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Let $(X,d_{X})$ be a metric space, and let $(Y,d_{Y})$ be a complete metric space. The subspace $(C(X\to Y),d_{\infty})$ 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 $C(X\to Y)$.

MY ATTEMPT

Here, $B(X\to Y)$ represents the set of bounded functions from $X$ to $Y$, $C(X\to Y)$ is the set of bounded and continuous functions from $X$ to $Y$ and $d_{\infty}$ is the sup norm distance.

I was thinking about proving that $B(X\to Y)$ is complete, whence we conclude that $C(X\to Y)$ is complete, since it is already closed.

However I am not able to do so. Could someone provide a proof based on such line of reasoning? Any other approach is welcome as well.

Sahiba Arora
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user0102
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    The proof can be done as in https://math.stackexchange.com/a/168167/450140 just replacing $\mathbb{R}$ with $(Y,\operatorname{d}_Y)$ as only the completeness is needed. – Jonas Lenz May 19 '20 at 19:20
  • @SahibaArora Thanks for the reference, but I am not restricting $f_{n}$ to be a linear operator. Thus I don' think it answers my queston. – user0102 May 19 '20 at 19:20

1 Answers1

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Yes, it works to show that $B(X, Y)$ is complete and $C(X, Y)$ is a closed subset of $B(X, Y)$.

Let $\epsilon > 0$ and let $(f_{n})_{n = 1}^{\infty}$ be a Cauchy sequence of functions in $B(X \to Y)$. For all $x \in X$ and $m, n \geq N$, $$d_{Y}(f_{m}(x), \: f_{n}(x)) \leq \sup\{d_{Y}(f_{m}(x), \: f_{n}(x)) : x \in X\} = d_{\infty}(f_{m}, \: f_{n}) < \frac{\epsilon}{2}$$

Thus, for each fixed $x \in X$, the sequence $(f_{n}(x))_{n = 1}^{\infty}$ is a Cauchy sequence, which converges to some $y_{x} \in Y$. Define a function $f: X \to Y$ such that $f(x) = y_{x}$ for all $x \in X$.

Now recall that $d_{Y}(f_{m}(x), \: f_{n}(x)) < \frac{\epsilon}{2}$ for all $m, \: n \geq N$ and $x \in X$. Taking limits $m \to \infty$ gives $$ \lim_{m \to \infty} \: d_{Y}(f_{m}(x), \: f_{n}(x)) = d_{Y}(\lim_{m \to \infty} \: f_{m}(x), \: f_{n}(x)) = d_{Y}(f(x), \: f_{n}(x))\leq \frac{\epsilon}{2} < \epsilon$$ for all $n \geq N$ and $x \in X$. Taking suprema in $x \in X$, it follows that $$\sup\{d_{Y}(f_{n}(x), \: f(x)) \: : \: x \in X\} = d_{\infty}(f_{n}, \: f) \leq \frac{\epsilon}{2} < \epsilon$$ for $n \geq N$.

user156205
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