# Complete metric space of bounded functions

Consider the metric spaces $(X,d_{X})$ and $(Y,d_{Y})$. We call the function $f:X \rightarrow Y$ bounded if $f(X)$ is bounded in $Y$. Consider $C_{b}(X,Y)= \{f:X \rightarrow Y |f$ is continuous and bounded$\}$ with the metric $d_{\infty}(f,g)= \sup \{d_{Y}(f(x),g(x)) | x \in X \}$. Show that $(C_{b}(X,Y),d_{\infty})$ is complete if $(Y,d_{Y})$ is complete.

Now I started writing out a proof, but I thought that maybe it could be shortened since we proved this property earlier in our class: "$(C_{b}(X, \mathbb{R}),d_{\infty})$ is complete with the Euclidian metric on $\mathbb{R}$". Is there then a simple method to generalize this for every complete metric space $(Y,d_{Y})$?

• Do you mean use the result for $C_b(X, \mathbb{R})$ to prove that $C_b(X,Y)$ is complete? I'm not sure if that's possible. But it's not so hard to carry over the proof to $C_b(X, Y)$. You can probably copy it almost verbatim. Jun 18 '17 at 12:18
• Yes that is what I meant indeed. And okay that's too bad. Thank you for your answer.
– simp
Jun 18 '17 at 13:01

## 1 Answer

You can actually use the real-valued result for a part of the proof, marked as Step 2 below. It does not make the proof a whole lot shorter, but perhaps saves some repetition.

### Step 1

A Cauchy sequence $\{f_n\}$ in $C_b(X,Y)$ is pointwise Cauchy, hence (by the completeness of $Y$) is pointwise convergent. Call its limit $f$.

### Step 2

We must show that $f_n\to f$ in $C_b(X,Y)$. To do this, consider the functions $g_n(x)=d(f_n(x),f(x))$ which are real-valued. By the triangle inequality, $d_\infty(g_n, g_m)\le d_\infty(f_n, f_m)$, so $\{g_n\}$ is Cauchy. Therefore, $g_n$ converge uniformly to some $g$. But we already know that $g_n\to 0$ pointwise. So $g\equiv 0$, which means $g_n\to 0$ uniformly, which means $f_n\to f$ uniformly.