# Brown measure of left shift operator

Let $L$ be the left shift operator on $\ell^2(\mathbb{Z})$ with trace $\tau(T) := \langle T \delta_0, \delta_0 \rangle$.

How can I show that the Brown measure of $L$ is the uniform measure on the unit circle?

The Brown measure is defined as follows: For each $z \in \mathbb{C}$ define $\nu_{z}$ to be the spectral measure of the (self-adjoint) operator $(L - z)^*(L-z)$. Then let $$f(z) = \frac{1}{2} \int_0^\infty \log x \,d\nu_z(x).$$ Then the Brown measure is defined as $$\mu_L := \frac{1}{2\pi} \Delta f,$$ where the Laplacian is taken in the sense of distributions. Motivation for the definition can be found at Terence Tao's notes on the circular law for random matrices.

First of all, as I commented over at MO, I assume that we define $\nu_z$ as the spectral measure of $(L-z)^*(L-z)$ of the vector $\delta_0$. Then, by functional calculus, $$f(z) = \frac{1}{2} \langle \delta_0, \log |z-L|^2 \delta_0 \rangle = \int_0^{2\pi} \log |z-e^{it}| \frac{dt}{2\pi} .$$ The second equality follows because Lebesgue measure on the circle is the spectral measure of $L$ and $\delta_0$ (well known, or take Fourier transforms to see this).
Now we've reduced matters to your other question. To see that $\Delta f$ is Lebesgue measure on the circle, we recall that $\Delta \log |z|=2\pi\delta$ (in other words, $(1/2\pi)\log |z|$ is the fundamental solution of the Laplacian; this is discussed in many places, and an internet search should work fine if you want more background). So if $\varphi$ is a test function, then $$\int f(z)\Delta\varphi(z)\, dz = \int_0^{2\pi}\frac{dt}{2\pi}\int dz\,\log|z-e^{it}|\Delta\varphi(z) = \int_0^{2\pi} \varphi(e^{it})\, dt ,$$ as required (I use the slightly unusual but convenient notation $dz$ for an area integral here).