseteq \bigcup_{j = 1}^n [a_j, b_j],
\end{equation}
then
\begin{equation}
        b - a \le \sum_{j = 1}^n (b_j - a_j).
\end{equation}
More generally, if $E_1, \ldots, E_n$ are bounded subsets of ${\bf R}$
such that
\begin{equation}
        [a, b] \subseteq \bigcup_{j = 1}^n E_j,
\end{equation}
then
\begin{equation}
        b - a \le \sum_{j = 1}^n \mathop{\rm diam} E_j.
\end{equation}
Indeed, each $E_j$ is contained in a closed interval of the same diameter.

        Let $(M, d(x, y))$ be a metric space, and let $A, E_1, \ldots, E_n$
be bounded subsets of $M$ such that
\begin{equation}
        A \subseteq \bigcup_{j = 1}^n E_j.
\end{equation}
If $A$ is connected, then
\begin{equation}
        \mathop{\rm diam} A \le \sum_{j = 1}^n \mathop{\rm diam} E_j.
\end{equation}
To see this, remember first that continuous mappings send connected
sets to connected sets.  If $f : M \to {\bf R}$ is continuous, then
$f(A)$ is an interval in the real line, which may be open, or closed,
or half-open and half-closed.  At any rate,
\begin{equation}
        {\mathop{\rm diam}}_{\bf R} f(A) \le \sum_{j = 1}^n {\mathop{\rm diam}}_{\bf R} f(E_j),
\end{equation}
where the subscripts indicate that these are diameters in ${\bf R}$.
If $f$ is $C$-Lipschitz, then
\begin{equation}
        {\mathop{\rm diam}}_{\bf R} f(A) \le C \, \sum_{j = 1}^n \mathop{\rm diam} E_j.
\end{equation}
The desired estimate follows by applying this to $1$-Lipschitz
functions of the form $f_p(x) = d(p, x)$, $p \in A$.

        The hypothesis that $A$ be connected is essential here.  If
$A$ is a finite set with at least two elements, then $\mathop{\rm diam} A > 0$,
but $A$ is contained in the union of finitely many sets with one
element and thus diameter $0$.  Cantor's middle-thirds set in the real
line has diameter equal to $1$, and is contained in the union of
$2^\ell$ intervals of length $3^{-\ell}$ for each $\ell \ge 1$.  A
compact set $A \subseteq {\bf R}$ has Lebesgue measure $0$ exactly if
for each $\epsilon > 0$ there are finitely many bounded sets $E_1,
\ldots, E_n \subseteq {\bf R}$ such that $A \subseteq \bigcup_{j =
1}^n E_j$ and $\sum_{j = 1}^n \mathop{\rm diam} E_j < \epsilon$.

        In particular, if $A \subseteq M$ is a bounded connected set
and $\rho > 0$, then $A$ is not covered by fewer than $\mathop{\rm diam} A / \rho$
bounded subsets of $M$ of diameter $\le \rho$.  Depending on the
situation, many more of these subsets may be required.  If $M$ is
${\bf R}^n$ equipped with the standard metric, for example, then a
bounded set $A$ can be covered by $O(\rho^{-n})$ sets of diameter $\le
\rho$.  One needs at least a positive multiple of $\rho^{-n}$ such
sets when $A$ has nonempty interior, because otherwise the
$n$-dimensional volume of $A$ would be too small.

\section{Domains in ${\bf R}^n$}
\label{domains, R^n}
\setcounter{equation}{0}

        A set $U$ in a metric space $M$ is an open set if for every $p
\in U$ there is an $r > 0$ such that $B(p, r) \subseteq U$.  Any norm
$N$ on ${\bf R}^n$ determines the same open sets as the standard
metric.  This is because $N$ is less than or equal to a constant times
the standard norm, and conversely the standard norm is less than or
equal to a constant times $N$.  The first statement can be checked
directly by expressing any element of ${\bf R}^n$ as a linear
combination of the standard basis for ${\bf R}^n$ and using the
triangle inequality.  As mentioned previously, this and the triangle
inequality imply that $N$ is continuous with respect to the standard
norm.  Hence the minimum of $N$ is attained on the standard unit
sphere in ${\bf R}^n$, since the latter is compact.  The standard norm
times the minimum of $N$ on the unit sphere is less than or equal to
$N$ on all of ${\bf R}^n$, by homogeneity, which implies the second
statement.  For explicit norms like $\|w\|_p$, $1 \le p \le \infty$,
the comparison with the standard norm can be verified directly.

        Suppose that $U$ is a connected open set in ${\bf R}^n$, which
is to say that $U$ is not the union of two disjoint nonempty open
sets.  It is well known that $U$ is then pathwise-connected, so that
for every $p, q \in U$ there is a continuous mapping $f : [a, b] \to
U$ such that $f(a) = p$ and $f(b) = q$.  More precisely, one can even
take $f$ to be piecewise-affine on $[a, b]$.  In particular, $f$ then
has finite length.

