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)

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

$$g(t) = \begin{cases} 0 & \text{if} \; t<a \\ \lim_{n \to \infty} \Gamma(\phi_{t,n}) & \text{if} \; t \in [a,b) \\ \Gamma(1) & \text{if} \; t \geq b \end{cases}$$

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 $t<a$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.