Example of analytic function in unit disk in no class $H^p$ but with non-tangential limits at almost every point of unit circle Given an example of an analytic function in the unit disk which is in no class $H^p$ but which has non-tangential limits at almost every point of the unit circle?
 A: Just create a really bad singularity at one point of the unit circle. For example, $f(z)=\exp(1/(1-z))$ is not in any $H^p$ class. I show this in two ways:


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*Integral estimate. Let $r=1-\epsilon$ where $\epsilon>0$ is small. Consider $z=re^{it}$ with $0<t<\epsilon$. Since $|\arg (1-z)|\le \pi/4$, we have $\operatorname{Re}(1/(1-z))\ge \frac{1}{\sqrt{2}|1-z|}$. Also, $|1-z|\le 2\epsilon$ by triangle inequality. It follows that 
$$\int_0^\epsilon |f(re^{it})|^p\,dt \ge \epsilon \,\exp(-c\,p/ \epsilon),\quad c=\frac{1}{2\sqrt{2}} $$
As $\epsilon\to 0$, the integral tends to infinity. 

*Power series expansion. If $f\in H^p$, then $f^{p/2}(z)=\exp\left(\frac{p}{2}  (1-z)^{-1}\right)$ is in $H^2$. Therefore the coefficients of its power series at $0$ must be square summable. This power series is obtained by plugging $(1-z)^{-1}=\sum_{n=0}^\infty z^n$ into the function $\exp(\frac{p}{2}\zeta)$. All resulting coefficients are nonnegative. And since
$$f^{p/2}(z)=1+ \frac{p}{2} \left(\sum_{n=0}^\infty z^n\right)+\dots $$
the coefficients are not square summable.