        However, it is not clear how small the length of $f$ can be.
Of course, the length of $f$ is at least the distance between $p$ and
$q$.  If $U$ is convex, then one can take $f$ to be affine, and the
length of $f$ is equal to the distance between $p$ and $q$.
Otherwise, the length of $f$ may have to be quite large compared to
the distance between $p$ and $q$.  It is easy to give examples where
this happens in the plane.  For instance, there may be elements of $U$
on opposite sides of the boundary locally.  The boundary of $U$ might
also be complicated, with spirals or other obstacles.

        Even if the boundary of $U$ is complicated, it may be that $U$
behaves well in terms of lengths of paths.  For example, if $U$ is the
region in the plane bounded by the von Koch snowflake, then every pair
of elements of $U$ can be connected by a path of length bounded by a
constant multiple of the distance between them.  The main idea is for
the path to avoid the boundary as much as possible, without going too
far away.  There can also be relatively small parts of the boundary
that only cause minor detours for paths in the domain.

\section{Lipschitz graphs}
\label{lipschitz graphs}
\setcounter{equation}{0}

        Let $k, l, n$ be positive integers such that $k + l = n$, and
let us identify ${\bf R}^n$ with ${\bf R}^k \times {\bf R}^l$, so that
an element $x$ of ${\bf R}^n$ may be expressed as $(x', x'')$, where
$x' \in {\bf R}^k$ and $x'' \in {\bf R}^l$.  Also let $A : {\bf R}^k
\to {\bf R}^l$ be a continuous mapping, and consider its graph
\begin{equation}
        \{(x', x'') \in {\bf R}^n : x'' = A(x')\}.
\end{equation}
This is a nice $k$-dimensional topological submanifold of ${\bf R}^n$.
If $k = n - 1$, then this hypersurface has two complementary
components $U_+$, $U_-$ consisting of $(x', x'') \in {\bf R}^n$ such
that $x'' > A(x')$ and $x'' < A(x')$, respectively.  If $k < n$, then
the complement of the graph in ${\bf R}^n$ is connected.  For any $k$,
\begin{equation}
        (x', x'') \mapsto (x', x'' + A(x'))
\end{equation}
defines a homeomorphism on ${\bf R}^n$ that sends the $k$-plane $x'' =
0$ to the graph of $A$.  If $f(t)$ is a continuous path in ${\bf R}^k$
parameterized by an interval $[a, b]$, then $\widehat{f}(t) = (f(t),
A(f(t))$ is a continuous path in the graph of $A$.  The graph of $A$
is itself a curve in ${\bf R}^n$ when $k = 1$.

        Suppose that $A$ is Lipschitz.  If $f$ has finite length, then
$\widehat{f}$ does too, and the length of $\widehat{f}$ is bounded by
a constant multiple of the length of $f$.  If the Lipschitz constant
of $A$ is small, then this constant multiple is close to $1$.
Using affine paths in ${\bf R}^k$, we get that every pair of elements
of the graph of $A$ can be connected by a continuous path in the graph
of $A$ of finite length bounded by a constant multiple of the distance
between them, where the constant multiple is close to $1$ when $A$
has small Lipschitz constant.

        A $k$-dimensional $C^1$ submanifold of ${\bf R}^n$ is locally
the same as the graph of a continuously-differentiable mapping on
${\bf R}^k$ with respect to a suitable choice of coordinate axes.  By
rotating the axes so that ${\bf R}^k$ is parallel to the tangent plane
of the submanifold at a particular point, the submanifold can be
represented near the point as the graph of a function with small
Lipschitz constant.  The Lipschitz constant tends to $0$ as one
approaches the point in question.  Thus distances on $C^1$
submanifolds are approximately the same as the infimum of lengths of
paths on the submanifold locally.

\section{Real analysis}
\label{real analysis}
\setcounter{equation}{0}

        For the sake of simplicity, we have so far avoided referring
to Lebesgue integrals and measure.  However, this more sophisticated
theory can be quite convenient in the present context.  Let us mention
some of the key points.

        A basic fact is that a monotone real-valued function on an
open interval in the real line is differentiable ``almost
everywhere'', which is to say on the complement of a set of Lebesgue
measure $0$.  Thus additional hypotheses of differentiability are
sometimes superfluous.  Unfortunately, even continuous monotone
functions cannot necessarily be recoved from their almost everywhere
derivative, as in the fundamental theorem of calculus, without an
extra condition of ``absolute continuity''.  Indeed, there are
examples of nonconstant continuous monotone increasing functions with
derivative equal to $0$ almost everywhere.

        It follows that a real-valued function of bounded variation on
an interval in the real line is differentiable almost everywhere,
since it can be expressed as the difference of two monotone increasing
functions.  In particular, a real-valued Lipschitz function on an
interval is differentiable almost everywhere.  Lipschitz functions are
absolutely continuous, and so there is a version of the fundamental
theorem of calculus for them.  As corollaries of this fact, a
Lipschitz function $f$ on an interval is constant if $f'(x) = 0$
almost everywhere, $f$ is monotone increasing if $f'(x) \ge 0$ almost
everywhere, and $f$ is $C$-Lipschitz if $|f'(x)| \le C$ almost
everywhere.

