$0\leq u\leq P_r\implies u=\lambda P_r$

Question

Let $u$ be a real-valued harmonic function on the open unit disk such that $0\leq u(r,\theta)\leq P_r(\theta)$ for all $0\leq r<1$ and $\theta$, where $P_r(\theta)$ denotes the Poisson kernel. How can I deduce from this that $u(r,\theta)=\lambda P_r(\theta)$ for some constant $\lambda$? If the claim is true, then certainly $\lambda=u(0,0)$, so we have a candidate value for $\lambda$. But I'm not sure how to deduce $u(r,\theta)=\lambda P_r(\theta)$ on the whole unit disk.

Answer 1

One can give a direct proof (without using Herglotz) as follows: for $z=re^{it}, r <1, t \in [-\pi, \pi]$ let $$u_r(t)=u(z)=a_0+\sum_{n \ge 1} (a_nz^n+\bar a_n\bar z^n), |z|<1$$ as any real harmonic function in the unit disc can be written like that (eg it is the real part of an analytic function).

Note that $|(1-z^n)\frac{1+z}{1-z}| \le 2n, |z| \le 1$ by maximum modulus so $|(1-z^{-n})\frac{1+z}{1-z}| \le 2^{n+1}n, 1/2 \le |z| \le 1$

For $n \ge 1$ we get $a_n-a_0=\frac{1}{2\pi}\int_{-\pi}^{\pi} u_r(t)(r^{-n}e^{-int}-1)dt$.

But since for $z=re^{it}, 1/2 \le r <1$ we have $|P_r(t)|=|\Re \frac{1+z}{1-z}|\le |\frac{1+z}{1-z}|$ we have that $|1-z^{-n}|u_r(t) \le |1-z^{-n}|P_r(t) \le 2^{n+1}n$ as above so in particular $|u_r(t)(r^{-n}e^{-int}-1)|$ is uniformly bounded in $1/2 \le r <1, t \in [-\pi, \pi]$.

But $0 \le u_r(t) \le P_r(t)$ means that $u_r(t) \to 0, r \to 1$ for all $t \ne 0, t \in [-\pi, \pi]$ and so does $u_r(t)(r^{-n}e^{-int}-1)$ and we can apply the bounded convergence theorem for integrals to conclude that $a_n-a_0= \lim_{r \to 1}\frac{1}{2\pi}\int_{-\pi}^{\pi} u_r(t)(r^{-n}e^{-int}-1)dt=0$ since the integrand goes to $0$ ae and is uniformly bounded.

Hence $a_n=a_0$ for all $n \ge 1$ and since $a_0 \ge 0$ so $\bar a_n=a_n$ so $u_r=a_0P_r$