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Suppose we have a pseudocompact, Hausdorff space $L$ (pseudocompact means that each continuous function $f\colon L\to \mathbb{R}$ is bounded). Consider the space $C(L)$ of continuous real-valued functions on $L$. It seems that it is a Banach space (the proof for compact spaces should carry on).

Assuming $L$ is locally compact, is $C(L)$ isomorphic to $C_0(L)$, where $C_0(L)$ stands for the Banach space of continuous functions vanishing at infinity?

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1 Answer 1

Edit: This answers a previous version of the question.

$C(L)$ is not equal to $C_0(L)$. For example, $C(L)$ contains the constant function 1, which is not in $C_0(L)$ unless $L$ is compact.

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Thank you. Apologies for my unclearly posed question. I understand the equality between Banach spaces isomorphically. Yet, I find your answer interesting for me and possibly other readers. –  Vitalis V. Jul 22 '12 at 17:57
@VitalisV.: I'll edit your question to clarify this. If you register your account, you'll be able to do this yourself in the future. –  Nate Eldredge Jul 22 '12 at 18:05
There can be other examples besides @NateEldredge 's constant functions. Suppose $L$ is a noncompact pseudocompact Hausdorff space. Then so is $M = L \sqcup L$, the disjoint union of two copies of $L$, and on $M$ we can have a continuous function that is $1$ on the first copy of $L$, $0$ on the second, so that this function has no limit at infinity. –  Robert Israel Jul 22 '12 at 18:06

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