Why is the linear functional $F:C[a,b] \rightarrow \mathbb{R}$ such that $F(f)=\int f d\mu$ continuous?

Question

Let $\mu$ be a finite, Borel measure, and let $[a,b]$ be a closed, bounded interval on $\mathbb{R}$.

Why is the linear functional $F:C[a,b] \rightarrow \mathbb{R}$ such that $F(f)=\int f d\mu$ continuous?

First of all, I see that:

$$|F(f)| = |\int f d\mu| \leq \int |f| d\mu \leq ||f||_{\infty} \mu([a,b]) $$

... where $||f||_{\infty} = max\{|f(x)|:x \in [a,b]\}$, and so this is just the M-L inequality.

It is said that a linear functional is continuous iff it is bounded. However, I don't see why this one is continous, since the above computation is just for a single function $f$, and this isn't a bound for all $f$. Am I missing something?

Thanks in advance!

Answer 1

You proved correctly that $\bigl|F(f)\bigr|\leqslant\|f\|_\infty\mu\bigl([a,b]\bigr)$. Therefore, $\|f\|_\infty=1\implies\bigl|F(f)\bigr|\leqslant\mu\bigl([a,b]\bigr)$. So, $F$ is a bounded linear map.

Answer 2

The operator norm is $$ \sup_{f\neq 0}\frac{|F(f)|}{\lVert f\rVert_{\infty}} $$ which is bounded by $\mu[a,b]$