Prove that : an operator $\ell$ satisfying, $\ell(f)\ge 0$ whenever $f\ge 0$ is bounded on $C_b(X)$ .

Question

While reading some functional analysis note I came across the following theorem.

Riesz-Markov: (for linear forms on Wiener spaces) If $X$ is locally compact Hausdorff space and $\ell : C_b(X)\to \Bbb R. $ is a linear and continuous form satisfying $\ell(f)\ge 0$ whenever $f\ge 0$. Then there exists a unique Borel measure $\mu$ on $X$ such that $$\ell(f) = \int_X f d\mu, ~~~~\forall~~f\in C_b(X).$$ Where $C_b(X)$ is the space of bounded functions on $X.$

The document says the following statement: Such operators $\ell$ satisfying: $\ell(f)\ge 0$ whenever $f\ge 0$ is automatically bounded.

How to prove that $\ell$ is bounded on $C_b(X)$. I though it could be a good idea to share this on MSE.

For anyone looking, a much more general statement holds, and there are also proofs for linear maps on C$^*$-algebras. — Martin Argerami, Dec 03 '17 at 00:47
Guy Fsone, I will point out that there is a new thread on meta which seems to be related to your activity on the site: Why so many Riesz Representation questions in such short time? — Martin Sleziak, Dec 05 '17 at 05:09

score 1 · Accepted Answer · answered Dec 02 '17 at 15:40

In fact: we have $$\|f\|_\infty\pm f\ge 0\implies \ell(1)\|f\|_\infty\pm \ell(f) \overset{\text{linearity}}{=} \ell(\|f\|_\infty)\pm \ell(f)\overset{\text{linearity}}{=}\ell(\|f\|_\infty\pm f) \ge 0$$

That is for all $f\in C_b(X)$ we have, $$ |\ell(f)| = \pm\ell(f) \le \ell(1)\|f\|_\infty.$$ this prove the continuity of $\ell$ and hence $\ell \in (C_b(X))^*$

Prove that : an operator $\ell$ satisfying, $\ell(f)\ge 0$ whenever $f\ge 0$ is bounded on $C_b(X)$ .

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