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Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces. Given $\emptyset \neq B\subseteq Y,$ let $\mathrm{diam}_Y(B) = \sup_{y_1, y_2\in B} d_Y(y_1, y_2).$ Let $C_b(X,Y) = \{f\in Y^X : \text{ f is continuous and } \mathrm{diam}_Y (f(X)) < \infty\}$ where $f(X) $ is the range of $f.$

  1. Show that $d_C : C_b(X,Y) \times C_b(X,Y) \to [0,\infty)$ given by $d_C(f,g) = \sup_{x\in X}d_Y (f(x), g(x))$ defines a metric on $C_b(X,Y).$
  2. Show that if $(Y,d_Y)$ is complete, then $(C_b(X,Y) , d_C)$ is also complete.

$1$ seems fairly straightforward; most of the properties of a metric (e.g. positive definiteness, symmetry, etc.) are fairly easy to verify. Also, the triangle inequality follows from properties of the supremum.

But I'm not sure how to justify why $d_C(f,g) < \infty$ for all $f,g$. I tried to come up with a contradiction, but it seems that that method isn't working too well.

For $2,$ I think the proof is similar to the one that $C([0,1])$ is a Banach space. So I need to show that every Cauchy sequence $(f_n)\subseteq C_b(X,Y)$ converges to some $f\in C_b(X,Y)$. Let $\epsilon>0.$ It'll be useful to use the fact that $(f_n(x))$ is Cauchy for each $x.$ So since $Y$ is complete, we can define $f(x):= \lim_{n} f_n(x)$ for each $x$. By the Cauchyness of $(f_n),\exists N\in\mathbb{N}$ so that $n,m\geq N$ implies $d_C(f_n, f_m) < \frac{\epsilon}3$. Then for $n\geq N, d_Y(f(x), f_n(x)) = \lim_{m\to \infty}d_Y(f_m(x), f_n(x)) \leq \lim_{m\to\infty} d_C(f_m, f_n) \leq \frac{\epsilon}3 <\frac{\epsilon}2 $.

Hence $d_C(f, f_n) \leq \frac{\epsilon}2 < \epsilon$ for $n\geq N$ so $f_n \to f.$ Also, $\mathrm{diam}_Y(f(X)) < \infty$ as $\sup_{x,y\in X}d_Y(f(x), f(y)) \leq \sup_{x\in X} d_Y(f_N(x), f(x)) + \sup_{x,y \in X}d_Y(f_N(x), f_N(y)) + \sup_{y\in X} d_Y(f_N(y), f(y)) = 2d_C(f_N, f) + \mathrm{diam}_Y(f_N(X)) < \infty$ and $f$ can be shown to be continuous as $f_n$ converges uniformly to $f$. Indeed if $(a_n)\subseteq Y, a_n\to x \in Y, $ since $f_n\to f,$ there exists $N_2\in \mathbb{N}$ so that $n\geq N_2$ implies $d_C(f_n, f) < \epsilon/3.$ Also, $\exists N_3\in\mathbb{N}$ so that $n\geq N_3\Rightarrow d_Y(a_n, x) < \epsilon/3.$ Then for $n\geq N, d_Y(f(a_n), f(x)) \leq d_Y(f(a_n), f_{N_2}(a_n)) + d_Y(f_{N_2}(a_n), f_{N_2}(a)) + d_Y(f_{N_2}(a), f_n(a)) < 2d_C(f,f_{N_2}) + d_Y(f_{N_2}(a_n), f_{N_2}(a)) < \epsilon.$ So $f$ is continuous.

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1 Answer 1

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For 1) fix any point $x_0 \in X$.We have $$d_Y(f(x),g(x))$$ $$ \leq d_Y(f(x),f(x_0))+d_Y(f(x_0),g(x_0))+d_Y(g(x_0), g(x))$$ $$ \leq diam (f(X))+d_Y(f(x_0,g(x_0))+diam (g(X))<\infty$$.

Your argument for 2) looks OK.

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