I have a question about a continuous linear functional.
$T>0$ : fix.
$C([0,T]):=\{w:[0,T]\to \mathbb{R}\,;\, w \,{\rm is\,conti.} \}$
$C_{0}([0,T]):=\{w \in C([0,T]) \,; \,w(0)=0 \}$
Then $C([0,T]),C_{0}([0,T])$ is Banach space with supremum norm $\|w\|=\sup_{0 \leq t \leq T}|w(t)|$
Let $\varphi $ is a continuous linear functional on $C_{0}([0,T])$. i.e. $\varphi\in C_{0}([0,T])^{*}$
Can $\varphi$ be represent as a limit of functionals of the form $\sum_{i=1}^{n}\xi_{i}w(t_{i})$ ?
Here $\xi_{1},\cdots\xi_{n} \in \mathbb{R},t_{1},\cdots,t_{n} \in [0,1]$
My idea:
By Riesz theorem, $\varphi$ is realized by an integral with respect to an certain finite signed measure on $[0,1]$
i.e. $\varphi(w)=\int_{[0,1]}w(s) \mu (ds)\cdots(1)$
If the right side of the equation $(1)$ is represent as a limit of Riemann sum, I think $\varphi$ is represent as a limit of functionals of the form $\sum_{i=1}^{n}\xi_{i}w(t_{i})$. Is this possible?
Yes, $\varphi$ can be approximated by such sums.
First of all, $\varphi(f)=\int_{[0,1]}f\,d\mu,$ where $\mu\in\mathscr M[0,1]$ - the set of signed or complex Borel measures on $[0,1]$.
Next, let $n\in\mathbb N$, $\,P=\Big\{t_k=\frac{k}{n}: k=0,1,\ldots,n\Big\}$, and $$ w_k=\mu\big((t_{k-1},t_k]\big),\,\, k=1,\ldots,n. $$ Then $$ \varphi(f)-\sum_{k=1}^n f(t_k)w_k=\varphi(f)-\varphi_n(f)=\sum_{k=1}^n \int_{(t_{k-1},t_k]} \big(f(x)-f(t_k)\big)\,d\mu(x), $$ and due to the uniform continuity of $f$, then for every $\varepsilon$, there exists an $n$ sufficiently large, such that $$ \lvert \varphi(f)-\varphi_n(f)\rvert \le\varepsilon \|\mu\|. $$ This shows that $\varphi_n\to\varphi$, weakly.