# the space of continuous functions is complete

Let $$(X, d_X)$$ be a metric space, and let $$(Y, d_Y)$$ be a complete metric space. The space $$(C(X \to Y), d_{\infty})$$ is a complete subspace of $$(B(X \to Y), d_\infty)$$. In other words, every Cauchy sequence of functions in $$C(X \to Y)$$ converges to a function in $$C(X \to Y)$$. ($$C(X \to Y)$$ is the space of continuous functions; $$B(X \to Y)$$ is the space of bounded functions; $$d_\infty$$ is a supnorm metric)

We have a complete metric space $$(Y, d_Y)$$. For each $$x \in X$$, let $$(f_n(x))_{n=1}^\infty$$ be any Cauchy sequence in $$C(X \to Y)$$, and it converges to $$f(x)$$ (i.e., for every $$\epsilon >0$$, there exists $$N$$ s.t. for $$n \ge N$$, $$d_Y(f_n(x), f(x)) < \epsilon$$).

I have the result that $$C(X \to Y)$$ is closed in $$B(X \to Y)$$. Thus, if I show that $$f_n$$ converges to $$f$$ with respect to $$d_\infty$$-metric, $$f$$ becomes automatically in $$C(X \to Y)$$.

The book suggests to use triangular inequality but I am not sure how to use this. I appreciate if you give me some help.

I will use the notation $$C$$ for $$(C(X\to Y), d_{\infty})$$. Let $$\{f_n\}$$ be a Cauchy sequence in $$C$$.

Then for each $$\epsilon \gt 0$$, there is $$N\in \mathbb Z^+$$ s.t. for all $$m,n\gt N$$, $$||f_n-f_m||_{\infty} \lt \epsilon$$.

Thus for all $$x\in X$$, $$\{f_n(x)\}$$ is a Cauchy sequence in $$Y$$. Since $$Y$$ is complete define $$f:X\to Y$$ as

$$f(x)=\lim\limits_{n\to\infty}f_n(x)$$. Then clearly $$f_n$$ converges to $$f$$.

The only thing we have to show is the continuity of $$f$$.

Let $$t\in X$$ and let $$\epsilon\gt0$$.

Since $$f(t)=\lim\limits_{n\to\infty}f_n(t)$$ there is $$N\in \mathbb Z^+$$ s.t for all $$n\gt N$$, $$||f_n-f||_{\infty}\lt\frac{\epsilon}{3} \qquad (*)$$

Choose $$n_0\gt N$$. Then since $$f_{n_0}$$ is continious there exist $$\delta\gt0$$ s.t for all $$x\in X$$ $$|x-t|\lt \delta\implies |f_{n_0}(x)-f_{n_0}(t)|\lt \frac{\epsilon}{3} \qquad (**)$$

then by $$(*),(**)$$

for all $$x\in X$$, if $$|x-t|\lt \delta$$ then

$$\begin{split}|f(x)-f(t)|& =|f(x)-f_{n_0}(x)+f_{n_0}(x)-f_{n_0}(t)+f_{n_0}(t)-f(t)|\\ &\lt |f(x)-f_{n_0}(x)|+|f_{n_0}(x)-f_{n_0}(t)|+|f_{n_0}(t)-f(t)|\\ &\lt \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3} =\epsilon\\ \end{split}$$.

Thus $$f$$ is continuous.

• Thanks for the answer. In my book, $d_{\infty}(f_n, f) = \sup \{d_Y(f_n(x), f(x)) : x \in X\}$, but in $(*)$, it seems that you write it as $d_{\infty} (f_n(t), f(t))$. Do we have such $d_\infty$ definition? – shk910 Mar 25 '20 at 11:30
• @shk910 edited the answer. thare is no deffirence – Gune Mar 25 '20 at 17:37
• I am still confusing with $(*)$. Why $f(t) = \lim_{n\to \infty} f_n(t) \implies ||f_n - f||_\infty<\epsilon/3$? I think it should be the opposite. – shk910 Mar 25 '20 at 22:13