Exercise 7.24

Question

Exercise 24: 
    Let $X$ be a metric space, with metric $d$. Fix a point $a\in X$. Assign to each $p\in X$ the function $f_p$ defined by $$f_p(x)=d(x,p)-d(x,a)\quad\quad(x\in X).$$ Prove that $\big|f_p(x)\big|\le
    d(a,p)$ for all $x\in X$, and that therefore $f_p\in\mathscr C(X)$. Prove that $$||f_p-f_q||=d(p,q)$$ for all $p,q\in X$.

If $\Phi(p)=f_p$ it follows that $\Phi$ is an {\it isometry} (a distance-preserving mapping) of $X$ onto $\Phi(X)$ in $\mathscr C(X)$.

Let $Y$ be the closure of $\Phi(X)$ in $\mathscr C(X)$. Show that $Y$ is complete. {\it Conclusion: $X$ is isometric to a dense subset of a complete metric space $Y$.}

analambanomenos · Accepted Answer

\def\eps{\varepsilon}

Note that the triangle inequality gives us, for all $x\in X$, $y\in X$, and $z\in X$,
    \begin{align*}
        d(x,z)-d(x,y) &\le d(y,z) \
        d(x,y)-d(x,z) &\le d(z,y)=d(y,z),
    \end{align*}
    so that $$\big|d(x,z)-d(x,y)\big|\le d(y,z).$$
    Hence, for all $x\in X$, $$\big|f_p(x)\big|=\big|d(x,p)-d(x,a)\big|\le d(a,p).$$

To show that $f_p$ is continuous, let $\varepsilon>0$ and let $x\in X$ and $y\in X$ such that $d(x,y)<\delta=\varepsilon/2$. Then
    \begin{align*}
        \big|f_p(x)-f_p(y)\big| &= \big|d(x,p)-d(x,a)-d(y,p)+d(y,a)\big| \
        &\le \big|d(x,p)-d(y,p)\big|+\big|d(y,a)-d(x,a)\big| \
        &\le d(x,y)+d(x,y) \
        &\le \varepsilon.
    \end{align*}
    (This shows that $f_p$ is uniformly continuous.)

Note that if $p\in E$ and $q\in E$, then for all $x\in E$ we have $$f_p(x)-f_q(x) = d(x,p)-d(x,a)-d(x,p)+d(x,a) = d(x,p)-d(x,q)$$ so that 
    \begin{align*}
        ||f_p-f_q|| &= \sup_{x\in E}\big|f_p(x)-f_q(x)\big| \
        &= \sup_{x\in E}\big|d(x,p)-d(x,q)\big| \
        &\le d(p,q).
    \end{align*}
    And since $f_p(q)-f_q(q)=d(p,q)$, we have $$||f_p-f_q||=d(p,q)$$ for all $p,q\in X$.

By Theorem 7.15, $\mathscr C(X)$ with the uniform convergence metric is complete. Since it is clear that any closed subset of a complete metric space is also complete (if a sequence of
    elements in the closed subset satisfies the Cauchy condition, then it must converge to an element of the complete metric space, which must be an element of the closed subset since it is
    closed), we see that the closure $Y$ if $\Phi(X)$ is complete.

Exercise 7.24

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Exercise 7.24

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