        At the same time, bounded variation and Lipschitz conditions
have natural extensions involving metric spaces, as we have seen.  The
composition of a path of finite length in a metric space with a
real-valued Lipschitz function on the metric space is a function of
bounded variation, which is Lipschitz when the path is.  There are
also a lot of real-valued Lipschitz functions on any metric space.
Even on ${\bf R}^n$, there are a lot of nice functions that are
Lipschitz and not continuously differentiable, such as the distance to
a point or to a set.

redpajama_set_name: RedPajamaArXiv


text:
ScholarWorks at University of Montana
Maureen and Mike Mansfield Library
Home > The Mathematics Enthusiast > Vol. 9 > No. 3 (2012)
Evaluating Mersenne Primes Using a Single Quadrant Expanding Square
Cory P. Daignault
Douglas Sprague
Joe L. Mott
By forming a table of sequential odd numbers using a single quadrant expanding square pattern it was observed that multiple Mersenne primes fall into the first column. We prove that the Mersenne primes which fall into the first column will be of the form p = 3 mod 4, where p is the Mersenne prime exponent. Focusing the search of primes to those which fall into the first column of this square may be a means to increase the speed at which large primes are discovered.
Daignault, Cory P.; Sprague, Douglas; and Mott, Joe L. (2012) "Evaluating Mersenne Primes Using a Single Quadrant Expanding Square," The Mathematics Enthusiast: Vol. 9 : No. 3 , Article 11.
Available at: https://scholarworks.umt.edu/tme/vol9/iss3/11
TME - 2020 Circulation Statistics
Call for Papers: Special Issue on Mathematics CURES (Course-Based Undergraduate Research Experiences)
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redpajama_set_name: RedPajamaCommonCrawl


text:
Ireland's mother and baby homes: 9,000 babies and children died in 18 mother and
Around 56,000 people — from girls as young as 12 to women in their 40s — were sent to the 18 institutions investigated, where some 57,000 children were born, according to the report.
One in seven of those children (15%) didn't survive long enough to leave the homes, yet no alarm was raised by the State over the high mortality rates, even though it was "known to local and national authorities" and was "recorded in official publications," the report found.
Prior to 1960, mother and baby homes "did not save the lives of 'illegitimate' children; in fact, they appear to have significantly reduced their prospects of survival," it said.
The report called the infant mortality rates the most "disquieting feature of these institutions."
Speaking on Tuesday, Taoiseach Micheál Martin said that the report "opens a window onto a deeply misogynistic culture in Ireland over several decades," and that the report "reveals significant failures of the state and of society."
The report, which runs to more than 2,800 pages, was released just days after its key findings were leaked to a national newspaper — compounding the pain and anguish of survivors who have waited years for the final report — and who had been promised a first view of it by the Minister of Children.
Susan Lohan, co-founder of the Adoption Rights Alliance and a member of a dedicated survivors group appointed to advise the government, told CNN on Tuesday that the leaked extracts of the report, seen on Sunday, shows that the Irish government may seek to "trivialize" the human rights abuses that took place on a "massive scale" inside of these homes.
Survivor Philomena Lee, who spent years searching for the son she was forced to give up for adoption said in a statement on Sunday that she had "waited decades for this moment — the moment when Ireland reveals how tens of thousands of unmarried mothers, such as I, and the tens of thousands of our beloved children, such as my dear son Anthony, were torn asunder, simply because we were unwed at the moment our children were born."
During her time at the Sean Ross Abbey mother and baby home, Lee said that she was "deprived" of her liberty, independence and autonomy, and was "subject to the tyranny of the nuns," who told mothers daily that they were to atone for their sins by "working for our keep and surrendering our children to the nuns for forced adoption."
Lee, whose story was told in an Oscar-nominated movie starring Judi Dench, added that she was "taunted" by the nuns during a difficult labor, who she says told her that the "pain was a punishment for my promiscuity."
The commission's final report reported that this practice was not unusual.
For many survivors and advocate groups, there is concern that the report fails to vindicate their experience.
Lohan told national broadcaster RTE that the institutions were a "form of social engineering," and that the "state and church worked in concert to ensure that women — unmarried mothers and girls who were deemed to be a threat to the moral tone of the country" were "incarcerated behind these very high walls to ensure that they would not impact or offend public morality."
On Tuesday, Minister for Children, Equality, Disability, Integration and Youth Affairs Roderic O'Gorman said: "The report makes clear that for decades, Ireland had a stifling, oppressive and brutally misogynistic culture, where a pervasive stigmatization of unmarried mothers and their children robbed those individuals of their agency and sometimes their future."
Survivors are expected to receive an official State apology from Martin, the Irish Prime Minister, on Wednesday.
But for many, that apology will not be enough.
Lohan told CNN that she disagrees with the planned state apology, saying that no apology should be issued until survivors have had a chance to read and digest the Commission's findings, which could take many weeks.
She also suggested that an apology should be the first of a series of several, noting that the commission's investigations only covered 18 institutions, while some 180 sites were part of an Irish system that facilitated child neglect, premature death, forced adoptions, enforced disappearances, enforced labor, the stripping of identities, the falsification of state documents and the forging of mothers' consents.
The report does not appear to fully address the allegations of forced or illegal adoptions, only stating that "many allegations have been made that large sums of money were given to the institutions and agencies in Ireland that arranged foreign adoptions. Such allegations are impossible to prove and impossible to disprove."
Lee also underlined the role that other state-run and private institutions played, saying in her statement that she "can only hope" that…
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U.S. equity markets fell sharply into the red in afternoon trading Tuesday as markets participants parsed through speeches from two Fed officials.
As of 1:18 p.m. ET, the Dow Jones Industrial Average fell 140 points, or 0.85%, to 16371, the S&P 500 dipped 13.5 points, or 0.71%, to 1872 and the Nasdaq Composite slipped 34.3 points, or 0.83%, to 4092.
The markets spent much of Monday lazily floating about the unchanged line. The business news calendar is similarly slow on Tuesday, which is amplified by light volume ahead of the Memorial Day holiday in the U.S.
Home Depot (NYSE:HD) revealed narrowly better-than-expected quarterly profits, but weak sales results drove sales of the world's biggest home improvement retailer into the red. Urban Outfitters (NASDAQ:URBN) revealed mixed results as well, pushing its shares lower.
Credit Suisse (NYSE:CS) pleaded guilty to criminal charges it helped Americans evade taxes. The big Swiss bank will pay $2.6 billion to settle the suit, but it doesn't expect the pact to materially affect its operations.
On the economic front, Philadelphia Federal Reserve chief Charles Plosser said he isn't too concerned about recent weakness in the housing market. At the same time, New York Fed President Bill Dudley said the central bank will pay close attention to Wall Street's response as it eventually hikes interest rates.
Elsewhere, U.S. crude oil futures rose 3 cents, or 0.03%, to $102.64 a barrel. Wholesale New York Harbor gasoline dipped 0.18% to $2.959 a gallon. Gold slumped $2.80, or 0.21%, to $1,291 a troy ounce.


