Show that $\exists L \in (l^\infty)'$ such that $\forall f \in C([0,1]): \int\limits_{0}^{1}f(t)dt = L(T(f))$.

Let $S = \{ s_1, s_2,... \}$ be a countable dense set in $[0,1]$. Define the linear map $T: C([0,1]) \to l^{\infty}$ as $T(f) = (f(s_1), f(s_2), ...)$. I already solved the first part of this problem in Show that $T$ is isometric and not surjective. Proof validation., where I showed that $T$ is isometric and not surjective. But the following has me completely lost:

Problem

Show that $\exists L \in (l^\infty)'$ such that $\forall f \in C([0,1]): \int\limits_{0}^{1}f(t)dt = L(T(f))$.

Hint: Define first such $L$ on the range of $T$.

My Thoughts

Using the hint, for each $y \in T(C([0,1]))$, let $f_y$ denote the function that generates it, i.e. $T(f_y) = y = (f_y(s_1), f_y(s_2), ....)$. I can do this, i.e. $y \to f_y$ is bijective, since an isometry is injective. Now I need to construct a linear and continuous $L$ such that $L(y) = \int f_y(t)dt$. The only thing that comes to my mind is a Riemann sum. But it gets really strange.

Question

Seing $S$ as sequence, can I construct a sorted sequence $S_s = (t_1, t_2,...)$ such that $\forall i: t_i < t_{i+1}?$. In the sense that $s \in S \iff s \in S_s$. I've never thought of this before. If it is possible, then if $S = \mathbb{Q}\cap [0,1]$, which is countable and dense, what is the value of $t_2 - t_1$? If it is not possible, why not?

Assuming that such $S_s$ exists, we can define $L$ as: $L(y) = \large\sum\limits_{n\ge 1} \frac{f_y(t_i)}{t_{i+1}- t_i}$. Is it worth continuing from here? I'm very skeptic.

Solution 1:

Define $l \colon \operatorname{range}T \to \Bbb R$ as follows:

If $s \in \operatorname{range}T$, pick $f \in C([0,1])$ such that $T(f) = s$, and write $l(s) := \int_0^1 f(t)dt$.

Since $T$ is injective, for each $s \in \operatorname{range}T$ there is a unique $f \in C([0,1])$ such that $T(f) = s$; meaning that $l$ is well-defined.

Exercise: Prove that $l \in (\operatorname{range}T)'$, in other words, $l$ is linear and continuous.

Now, by the Hahn-Banach theorem we know that there exists $L \in (\ell^\infty)'$ that extends $l$, so that $$\forall f \in C([0,1]), \quad L(T(f)) = l(T(f)) = \int_0^1 f(t)dt.$$