If $J=[0,T]\subset \mathbb{R}$, and $X$ is a Banach space? Then, is $C[J,X]$ (Banach space of all continuous functions from $J$ to $X$) a compact metric space with respect to supnorm defined as $\|x\|_{\infty}=\sup \{x(t): t\in[0,T]\}$?
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1$\begingroup$ I don't think so. Any compact metric space is separable, and $C[J,X]$ is not necessarily separable. $\endgroup$– OscarCommented Mar 8 at 17:02
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1$\begingroup$ Its not even true for $J=[0,1]$ and $X=\mathbb{R}$: The sequence $f_n=x^n$ is bounded by $||f_n|| \leq 1$, but has no convergent subsequence. $\endgroup$– F. ConradCommented Mar 8 at 17:11
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2$\begingroup$ No nontrivial normed space is compact. So except in the trivial situation where $X = \{0\}$, the space you wrote down is never compact. Though, of course, it is a Banach space, so it’s complete. $\endgroup$– David GaoCommented Mar 8 at 17:16
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2 Answers
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The only normed space that is compact is $\{0\}$. Any other Banach space is unbounded (as soon as $x\ne0$, $\|nx\|$ can be made arbitrarily large for $n\in\mathbb N$), so not compact.
answered Mar 8 at 18:05
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No unbounded metric space can be compact. In particular $C(X,Y)$ is not compact when $Y$ is not bounded. Just pick a sequence of constant functions, with unbounded values. And every normed space is unbounded, except for the zero space.
But even if both $X$ and $Y$ are compact, $C(X,Y)$ does not have to be. Consider $X=Y=[0,1]$ and the sequence $f_n(t)=t^n$.