text:
Project iMlango Hits Ground Running
by Cormac Lynch | Oct 16, 2014 | Blog, Press Release | 0 comments
Volunteers and staff get a chain going following the arrival of the first iMlango container in July
Back in July, we told you about Camara's new strategic partnership with the UK Department for International Development (DFID) and corporate partners Avanti, sQuid and Whizz Education. Derived from the Swahili word for door or portal, iMlango seeks to improve learning and educational outcomes for 25,675 marginalised girls, across 195 primary schools in Kenya.
Project iMlango delivers:
*High-speed broadband connectivity to schools
*Personalised maths tuition with a virtual online tutor, alongside digital learning content for maths, literacy and life skills
*Tuition and support to teachers to use ICT in their teaching
*Electronic attendance monitoring with conditional payments – to incentivise families to send their daughters to school – for use with local merchants
*In-field capacity in IT, technology and support resources
*Real-time project monitoring and measurement
Speaking at the time of the announcement, Camara CEO John Fitzsimons said, "Project iMlango is hugely exciting for Camara Education. It is an honour to work with the Department for International Development for the first time. It is also exciting to work with such a diverse and successful set of companies in the consortium."
It has now been over two months since the iMlango initiative was announced, and there has been plenty of activity in the interim. The first of four containers earmarked for the iMlango project arrived in Kenya in mid-July, at which point the hard work really began. Camara's Africa CEO, Philip McAllister was on the ground in Kenya to see the work first-hand;
"Camara Kenya received the container to the hub this morning, the first of four containers earmarked for the iMlango project being rolled out this year to 195 schools across Kenya. Every time another container arrives the team get quicker at unloading. It took them just 90 minutes this time around to unload a 40-foot container of over 1,000 PC's. Now the fun really begins with the busy work of processing and checking all the equipment before dispatch."
Once all the equipment is checked, the hard work of dispatching the computers to the schools begins. With 195 schools being serviced, the task of distributing the equipment is a demanding one, but one which everybody in the hubs is well used to by now. As well supplying the schools with computers, training is provided to teachers who will ultimately be responsible for imparting that knowledge to their students.
The second container arrived in mid-August and work has been ongoing to distribute the equipment from that shipment to schools all over Kenya, a country of 580,367 square kilometres – a land mass six times the size of the island of Ireland.
Volunteers in Mombasa hard at work!
This week, the team in Kenya has been dispatching to the last region to be covered – Uasin Gishu. The third container is due to arrive by the end of October, with the fourth and final container landing in Kenya in December. There is plenty of work left to be done, but the impetus provided by the iMlango project – which will result in 3,250 computers, 520 laptops with headphones and 520 projectors being put into schools in Kenya in just six months – should not be underestimated. 95 schools will be serviced by the end of 2014, with the remaining 100 to be supplied with their equipment in early 2015.
This kind of partnership is an example of what can be done when charities and businesses come together, and is one that we in Camara are extremely proud to be a part of.
You can read about iMlango in more detail here.


