# What are non-obvious examples of measures obtained from linear functionals by the Riesz representation theorem?

In chapter two of Rudin's "Real and Complex Analysis" there is a "Riesz Representation Theorem" that dominates the chapter. My understanding of the statement of the thm. is that given a complex-valued linear functional, $\lambda$ defined on the set of continuous functions with compact support in a "nice" topological space (Hausdorff and locally compact) the theorem states there is a corresponding unique positive complete measure $\mu$ defined on a sigma algebra containing all borel sets in X so that the integral of $f$ w.r.t $\mu$, $\int_{X}{f}d\mu$= $\lambda(f)$.

There are a few other parts of the theorem's statement, but I stop here to ask: "What are some examples of corresponding measures to linear functionals that are not immediately obviously translatable to a measure without the theorem?"

An example I thought of first was a linear functional that gives the n'th coefficient of some function $f$ represented by a polynomial, possibly trigonometric. But I also wondered how differentiation at a point would correspond to a measure. Are these questions well-posed? If so could you help me think about them?

• I would guess that the value of the theorem is that it gives the form of the functional. Once you know the form, determining the measure is fairly straightforward. To wit, if you choose $f$ to approximate $1_A$ in some sense, you get $\lambda(f) \approx \mu A$, so you can 'recover' $\mu$ 'easily'. Commented Oct 11, 2014 at 3:25
• thanks, I'll think on that.
– asd
Commented Oct 11, 2014 at 3:31
• Differentiation will not give you a measure. First of all, not all continuous functions with compact support can be differentiated. Secondly, you have to have some form of continuity, e.g. $|\phi(f)|\leq C_K \Vert f \Vert_\sup$ for all $f$ supported in $K$. You should check that this does not hold for differentiation. One important example (which Rudin uses the theorem for) is to construct Lebesgue measure. The existence of Lebesgue measure is not immediately clear (without Riesz representation), so this has some value. Commented Oct 11, 2014 at 4:59
• @PhoemueX if you think about it, integrating a function is, loosely speaking, measuring the area under the function, so you're measuring something. Likewise, by differentiating a function your are measuring the ratio of change of the function at a point so I guess that's why asd came up with that question, but as you said, it seems to be unfruitful since when differentiating a function you usually work with a lot less functions than in the Riesz Theorem. Now, since the derivative of a function requires, a point and a neighbourhood, I'd guess you may get some local meaning for what he asks. Commented Aug 11, 2016 at 23:34

## 4 Answers

The spectral theorem for bounded selfadjoint operators is a surprising consequence of the Riesz Representation Theorem for continuous functions $C[a,b]$ on a finite interval $[a,b]\subset\mathbb{R}$.

If $A$ is a bounded selfadjoint operator on a Hilbert space $X$ with spectrum $\sigma(A)$, it is not so difficult to show that $\|p(A)\| \le \sup_{\lambda\in\sigma(A)}|p(\lambda)|$ for polynomials $p$. This is because the norm and spectral radius of a selfadjoint are found to be the same. Therefore, the map $p\mapsto p(A)$ then extends to continuous functions $p$ on $\sigma(A)$, and gives the existence of unique finite complex Borel measures $\mu_{x,y}$ such that $$(p(A)x,y) = \int_{\sigma(A)}p(\lambda)d\mu_{x,y}(\lambda).$$ And $\mu_{x,x}$ is a positive Borel measure with $\|x\|^{2}=\mu_{x,x}(\sigma(A))$. For a fixed Borel subset of $\sigma(A)$, this leads to a bounded sesquilinear form $\mu_{x,y}(S)$ with $|\mu_{x,y}(S)|\le \|x\|\|y\|$ and gives the existence of a unique positive Borel operator measure $E$ such that $$\mu_{x,y}(S) = (E(S)x,y).$$ After a little study, one finds that $E$ is a spectral measure, i.e., satisfies $$E(\sigma(A))=I,\;\; E(S)^{\star}=E(S)=E(S)^{2},\;\; E(S)E(T)=E(S\cap T).$$ This then yields the spectral theorem $$A = \int_{\sigma(A)}\lambda dE(\lambda).$$ For every polynomial $p(\lambda)$, one automatically has $$p(A) = \int_{\sigma(A)}p(\lambda)\,dE(\lambda),$$ which extends to continuous functions $p$, and satisfies $$p(A)q(A) = (pq)(A).$$

In this light, the Spectral Theorem for selfadjoint linear operators is a lifting of the Riesz Representation to an operator measure $E$ representation.

My favorite example is harmonic measure. Let $\Omega$ be a domain in $\mathbb R^n$ (with smooth boundary $\Gamma=\partial \Omega$, to avoid technicalities). For every $\phi\in C(\Gamma)$ there is a unique harmonic function $u$ on $\Omega$ such that $u(x)\to \phi(y)$ as $x\to y$, for every $y\in \partial\Omega$.

Fix $z\in\Omega$. By the maximum principle, $|u(z)|\le \max_\Gamma|\phi|$. Thus, the map $\phi\mapsto u(z)$ is a bounded linear functional on $C(\Gamma)$. By Riesz representation, there is a measure $\omega_z$ on $\Gamma$ such that $$u(z)= \int_{\Gamma} \phi(y)\, d\omega_z(y)$$ (Note that $\omega_z$ is a probability measure.)

While for smooth domains one can relate $\omega_z$ to a familiar object (Green's function), for general domains, where $\omega_z$ still makes sense, its structure is opaque. For $n>2$, it is unknown how large the support of $\omega_z$ can be: that is, what is the smallest $d=d(n)$ such that every $\omega_z$ gives mass $1$ to some set of Hausdorff dimension $d$? In 1980s Bourgain proved $d(n)<n$; in 1990s Wolff proved that $d(3)>2$... and I think this is where it remains now.

• All of the answers qualified.
– asd
Commented Sep 2, 2018 at 7:41

My favourite examples are the Riesz products, that is, weak*-limits of

$$\prod_{k=1}^n \big(1+\cos(3^kt)\big)$$

in $C(\mathbb{T})^*$. Colin C. Graham used the Riesz products to give a slick proof of the Wiener–Pitt phenomenon in $M(\mathbb{T})$.

A good example is the average of $f\in C[0,1]$ on the Cantor set.

If $C=\bigcap_{n\in\mathbb N}C_n$, where $C_1=[0,1/3]\cup [2/3,1]$, $C_2=[0,1/9]\cup[2/9,1/3]\cup[6/9,7/9]\cup[8/9,1]$, etc.

As $\mu(C_n)=2^n/3^n$, we define $$\varphi_n(f)=\frac{1}{m(C_n)}\int_{C_n}f(x)\,dx,$$ and show that the sequence $\varphi_n(f)$ converges, for every such $f$, to a limit $\varphi(f)$, with $\lvert\varphi(f)\rvert\le \|f\|_\infty$. This defines a weak$^*$ convergence of $\varphi_n\to\varphi$ which is a unit measure, singular to Lebesgue measure and containing no atomic (Dirac) measures.