<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Consider the function function $f: {\bf H}\times {\bf H}
\rightarrow {\bf R}$ defined by
\[ f(x,y) = \frac{{\rm d}_{\bf H}(x,y)}{1 +{\rm d}_{\bf H}(x,y)}. \]

\medskip
\noindent Prove that $f$ is a metric on ${\bf H}$.  Also, prove
that the {\em diameter} of ${\bf H}$ with the metric $f$ is
finite, where the diameter ${\rm diam}({\bf H}, f)$ of the metric
space $({\bf H}, f)$ is defined by
\[ {\rm diam}({\bf H}, f) = \sup\{ f(x,y)\: |\: x,\: y\in {\bf H}\}. \]

\medskip

{\bf Answer}

Note that the first two conditions of a metric, that $f(x,y) \geq
0$ with equality if and only if $x=y$ and that $f(x,y)=f(y,x)$,
are satisfied since they hold true for $d_{\bf{H}}(\cdot, \cdot)$.
To check the triangle inequality, assume that it fails for
$f(\cdot, \cdot)$, so that there are parts $x,y,z$, so that
$f(x,y)&gt;f(x,y)+f(y,z)$. Then,

$$\ds\frac
{d_{\bf{H}}(x,z)}{1+d_{\bf{H}}(x,z)}&gt;\ds\frac{d_{\bf{H}}(x,y)}
{1+d_{\bf{H}}(x,y)}+\ds\frac{d_{\bf{H}}(y,z)}{1+d_{\bf{H}}(y,z)}$$

$d_{\bf{H}}(x,z)(1+d_{\bf{H}}(x,y))(1+d_{\bf{H}}(y,z))&gt;$

$d_{\bf{H}}(x,y)(1+d_{\bf{H}}(x,z))(1+d_{\bf{H}}(y,z))+
d_{\bf{H}}(y,z)(1+d_{\bf{H}}(x,z))(1+d_{\bf{H}}(x,y))$

\bigskip
\un{Simplifying}:

\bigskip
$d_{\bf{H}}(x,z)+d_{\bf{H}}(x,z)d_{\bf{H}}(x,y)+d_{\bf{H}}(x,z)d_{\bf{H}}(y,z)&gt;$

\bigskip
$d_{\bf{H}}(x,y)+d_{\bf{H}}(x,y)d_{\bf{H}}(x,z)+d_{\bf{H}}(x,y)d_{\bf{H}}(y,z)$

\bigskip
$+d_{\bf{H}}(y,z)+d_{\bf{H}}(y,z)d_{\bf{H}}(x,z)+d_{\bf{H}}(y,z)d_{\bf{H}}(x,y)$

\bigskip
$+d_{\bf{H}}(x,y)d_{\bf{H}}(x,z)d_{\bf{H}}(y,z)$

\bigskip
\un{Thus}:

\bigskip
$d_{\bf{H}}(x,z)&gt;d_{\bf{H}}(x,y)+d_{\bf{H}}(y,z)+$ stuff
$&gt;d_{\bf{H}}(x,y)+d_{\bf{H}}(y,z)$

\bigskip
But this is a contradiction, since $d_{\bf{H}}(\cdot, \cdot)$ is a
metric. Hence, $f(\cdot, \cdot)$ satisfies the triangle inequality
and hence is a metric.

\bigskip
\begin{eqnarray*} {\rm{diam}}({\bf{H}},f) &amp; = &amp;
{\rm{sup}}\{f(x,y)|x,y \in {\bf{H}}\}\\ &amp; = &amp;
{\rm{sup}}\left\{\left.\ds\frac{d_{\bf{H}}(x,y)}{1+d_{\bf{H}}(x,y)}\right|x,y
\in{\bf{H}}\right\} \end{eqnarray*}

\bigskip
Note that $\ds\frac{d_{\bf{H}}(x,y)}{1+d_{\bf{H}}(x,y)}&lt;1$ for all
$x,y \in {\bf{H}}$. Moreover,

\bigskip
$f(i,\lambda i)=\ds\frac{d_{\bf{H}}(i,\lambda
i)}{1+d_{\bf{H}}(i,\lambda
i)}=\ds\frac{\ln(\lambda)}{1+\ln(\lambda)} \to 1$ as $\lambda \to
\infty$,

\bigskip
and so diam$({\bf{H}},f)=1$.
\end{document}
</pre></body></html>