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.
 A: 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).