text:
TikTok Is Suing The Trump Administration
Story from TikTok
TikTok Sues The Trump Administration Over Ban
Michelle Santiago Cortés
Photo: Stefani Reynolds/Bloomberg/Getty Images.
TikTok is fighting back.
Ever since TikTok's Chinese origins became common knowledge, claims (both founded and unfounded) that the app is a vehicle for Chinese government surveillance haven't taken a rest. It gave President Trump — the man who coined the term "China flu" — an occasion to undermine China and liberal teens in one fell swoop. When the app was first banned, TikTok was given 30 days to sell to a U.S.-based company with Trump demanding that a cut of the sale go into the pockets of the U.S. government. Then, the 30-day deadline became a 45-day deadline.
In that time, TikTok has fought hard to save the youth's favorite app, announcing its $100 million creator fund, turning that into a $2 billion fund, and creating 10,000 jobs. Now, TikTok is suing the Trump administration for attempting a ban.
In a statement published to its website, TikTok explained, "We do not take suing the government lightly, however, we feel we have no choice but to take action to protect our rights and the rights of our community and employees." TikTok claims that by banning the app with no notice, the administration is violating the Fifth Amendment, which guarantees the right to due process.
TikTok is arguing that the Administration is stripping TikTok's users of their rights with no evidence that the platform presents any real risk to national security. It also claims the administration ignored the company's exhaustive efforts to prove it poses no significant threat to national security. In other words, TikTok says it's done everything it can to work with the government to prove it's a safe platform, but the administration doesn't care. In fact, according to TikTok, some of the cited national security concerns are sourced from "outdated news articles."
TikTok didn't just roll up to the U.S. — it landed on our phones after its parent company bought Musical.ly. For a transaction of that size, a lot of paperwork had to be run through the U.S. government, including detailed descriptions of TikTok's security practices.
So, what comes next? Today, TikTok filed a complaint against the Trump administration in federal court. As many experts have said, TikTok doesn't seem to present any obviously exceptional or significant national security threat. But what does seem obvious, is how political this all is. Never has TikTok's Chinese ownership been such a big deal.
TikTok Suing Trump Over Executive Order To Ban In US
TikTok • The Latest • Living • Tech • Lifestyle
written by Michelle Santiago Cortés
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Bonus Reviews: Metropolis (1927) and Dr. Jekyll and Mr. Hyde (1931)
Horror Bulletin Bonus for Week 186
For this week's bonus films, we've got a couple of really old classics for you. 1927's Uber-famous, silent "Metropolis" and the Academy-award winning "Dr. Jekyll and Mr. Hyde" from 1931. Good stuff!
Don't forget, the first week of each month, we publish ALL our reviews, including the bonus content, in our monthly "Horror Bulletin" print magazine (also available as an ebook). If you don't have time to read the website or email, here's one more option for you! The new issue is out now!
Thanks for reading Horror Bulletin! Subscribe for free to receive new posts and support my work.
https://www.amazon.com/dp/B09JN2WBJ8
1927 Metropolis
Director: Fritz Lang
Writers: Thea von Harbou
Stars: Brigitte Helm, Alfred Abel, Gustav Fröhlich
Run Time: 2 Hours, 33 Minutes
We begin with a montage of industrial machines. Very non-cheerful people march into work for a shift change. They work deep underground in a huge facility.
Meanwhile, in the "Club of the Sons," men run track races, play games, and generally have fun. This places was created by the very rich for their children. Freder, the son of Joh, the leader of the city, chases a girl around a fountain. He grabs the girl, and they get ready to "go to it" when a teacher and a bunch of children come in. The teacher and Freder stare at each other from across the courtyard, and Freder loses interest in the girl he was playing with.
Freder goes to inspect the machine. Meanwhile, in the boiler room, the pressure is rising, and it explodes, revealing Moloch, a giant demon head that takes the pace of the machine. Bound workers are leading its mouth by guards, followed by willing workers who march in without force. Freder blinks, and the machine reappears, with many injured and dead workers on the floor.
Freder runs out and demands to go to the New Tower of Babel to see his father, the master of the city. We see a panoramic city from many angles, with movement everywhere. Meanwhile, Joh Frederson dictates a letter to his assistants while a stock ticker shows the events of the day behind him. His assistant Josaphat didn't tell Joh about the explosion, but Freder tells him everything. Freder and Joh disagree on the value of workers. "What if one day those down below rise up to stop you?" He asks.
The foreman of the Heart Machine is Grot. He's found plans on the bodies of two of the dead workers— sabotage? Joh gets angry at this and fires Josaphat. Freder claims that being fired by Joh is the same as being fed to the machine. Freder chases after Josaphat, who is very upset. Freder explains that he'll fix Josaphat up with a job tomorrow morning.
Meanwhile, Joh is unhappy with Freder's weakness and orders that he be followed everywhere. Freder wants to work with the machines, so he trades clothes with an 11811, an exhausted worker, and takes over. He tells 11811 to go hang out with Josaphat for tonight.
Rotwang the inventor lives in a weird old house somewhere in the city. He's got a metal hand. Joh goes to visit the grave of Hel, his dead wife, and encounters Rotwang there. Rotwang claims to have re-created Hel and invites Joh in to see it. It's a metal woman. It stands up and walks!
Rotwang says this is the man of the future, the Machine Man. In another 24 Horus of work, no one will be able to tell this machine from a real person. Joh hands the plans Grot found to Rotwang and asks what they are. Rotwang explains that it's a map of the catacombs deep beneath the city. The two men head down Rotwang's secret staircase to the catacombs.
Meanwhile, Freder finds another copy of those plans and is invited to a secret meeting later today. Finally, his shift ends and he follows the other workers… down to the catacombs. It's like a big church down there, with crosses and candles and everything. Maria, the teacher from before, is on the altar and giving a sermon and a prayer. Rotwang and Joh also watch the service from a hole high in the wall.
