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Prove that : and $\ell$ satisfying: $\ell(f)\ge 0$ whenever $f\ge 0$ is bounded .
While reading some functional analysis note I came across the following theorem.
RieszMarkov: (for linear forms on Wiener spaces) If $X$ is locally compact topological 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.
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)

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 $\ell \in (C_b(X))^*$
20171203 03:47:26