Showing that a certain function cannot be represented as a Poisson integral

Question

Let $\Bbb T$ denote the unit circle in $\Bbb C$. A function on $\Bbb T$ can be identified with a $2\pi$-periodic function on $\Bbb R$. For a function $f$ on $\Bbb T$, we define its integral on $\Bbb T$ by $\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi} f(e^{it})~dt$ (if the integral is well-defined in Lebesgue sense).

For a function $f\in L^1(\Bbb T)$, let us define its Possion integral $u = P[f]$ by $u(z)=\dfrac{1}{2\pi}\displaystyle\int_0^{2\pi} P(z,e^{it})f(e^{it})~dt$ for $|z|<1$, where $P(z,e^{it})=\dfrac{1-|z|^2}{|e^{it}-z|^2}$. Then it is known that $\displaystyle\sup_{0<r<1} \dfrac{1}{2\pi}\displaystyle\int_0^{2\pi} |u(re^{i\theta})|~d\theta \leq ||f||_1$.

Using this result how can we show that for $u(z) =\textrm{Im}\left[\left(\frac{1+z}{1-z} \right)^2 \right]$, $|z|<1$, there is no $f\in L^1(\Bbb T)$ such that $u = P[f]$? I tried to show the supremum is unbounded but the calculation is so complicated so I got stuck. Thanks in advance

Answer 1

If $f\in L^1(\mathbb{T})$, $P[f](re^{i\theta}) =\frac{1}{2\pi} \int_0^{2\pi}{\frac{1-r^2}{|e^{i(t-\theta)}-r|^2} f(e^{it})\,dt}:= \int_0^{2\pi} {K_r(t-\theta) f(e^{it})\,dt}=C_r f(\theta)$, where $K_r$ is a function $\mathbb{R}/2\pi\mathbb{Z} \rightarrow [0,\infty)$ defined by $2\pi K_r(t)$ being the real part of $\frac{1+re^{it}}{1-re^{it}}$.

Complex analysis easily proves that each $K_r$ has integral $1$. Moreover, one can also see that $K_r$ converges uniformly to zero on compact subsets of $\mathbb{R}/2\pi\mathbb{Z}$ not containing zero. Thus, if $f \in L^1(\mathbb{T})$ is continuous, $C_rf(\theta)$ goes uniformly to $f(e^{i\theta})$ as $r \rightarrow 1$.

Let $V \subset L^1(\mathbb{T})$ be the subspace of $f$ such that $C_rf(\theta)$ converges to $f(e^{i\theta})$ in $L^1$ norm as $r \rightarrow 1$. $V$ contains all continuous functions; moreover, if $f \in V$, $g \in L^1(\mathbb{T})$, then (all norms being $L^1(\mathbb{R}/2\pi\mathbb{Z})$) $\|C_rg-g\| \leq \|C_r(g-f)+C_rf-f+f-g\| \leq 2\|f-g\|+\|C_rf-f\|$ (the $L^1$ norm of $K_r$ is $1$ so the $L^1 \rightarrow L^1$ operator norm of $C_r$ is at most $1$), so that the limsup of $\|C_rg-g\|$ is dominated by twice the distance of $g$ to $V$. But $V$ is dense, so that limsup is zero, so $g \in V$.

In particular, if $u=P[f]$, then the radial limit of $u$ goes in $L^1$ to $f$. Except that the radial limit of $u$ pointwise (and uniformly away from $\theta=0$) is $0$. So $f$ would be zero, a contradiction.

Answer 2

A different and possibly easier solution is to consider $2\pi F(z)=\int_{0}^{2\pi}\frac{e^{it}f(e^{it})}{e^{it}-z}dt, |z| <1$ the Cauchy transform of $f$ which is an analytic function in the unit disc; pretty much by definition $F(z)=-i(\frac{1+z}{1-z})^{2}$

But now it is an easy exercise to show that $F \in H^{p}(\mathbb D), 0<p<1$

(decompose $f=f_{+}-f_{-}$ and use that the conjugate Poisson kernel is in $L^p, p<1$ or directly that an analytic function with a positive real part is in $H^p, 0<p<1$ by using that $\Re G >0, G(0)=1$ implies $G$ subordinated to $(1+z)/(1-z)$ and the classic results on means of subordinated functions)

However it is obvious that $-i(\frac{1+z}{1-z})^{2} \in H^p$ only for $p < 1/2$ since $1/(1-r)^{2p}$ is integrable near $1$ only for $p < 1/2$ so contradiction!