Given a polynomial 
, where the coefficients are random, what can we say about the distribution of the roots (on 
)? Of course, it would depend on what “random” means. Here, “random” means that the sequence 
is an i.i.d. sequence of complex random variables.
It turns out that under a rather weak condition on 
, then as 
, the roots will tend to be distributed on the unit circle! (There are lots of interesting discussions here, which explain why this should be true intuitively.)
I will give a rigorous proof (and a rigorous formulation) of this result. For simplicity, we will assume that are complex standard Gaussians. That is, for any Borel set 
, we have
,
where 
is the Lebesgue measure on the complex plane. We will also assume two basic potential theoretic results without proofs, namely
For 
, we write 
, the normalized counting measure. We define the expected normalized counting measure, ![\mathbf{E}[Z_{p_n}]](https://s0.wp.com/latex.php?latex=%5Cmathbf%7BE%7D%5BZ_%7Bp_n%7D%5D&bg=ffffff&fg=000000&s=0 "\mathbf{E}[Z_{p_n}]")
, as follows. For any 
,
![\displaystyle \begin{array}{rl} \displaystyle (\mathbf{E}[Z_{p_n}],\psi) &:= \mathbf{E}[(Z_{p_n}, \psi)]\\ &\displaystyle =\frac{1}{\pi^{n+1}}\int_{\mathbb{C}^{n+1}} \frac{1}{n}\sum_{j=1}^n \psi(\zeta_j)e^{-\sum_{j=0}^n |a_j|^2} \,\text{d}m(a_0)\cdots\text{d}m(a_n). \end{array}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Barray%7D%7Brl%7D+%5Cdisplaystyle+%28%5Cmathbf%7BE%7D%5BZ_%7Bp_n%7D%5D%2C%5Cpsi%29+%26%3A%3D+%5Cmathbf%7BE%7D%5B%28Z_%7Bp_n%7D%2C+%5Cpsi%29%5D%5C%5C%C2%A0+%26%5Cdisplaystyle+%3D%5Cfrac%7B1%7D%7B%5Cpi%5E%7Bn%2B1%7D%7D%5Cint_%7B%5Cmathbb%7BC%7D%5E%7Bn%2B1%7D%7D+%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bj%3D1%7D%5En+%5Cpsi%28%5Czeta_j%29e%5E%7B-%5Csum_%7Bj%3D0%7D%5En+%7Ca_j%7C%5E2%7D+%5C%2C%5Ctext%7Bd%7Dm%28a_0%29%5Ccdots%5Ctext%7Bd%7Dm%28a_n%29.+%5Cend%7Barray%7D+&bg=ffffff&fg=000000&s=0 "\displaystyle \begin{array}{rl} \displaystyle (\mathbf{E}[Z_{p_n}],\psi) &:= \mathbf{E}[(Z_{p_n}, \psi)]\\ &\displaystyle =\frac{1}{\pi^{n+1}}\int_{\mathbb{C}^{n+1}} \frac{1}{n}\sum_{j=1}^n \psi(\zeta_j)e^{-\sum_{j=0}^n |a_j|^2} \,\text{d}m(a_0)\cdots\text{d}m(a_n). \end{array}")
Intuitively, this tells us how the zeros of 
are distributed “on average”.
Theorem. We have
![\displaystyle \lim_{n\to\infty} \mathbf{E}[Z_{p_n}] = \frac{1}{2\pi}\text{d}\theta](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Clim_%7Bn%5Cto%5Cinfty%7D+%5Cmathbf%7BE%7D%5BZ_%7Bp_n%7D%5D+%3D+%5Cfrac%7B1%7D%7B2%5Cpi%7D%5Ctext%7Bd%7D%5Ctheta+&bg=ffffff&fg=000000&s=0 "\displaystyle \lim_{n\to\infty} \mathbf{E}[Z_{p_n}] = \frac{1}{2\pi}\text{d}\theta")
in weak*-topology. |
We will write 
. The key ingredient of the proof will be the orthonormality of 
with respect to 
. Define
.
It is clear that 
. Also, note that
locally uniformly in 
(an easy calculus exercise). Why is this important? It is because using some basic potential theory, this implies
as in weak*-topology.