Maria tells the story of the Tower of Babel. The men thought up the tower and then hired huge numbers of men to build it. The tower was built and destroyed. "The mediator between head and hands must be the heart," Maria explains. Maria and Freder stare longingly at each other. She thinks he will be their mediator.
Joh commands Rotwang to make his Machine Man in Maria's likeness to sow discord among the workers. Rotwang follows Maria deep into the catacombs and scares her into going up the wrong staircase, to his secret lab.
Intermission.
Freder goes to church and hears an apocalyptic prophecy. Elsewhere, Rotwang orders his Machine Man to cause trouble.
Meanwhile, 11811 makes it to Yoshihara, but is grabbed by the Thin Man, the one who has been following Freder. Freder finally goes to see Josaphat, who hasn't seen 11811/Georgie yet. The Thin Man orders 11811 to go back to work and forget everything. The Thin Man then pays a visit to Josaphat and offers to give him lots of money to leave town.
Rotwang and Maria struggle inside the lab, and she screams. Freder hears her, as he just happens to be walking by outside. Rotwang hooks Maria up to the machine and copies her likeness onto the robot. The machine opens its eyes…
The door opens, and Freder rushes inside to find Maria's gone. The robot Maria goes to visit Joh, who orders her to destroy the rebels in the tunnels. Freder rushes in to interrupt, but instead has some kind of mental attack and passes out.
Rotwang and Joh go to the club and watch robot-Maria dance. She drives all the young men wild with her sexy shenanigans. Freder has a vision of the Whore of Babylon, who happens to resemble dancing Maria a lot. Death raises his scythe, and Freder says "Death descends upon the city!"
Josaphat catches up with Freder. The only thing keeping the workers in line is the hope of the promised "mediator" who is prophesied to be coming. Young men are fighting and killing each other over this new Maria, which is causing all kinds of trouble with the men. Freder accepts that he will be the mediator and heads to the big meeting.
Joh wants the workers to forcefully revolt, so he can use force against them. Fake Maria's out there nonstop spreading the word of the revolution. "Kill them— the machines!" The men start to forget about the promised mediator.
Rotwang admits that Maria works for him and not Joh, and that everything may not be Joh thinks it is. Joh, however, is listening outside and the two fight. As they do battle, the real Maria escapes.
The workers revolt. They all load up into the elevators and head for the surface world. The workers leave the machines, so robot Maria leads them on to the Heart Machine. Grot locks them out with shield doors. Joh gets on the videophone with Grot and gets an earful. Joh orders him to open up and let the workers in. Joh wants the workers to destroy the Heart Machine. Grot explains to the workers that if they destroy the machine, the whole city will be flooded. Maria turns the machine up to eleven, and the gauges start to rise…
The machine explodes, and the lower city starts to fill with water. The real Maria works to evacuate all the children. Freder and Josaphat climb the ladders to get out.
Joh sits in his office and waits for the lights to go out. The Thin Man comes in and tells Joh that his son is with the workers. Freder finds the real Maria, and they kiss and tell the children which way to go to escape the flood. Before long, the underground city is completely destroyed.
The Thin Man accuses Joh of murdering the city and thousands of workers. Grot points out to the workers that their children are all downstairs drowning in the flooded city. They hadn't thought of that. Grot turns them all against Maria— but which Maria will they catch first, the real or the robot?
They catch up to the robot first and tie her to a stake like a witch. They set fire to the wood and wait for her to burn. Freder sees this, and he tries to rescue her, but the men hold him back. They watch as "Maria" turns back into a robot.
Rotwang chases the real Maria around, thinking she's his long, lost Hel. Freder sees them running around and starts to chase them as well. Freder and Rotwang fight up on the tower as Joh regrets his decisions. Rotwang takes the fast way off the tower, and Joh runs into the church to find his son. Joh, Freder, and Maria come out the base of the church and encounter Grot and the crowd outside. Grot and Joh refuse to shake hands; management and the workers need… a mediator! Freder gets between the two and shows them how to be friends again!
I'd only seen clips of this prior to this viewing, and there was a lot more here than I thought there would be. This is really long (2.5+ hours), and a lot happens without getting slow or boring. If filmed today, this would probably be expanded into a full miniseries. There were more than 37,000 extras in this, and the crowd scenes are not only real, they're huge.
The sets here are incredibly impressive for their time. They're huge and quite intricate, and even the cityscapes are much more than simple matte paintings (well, some of them are matte paintings). The various special effects are all really well done, and they don't really stand out for the most part, they're just part of the story.
I'm not sure what Joh's plan was. He wanted the workers to revolt so he could use force against them, but he never even tried to stop them. He just incited them all to destroy the city.
Much like many more modern tales, such as Soylent Green, the horror here isn't so much the evil robot or any individual characters, but the dystopian world-system itself.
#sci-fi #silent #dystopia #unposted
Dr. Jekyll and Mr. Hyde (1931)
Directed by Rouben Mamoulian
Written by Samuel Hoffenstein, Percy Heath
Stars Fredric March, Miriam Hopkins, Rose Hobart
Run Time: 1 Hour, 46 Minutes
A classic horror film, really well done for its time, and it's still entertaining and interesting today. The make-up effects are cool, the story is good, the acting is excellent. Frederic March co-won the best actor Oscar that year, and we can see why.
The butler, Poole, nags Dr. Jekyll about his upcoming lecture; he's going to be late if he doesn't quit playing Bach on the pipe organ in his living room. We get a POV shot as Jekyll plays, gets up, and puts on his cloak for the evening. He even sees himself in the mirror. This was probably an innovative shooting style for the time.
Jekyll begins his lecture. He wants to talk about man's soul. Man is two beings, one striving for nobility, his good self. The other ties him to the Earth– the bad self. If these two selves could be separated from each other, how great could the good half be? He thinks this separation may be possible.
Later on, we see that Dr. Jekyll is pretty good at his job as surgeon and all-around physician. Dr. Lanyon wants Jekyll to go with him to the big ball at the General's house, but Jekyll would rather do more operations that are needed. Lanyon makes excuses for Jekyll to Muriel, who is in love with Jekyll. Jekyll finally arrives, and he dances with Muriel. They are engaged, and he wants to get married now, but her father, the General, wants them to wait eight more months– decency must be observed! The general calls Jekyll eccentric.
