# Riesz representation theorem for $C([0,1])$

i’m trying to prove the special case of Riesz representation theorem: Every positive (non-negative on non-negative functions) linear continuous functional $$\phi$$ on the normed space $$C([0,1])$$ is given by some measure $$\mu$$ by the rule: $$\phi\left(f\right)=\int_{\left[0,1\right]}fd\mu$$

I want to do it with using measure extension theorem: First I need to build $$\mu$$ on elementary sets. But I don't know what it should be like. Can you help me with this? (for open interval, for example)

• Possibly, you could to define $\mu^\star(A) = \phi(1_A)$ where $A$ is an interval on $[0,1]$. Did you try that? – Yanko Dec 9 '18 at 11:09
• it would be great, but $\chi_{A}$ is not continuous ($\notin C([0,1])$) – Ilya Dec 9 '18 at 11:19
• i think $\mu$ has to somehow be consistent with the norm on $C([0,1])$ – Ilya Dec 9 '18 at 11:21
• The idea would be to find a sequence of continuous functions $f_n$ that approximates $1_A$ (say, converging pointwise and boundedly), and then define $\mu(A) := \lim_{n \to \infty} \phi(f_n)$. – Nate Eldredge Dec 9 '18 at 16:51
• @NateEldredge It's already done below. – Rebellos Dec 9 '18 at 16:53

Proving it generally for $$C[a,b]$$ :

First, assume that $$\Gamma$$ is positive. For $$a \leq t < b$$ and for $$n$$ large enough so that $$t + \frac{1}{n} \leq b$$, let :

$$\phi_{t,n}(x) = \begin{cases} 1 & \text{if} \; x \in [a,t] \\ 1- n(x-t) & \text{if} \; x \in (t,t + \frac{1}{n}] \\ 0 & \text{if} \; x \in (t + \frac{1}{n}, b]\end{cases}$$

If $$n \leq m$$, then $$0 \leq \phi_{t,m} \leq \phi_{t,n} \leq 1$$. It follows that $$\{\Gamma(\phi_{t,n})\}$$ is decreasing and bounded below by $$0$$. Therefore, we can define :

$$g(t) = \begin{cases} 0 & \text{if} \; t

Moreover, if $$t_1 > t$$, we have : $$\phi_{t,m} \leq \phi_{t_1,n}$$.

Since $$\Gamma$$ is positive, $$g(t)$$ is monotonically increasing. It is clear that $$g(t)$$ is right continuous if $$t or if $$t\geq b$$. Assume that $$t \in [a,b)$$. Let $$\varepsilon >0$$ and choose $$n$$ large enough so that :

$$n > \max\left(2, \frac{\|\Gamma\|}{\varepsilon}\right)$$

and also that : $$g(t) \leq \Gamma(\phi_{t,n}) \leq g(t) + \varepsilon$$.

Let :

$$\psi_n(x) = \begin{cases} 1 & \text{if} \; x \in [a, t + \frac{1}{n^2}] \\ 1 - \frac{n^2}{n-2}\left(x-t-\frac{1}{n^2}\right) & \text{if} \; x \in (t + \frac{1}{n^2}, t + \frac{1}{n} - \frac{1}{n^2}] \\ 0 & \text{if} \; x \in (t + \frac{1}{n} - \frac{1}{n^2}, b] \end{cases}$$

It then is : $$\| \psi_n - \phi_{t,n}\|_\infty \leq 1/n$$. That means : $$\Gamma(\psi_n) \leq \Gamma(\phi_{t,n}) + \frac{1}{n}\|\Gamma\| \leq g(t) + 2\varepsilon$$

