# Bounded analytic function on the real axis

I have an analytic complex function $f(z)$ in the upper half-plane, $z=x+iy$, where $\int_{-\infty}^{\infty}|f(x)|^{2}dx<\infty$, and $f$ is continuous on the real axis. Is it true that $f$ is bounded on the real axis, and why?

( i.e, there exists $M>0$, such that $|f(x)|\leq M$ for all $x\in \mathbb R$).

Not necessarily. Let $g\colon\mathbb{R}\to\mathbb{R}$ be such that $g\in L^2(\mathbb{R})$ but $g\not\in L^\infty(\mathbb{R})$. For instance, we could take $$g(x)=\sum_{n=0}^\infty n\,\chi_{[n,n+2^{-n}]}(x),$$ where $\chi_{A}$ is the characteristic function of the set $A$. Let $$P_y(x)=\frac{1}{\pi}\,\frac{y}{x^2+y^2},\quad x\in\mathbb{R},\quad y>0,$$ be the Poisson kernel. Then $u(x,y)=P_y\ast g(x)$ is a harmonic function on the upper half-plane such that $u(x,0)=g(x)$. This last equality holds:
1. In the $L^2$ sense: $\lim_{y\to0}\int_{-\infty}^\infty|u(x,y)-g(x)|^2\,dx=0$
2. In the non tangential sense: given $\alpha>0$, for almost all $\xi\in\mathbb{R}$, $\lim_{(x,y)\to(\xi,0),x\in \Gamma(\xi)}u(x,y)=g(\xi)$, where $\Gamma_\alpha(\xi)=\{(x,y):|x-\xi| \le\alpha\,y\}$.
Let $v(x,y)$ be the conjugate function of $u$. Then $f(z)=u+i\,v$ is analytic on the upper half-plane, and $f(x)=g(x)+Hg(x)$, where $Hg$ is the Hilbert transform of $g$. Since $H$ is a bounded operator on $L^2$, it follows that $\int_{-\infty}^\infty|f(x)|^2\,dx<\infty$.