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Let be E a compact topologycal space. I know that $C(E)\subset B(E)$ with $C(E)=\{f:E\rightarrow \mathbb C \quad$continuous$\}$ and $B(E)=\{f: E\rightarrow \mathbb C \quad$continuous and bounded$ \}$.

$B(E)$ is a Banach space so $C(E)$ is a Banach space because subset of a Banach space?

Giulia B.
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  • Subset of a Banach space is not necessarily a Banach space. – WhatsUp Mar 23 '21 at 22:35
  • and if is it a closed subset? – Giulia B. Mar 23 '21 at 22:36
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    Yes, if it is a closed sub-vector space, then the induced norm gives it a Banach space structure. However, usually it should be $B(E) \subseteq C(E)$. And in your case, since $E$ is compact, any continuous function is necessarily bounded, hence one has $B(E) = C(E)$. – WhatsUp Mar 23 '21 at 22:38

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You can verify, without too many difficulties, that $C(E)$ is a vector space (just like $B(E)$), according to the definition of vector space. Furthermore, with the supremum norm, $C(E)$ is also a normed vector space (as $B(E)$). As $\mathbb{C}$ is a Banach space, with the norm sup $C(E)$ it is also a Banach space (just like $B(E)$ is also).

In addition, you can prove the following proposition (with a 3-$\epsilon$ argument, like in tis post):

Proposition. The $B(E)$ space is a closed subspace of $C(E)$, that is, the uniform limit of continuous bounded functions is continuous.

As the E space is compact, so the reciprocal of the proposition is valid, in the sense that we also have $C(E)\subset B(E)$.

Ozni
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