But, this yields that :

$$g(t) \leq g\left(t + \frac{1}{n^2}\right) \leq g(t) + 2 \varepsilon$$

Since $$g(t)$$ is increasing, it is sufficient to show that $$g(t)$$ is right continuous. The Hahn-Banach Extension Theorem gives a Borel measure $$\mu$$ such that $$\mu((\alpha,\beta]) = g(\beta) - g(\alpha)$$. In particular, if it is $$a \leq c \leq b$$, then it is :

$$\mu([a,c]) = \mu((a-1,c]) = g(c)$$

Let $$f \in C([a,b])$$ and let $$\varepsilon >0$$. Let $$\delta$$ be such that if $$|x-y| < \delta$$ and $$x,y \in [a,b]$$, then : $$|f(x) - f(y)| < \varepsilon$$

Now, let $$P =\{a=t_0,t_1,\dots,t_m=b\}$$ be a partition with $$\sup(t_k - t_{k-1}) < \delta/2$$. Then choose $$n$$ to be large enough so that : $$\frac{2}{n} < \inf(t_k-t_{k-1})$$ $$\text{and}$$ $$g(t_k) \leq \Gamma(\phi_{t,n}) \leq g(t_k) + \frac{\varepsilon}{\mu\|f\|_\infty}$$ Next, let : $$f_1(x) = f(t_1) + \phi_{t_1,n} + \sum_{k=1}^m f(t_k)(\phi_{t_k-n} - \phi_{t_{k-1},n})$$ $$\text{and}$$ $$f_2(x) = f(t_1)_{\mathcal{X}[t_0,t_1]} + \sum_{k=2}^m f(t_k)_{\mathcal{X}[t_{k-1},t_k]}$$ It can be seen that $$f_1$$ is continuous and piecewise linear, while $$f_2$$ is a step function. Both $$f_1$$ and $$f_2$$ agree with $$f(x)$$ at each point $$t_k$$ for $$k \geq 1$$. Moreover the function $$f_1$$ takes values between $$f(t_{k-1})$$ and $$f(t_k)$$ on the interval $$[t_{k-1},t_k]$$ of course.

It is :

$$\|f_1-f\|_\infty \leq \varepsilon$$ $$\text{and}$$ $$\sup\{|f_2(x)-f(x)| : x \in [a,b]\} \leq \varepsilon$$

From the above, we can conclude that : $$|\Gamma(f) - \Gamma(f_1)| \leq \varepsilon\|\Gamma\|$$

Now, for $$2\leq k\leq m$$, it is :

$$|\Gamma(\phi_{t_k,n} - \phi_{t_{k-1},n}) - (g(t_k)-g(t_{k-1}))| \leq \frac{\varepsilon}{m\|f\|_\infty}$$

Now, applying $$\Gamma$$ to $$f_1$$ and integrating $$f_2$$ with respect to $$\mu$$ : $$\Bigg| \Gamma(f_1) - \int_{[a,b]} f_2\mathrm{d}\mu \Bigg| \leq \varepsilon$$ But, it also is : $$\int_{[a,b]} f_2\mathrm{d}\mu - \int_{[a,b]} f \mathrm{d}\mu \leq \varepsilon \mu([a,b])$$ Thus : $$\Bigg|\Gamma(f) - \int_{[a,b]}f\mathrm{d}\mu\Bigg| \leq \varepsilon(2 \| \Gamma \| + \mu([a,b])$$ But $$\varepsilon$$ is arbitrary and we can yield :

$$\Gamma(f) = \int_{[a,b]} f \mathrm{d}\mu$$ for every $$f \in C[a,b]$$. It also is $$\|\Gamma\| = \Gamma(1) = |\mu|([a,b])$$.

Now, recall the Jordan Decomposition Theorem, which states that :

Let $$\Gamma \in C([a,b])^*$$. Then there exist positive linear functionals $$\Gamma^+, \Gamma^- \in C([a,b])^*$$ such that : $$\Gamma = \Gamma^+ - \Gamma^-$$ $$\text{and}$$ $$\|\Gamma\| = \Gamma^+(1) + \Gamma^-(1)$$

The general result now follows from that theorem and the proof is completed.