Proof of Theorem: Write 
and 
. Also write 
, the normalized vector. Then
Note that 
. Thus
Recall that 
. Also, note that
Therefore,
![\displaystyle \begin{array}{rl} (\mathbf{E}[Z_{p_n}],\psi) & \displaystyle = \mathbf{E}[(Z_{p_n}, \psi)]\\ &\displaystyle = \mathbf{E}\left[\left(\frac{1}{2\pi}\Delta\frac{1}{2n}\log S_n(z,z),\psi\right)\right]+\frac{1}{2\pi n} \mathbf{E}[(\Delta \log|\langle a,u(z)\rangle|,\psi )] .\end{array}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Barray%7D%7Brl%7D+%28%5Cmathbf%7BE%7D%5BZ_%7Bp_n%7D%5D%2C%5Cpsi%29+%26+%5Cdisplaystyle+%3D+%5Cmathbf%7BE%7D%5B%28Z_%7Bp_n%7D%2C+%5Cpsi%29%5D%5C%5C+%26%5Cdisplaystyle+%3D+%5Cmathbf%7BE%7D%5Cleft%5B%5Cleft%28%5Cfrac%7B1%7D%7B2%5Cpi%7D%5CDelta%5Cfrac%7B1%7D%7B2n%7D%5Clog+S_n%28z%2Cz%29%2C%5Cpsi%5Cright%29%5Cright%5D%2B%5Cfrac%7B1%7D%7B2%5Cpi+n%7D+%5Cmathbf%7BE%7D%5B%28%5CDelta+%5Clog%7C%5Clangle+a%2Cu%28z%29%5Crangle%7C%2C%5Cpsi+%29%5D+.%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\displaystyle \begin{array}{rl} (\mathbf{E}[Z_{p_n}],\psi) & \displaystyle = \mathbf{E}[(Z_{p_n}, \psi)]\\ &\displaystyle = \mathbf{E}\left[\left(\frac{1}{2\pi}\Delta\frac{1}{2n}\log S_n(z,z),\psi\right)\right]+\frac{1}{2\pi n} \mathbf{E}[(\Delta \log|\langle a,u(z)\rangle|,\psi )] .\end{array}")
The first term is actually nonrandom, and it converges to 
. We will show that the second terms goes to 
and this will finish the proof. It suffices to show that the expectation in the second term is bounded.
Consider
![\displaystyle \begin{array}{rl} \displaystyle \mathbf{E}[(\Delta \log|\langle a,u(z)\rangle|,\psi )] & \displaystyle = \mathbf{E}[(\log|\langle a,u(z)\rangle|, \Delta\psi)] \\ & \displaystyle = \int_\mathbb{C} \Delta \psi(z) \mathbf{E}[(\log|\langle a,u(z)\rangle|)]\,\text{d}m(z),\end{array}](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cbegin%7Barray%7D%7Brl%7D+%5Cdisplaystyle+%5Cmathbf%7BE%7D%5B%28%5CDelta+%5Clog%7C%5Clangle+a%2Cu%28z%29%5Crangle%7C%2C%5Cpsi+%29%5D+%26+%5Cdisplaystyle+%3D+%5Cmathbf%7BE%7D%5B%28%5Clog%7C%5Clangle+a%2Cu%28z%29%5Crangle%7C%2C+%5CDelta%5Cpsi%29%5D+%5C%5C+%26+%5Cdisplaystyle+%3D+%5Cint_%5Cmathbb%7BC%7D+%5CDelta+%5Cpsi%28z%29+%5Cmathbf%7BE%7D%5B%28%5Clog%7C%5Clangle+a%2Cu%28z%29%5Crangle%7C%29%5D%5C%2C%5Ctext%7Bd%7Dm%28z%29%2C%5Cend%7Barray%7D&bg=ffffff&fg=000000&s=0 "\displaystyle \begin{array}{rl} \displaystyle \mathbf{E}[(\Delta \log|\langle a,u(z)\rangle|,\psi )] & \displaystyle = \mathbf{E}[(\log|\langle a,u(z)\rangle|, \Delta\psi)] \\ & \displaystyle = \int_\mathbb{C} \Delta \psi(z) \mathbf{E}[(\log|\langle a,u(z)\rangle|)]\,\text{d}m(z),\end{array}")
where in the second inequality we used Fubini. Now, let’s recall that
![\displaystyle \mathbf{E}[\log|\langle a,u(z)\rangle|] = \frac{1}{\pi^{n+1}}\int_{\mathbb{C}^{n+1}}\log|\langle a,u(z)\rangle| e^{-\sum_{j=0}^n |a_j|^2}\,\text{d}m(a_0)\cdots \text{d}m(a_n).](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+%5Cmathbf%7BE%7D%5B%5Clog%7C%5Clangle+a%2Cu%28z%29%5Crangle%7C%5D+%3D+%5Cfrac%7B1%7D%7B%5Cpi%5E%7Bn%2B1%7D%7D%5Cint_%7B%5Cmathbb%7BC%7D%5E%7Bn%2B1%7D%7D%5Clog%7C%5Clangle+a%2Cu%28z%29%5Crangle%7C+e%5E%7B-%5Csum_%7Bj%3D0%7D%5En+%7Ca_j%7C%5E2%7D%5C%2C%5Ctext%7Bd%7Dm%28a_0%29%5Ccdots+%5Ctext%7Bd%7Dm%28a_n%29.&bg=ffffff&fg=000000&s=0 "\displaystyle \mathbf{E}[\log|\langle a,u(z)\rangle|] = \frac{1}{\pi^{n+1}}\int_{\mathbb{C}^{n+1}}\log|\langle a,u(z)\rangle| e^{-\sum_{j=0}^n |a_j|^2}\,\text{d}m(a_0)\cdots \text{d}m(a_n).")
Note that the integral is invariant under unitary transformations. Therefore, by applying an unitary transformation, we may assume that 
(recall that 
is an unit vector). Hence, the integral equals
which is a constant independent of 
and 
. Using the fact that 
is bounded, we are done.
Remark If you write  , where  is a sequence of orthonormal polynomials with respect to some  , where  supports on some “nice” compact subset of the complex plane, then the normalized counting measure of the roots of  will converge to the “equilibrium measure” on the support of . This requires more potential theory and so I am not putting the general result here.
|