Jekyll comes to the rescue of Ivy, a woman being beaten on the street, and the two flirt a little bit; well, at least she does. We see that yeah, he's into it too, but gets caught in the act by Lanyon. She tells him to "come back soon." Lanyon criticizes Jekyll all the way home. This gets the two men talking about Jekyll's theories again.
Jekyll works feverishly in his lab on his side project. Finally finished, he locks the door and drinks it. He grabs his throat and changes before our eyes. He dreams about people scolding him about indecency, impropriety, and the way things are done. Afterward, he looks more like a monkey than a man, but he seems to enjoy it.
The serum only works for a few minutes, and when Poole asks about hearing a strange voice, Jekyll says that it was his friend, Mr. Hyde that was visiting. Later, Muriel says that she and her father are going away for some time, and Jekyll will have to wait many weeks before seeing her again.
Jekyll makes more of his formula and downs it once again. He goes looking for Ivy and cackles like a madman. He's an ugly monster, but he's confident, well dressed, and has money, so he's OK in public. He goes to the music hall and watches the show. He runs into Ivy there, and the two sort of hit it off. He's lacking all the social niceties that Jekyll resents so much, "I'm no gentleman, no beauty, but I have money!" She soon decides he's too creepy and wants to leave, but he's not having it.
Hyde buys Ivy an apartment and keeps her there. He abuses her, but she's well fed and it's a nice place. He terrorizes her, but at least he's having a good time. He's pretty awful, but she has nowhere else to go.
Muriel and the general return to town, and Jekyll returns to himself. He sends fifty pounds to Ivy. Muriel is angry because Jekyll never answered her letters. He wasn't quite himself; he was ill, he explains. The two of them finally get the general to allow them to marry in a month. Jekyll's happy, so he plays more Bach on the pipe organ.
And then Ivy shows up, wanting to thank Jekyll for the money he sent. She complains that Hyde keeps beating her and asks for poison. She's do … anything for his help. Jekyll says Hyde will never come back, but she doesn't believe it.
He walks to the party where his marriage is going to be announced. He stops to watch a bird being eaten by a cat. The shock makes him change into Hyde without drinking anything.
Back at Ivy's place, Ivy celebrates never seeing Hyde again. Hyde comes in, and her celebrations are short-lived; so is she. Soon, the police are out looking for Hyde. Poole won't let him in the front door, and Jekyll threw away the key to the back door. Jekyll never shows up to his party, and the general is not pleased. "I forbid you to see this man again!"
Hyde sends a messenger to Lanyon, asking for help. Lanyon goes to his lab to get the chemicals the note asks for, but won't give up the drugs without seeing Dr. Jekyll. Hyde swears him to secrecy and drinks the potion; he changes back to Jekyll, who explains everything. Lanyon says it's going to happen again and again; it's conquered him already. He promises to release Muriel from their engagement.
Jekyll goes over to the general's house to break things off with Muriel. The general is not so forgiving this time. He apologizes and makes a speech about how evil he has been.
On the way out, he changes into Hyde once more. He breaks back in and attacks the butler and the general. He beats the general to death with his cane. The cops show up, and Hyde is on the run again. Lanyon investigates the crime scene and tells the cops about Jekyll.
The police chase Hyde to Jekyll's lab, and Jekyll tells them Hyde just went out the back way. Lanyon shows up and sets the policemen straight about who they're there for. After a short fight, one of them shoots Hyde.
This was the first horror film to win an Academy Award, three of them in fact. Best actor, best adapted screenplay, and best cinematography black-and-white.
There are a lot of point-of-view shots here as if the camera takes the place of one of the characters. We've seen this done hundreds of times, but this may be one of the earliest occurrences.
The actual transformation scene is a classic, and it was all done in one take. He really looks like he changes into a different person, but it was all done with makeup and lighting. He ends up looking like a cross between Jerry Lewis and a chimpanzee, but it's a good transformation at least.
It's pretty long, but it never drags; there's always something new going on.
Really classic!
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Buying The Cyberiad by Stanislaw Lem broke one of my rules - never acquire more than one book by an unread author. That's the tipping point, in my mind, when the stacks begin to grow for no reason whatsoever. That and the oh-so-dangerous sales at Hebrew Book Week. But something about The Cyberiad made me forget my rule, even though Solaris is still on my shelf, waiting for me. First was the fact that it was available at Border's going-out-of-business sale (bizarre that Borders had it in the first place, awesome that it made its way to my hands). Then the second reason nailed it home: the first story entertained the crap out of me.
The Cyberiad is a collection of short stories. Sort of. I think. Because it looks like the characters are going to be consistent throughout. But these are definitely stories, individual bites of brilliance. Even from the first pages, it's clear to me that Lem has a sharp, wonderful mind. It's everything I love about old-school science fiction - the wit, the intelligence, the quick drama. Quite refreshing.
I'm not going to read The Cyberiad any time soon. First Solaris (after all, I did buy it a few months earlier) and then I'll be able to devote my attention to these stories. And knowing my flightiness and general impatience with short stories, I wouldn't be surprised if I read the book sporadically and in a most disorganized manner. Still, I'm pleased I broke my rule for once - I'd rather like to reread the opening story - How the World was Saved. That's already worth it.
Coincidentally, I fingered a copy of The Cyberiad this weekend myself, and considered buying it but didn't, because it violated one of my own rules (I have a book of his unfinished book [but for a very good reason]). I probably would've bought it if it were on sale though.
Reading Lem is at times more like reading a philophy text than a story (which is fine by me). I highly recommend His Master's Voice. I haven't read any of the short stories — I look forward to your opinion.
Is that even a real name?


