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Let $X$ be a locally compact Hausdorff space. What kinds of inner product or norm or metric or topology are defined on $C_0(X)$ and on $C_c(X)$ for their Riesz representation theorems respectively, so that we can talk about positive linear functionals on $C_c(X)$ and continuous linear functionals on $C_0(X)$?

Wikipedia says for the relations between the two representation theorems for $C_c(X)$ and $C_0(X)$

Remark. One might expect that by the Hahn-Banach theorem for bounded linear functionals, every bounded linear functional on $C_c(X)$ extends in exactly one way to a bounded linear functional on $C_0(X)$, the latter being the closure of $C_c(X)$ in the supremum norm, and that for this reason the first statement implies the second. However the first result is for positive linear functionals, not bounded linear functionals, so the two facts are not equivalent.

In fact, a bounded linear functional on $C_c(X)$ need not remain so if the locally convex topology on $C_c(X)$ is replaced by the supremum norm, the norm of $C_0(X)$. An example is the Lebesgue measure on R, which is bounded $C_c(X)$ but unbounded on $C_0(X)$. This fact can also be seen by observing that the total variation of the Lebesgue measure is infinite.

So is the topology on $C_0(X)$ the supremum norm and its topology, and the topology on $C_c(X)$ the locally convex topology? How is the latter topology defined?

Thanks and regards!

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Unless otherwise stated, the norm on (subspaces of) $BC(X\to \mathbb K)$, the set of bounded continuous $\mathbb K$-valued functions on a Hausdorff space $X$, is always the uniform norm.

At least I can't see any reason why it should default to anything else…

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Thanks! In, "But positivity does not imply continuity. The Lebesgue integral is positive but discontinuous on $C_c(\mathbb R)$. – scineram", "@scineram: I was allowing myself to be a bit sloppy in my answer. Sorry about the confusion. It depends of course on what topology you put on $C_c(\mathbb R)$ and I decided to sweep that under the rug. – t.b." – Tim Feb 21 '13 at 2:57
Hi Kahen, I just updated my post to add more of my confusion. – Tim Feb 21 '13 at 18:04

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