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[ARCHIVED] City of Rifle Uses Technology to Engage with Public on Best Allocation of American Rescue Plan Funds
The City of Rifle Uses Technology to Increase Public Engagement on Best Allocation of American Rescue Plan Funds
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The City of Rifle has additional funds this year and wants citizen input on how to spend that money.
The American Rescue Plan Act of 2021 (ARPA) was a $1.9 trillion-dollar economic stimulus bill passed last year in an effort to speed recovery from the Covid-19 pandemic. That money was channeled into numerous funds including the Coronavirus State and Local Fiscal Recovery Funds (SLFRF). The City of Rifle was awarded $2.4 million dollars of SLFRF money which may be utilized to support government services.
City staff has discussed the potential allocation of these funds with projects ranging from supplementing necessary expenditures to spending on items that otherwise might not be funded. The question of whether to use these one-time funds to supplement the budget for existing costs or for the purchase of something that would not otherwise be possible is the issue being presented to citizens through Balancing Act.
Balancing Act is a tool for engaging the public in various types of spending scenarios including the annual budget. Balancing Act can target specific categories such as this grant funding, recreation, or economic development, or can present complex scenarios like an annual budget. Possible projects with dollar amounts assigned will be listed. Final decisions will be made by City Council. Staff is hopeful this simulation will inform and engage citizens and be demonstrative of the difficulty staff and elected officials experience meeting community needs and providing necessary services with limited resources.
The City of Rifle has until 2026 to spend the funds, but all monies must be committed to projects by the end of 2024. This will allow ample time for the evaluation of options and analysis of citizen input.
Balancing Act will be available on all City platforms including the website, rifleco.org, and social media sites Facebook and Twitter. QR codes will also be posted around town to make access quick and easy. The site is expected to go live on Friday, March 4, 2022.
For updated information on all City matters, please go to our website, www.rifleco.org or our Facebook and Twitter pages.
Other News in City News & Announcements
Rifle Community Foundation grants available for Community Assistance Programs
City of Rifle Downtown Improvement Project Garners Two Awards
Changes in Utility Billing & Cemetery Business Operations