Apparently, it is well known that $\text{Log}(1-z) \in H^p(\mathbb{D})$ for all $0<p<\infty,$ where $\text{Log}$ denotes the principal complex logarithm. How can this be proven?
$\text{Log}(1-z) \in H^p(\mathbb{D})$ for all $0<p<\infty$
complex-analysishardy-spaces
Best Answer
Solution 1 (general and applies to all $f=u+iv$ where either $u$ or $v$ is bounded on the unit disc)
It is enough to show the result for all even integers $2m \ge 2$ as the Hardy spaces are decreasing in $p$
$f(z)=\log(1-z)=\log|1-z|+i\arg (1-z)=u(z)+iv(z), |v(z)| \le \frac{\pi}{2}=A$ as $\Re (1-z) >0$
$\Re f(z)^{2m}=u^{2m}+\sum_{k=1}^{m}B_{k,m}u^{2m-2k}v^{2k}$ where $B_{k,m}$ are plus/minus binomial coefficients.
But $f(0)=0, f(0)^{2m}=0, (\Re f^{2m})(0)=0$ so by the harmonic mean property:
$\int_0^{2\pi} u^{2m}(re^{it})dt+\sum_{k=1}^{m}B_{k,m}\int_0^{2\pi} (u^{2m-2k}v^{2k})(re^{it})dt=0$
Holder Inequality implies:
$\int_0^{2\pi} u^{2m-2k}(re^{it})v^{2k}(re^{it})dt \le (\int_0^{2\pi} u^{2m}(re^{it})dt)^{\frac{2m-2k}{2m}}(\int_0^{2\pi} v^{2m}(re^{it})dt)^{\frac{2k}{2m}}$
($\int|ab| \le (\int|a|^p)^{1/p}(\int|b|^q)^{1/q}, 1/p+1/q=1, a=u^{2m-2k}, b=v^{2k}, p=\frac{2m}{2m-2k}, q=\frac{2m}{2k}$)
But now since $|v(z)| \le A$ we get that all integrals $(\int_0^{2\pi} v^{2m}(re^{it})dt)^{\frac{2k}{2m}} \le (2\pi A)^{2k}$ so denoting by $X_r(m)=(\int_0^{2\pi} u^{2m}(re^{it})dt)^{1/2m}$ and using the triangle inequality we get:
$X_r(m)^{2m} \le \sum_{k=1}^{m}|B_{k,m}|X_r^{2m-2k}(2\pi A)^{2k}$
In particular, this means that $X_r(m)$ is bounded for every $r<1$ by the positive root of the equation $R^{2m}-\sum_{k=1}^{m-1}|B_{k,m}|(2\pi A)^{2k}R^{2m-2k}=0, |B_{k,m}|>0$
(or one can use induction on $m$),
so anyway $\int_0^{2\pi} u^{2m}(re^{it})dt \le C_m, r <1$ which immediately implies $\int_0^{2\pi} |u(re^{it})|^pdt \le C_p, 0<p<\infty, r<1$ and hence
$\int_0^{2\pi} |f|^{2m}(re^{it})dt \le C_{m,1}, r <1$ by expanding $|f|^{2m}=(u^2+v^2)^m$ and we are done!
Note that we only used that $|v| \le A$ and nothing else; same proof works of course if $|u| \le A$
Solution 2 (sketch): Because $f \in H^2$ since $\log(1-z)=-\sum_{n \ge 1} {z^n/n}$ and Parseval, it is enough to show that $\int_0^{2\pi} |f|^{p}(e^{it})dt < \infty$ where $f(e^{it})$ is the non-tangential limit of $f$ (finite except at $t=0$ or $z=1$) since then (ie we already know apriori that $f$ is in some Hardy space, but maybe not in the one we want) $f \in H^p$ iff $f(e^{it}) \in L^p(dt)$
Note that in general even if an analytic function in the disc has non-tangential limits ae and $f(e^{it}) \in L^p(dt)$, it is far from true that $f$ is in a Hardy space; simple examples like $\exp \frac{1+z}{1-z}$ which is bounded on the unit circle ae so it is in $L^{\infty}(dt) \subset L^p(dt), p< \infty$ but it is highly unbounded and in no $H^p$ in the unit disc, show this, but it is possible for the function not to be even in the Nevanlinna space $N$ like the counterexample above and even worse, the zeroes may not satisfy the Blaschke condition that functions in $N$ do regardless if they are in a Hardy space or not.
A simple computation shows that $1-e^{it}=-2i\sin({t/2})e^{it/2}$ and hence the only problem is with $|\log \sin({t/2})|^p$ but that is roughly $ \log^p (1/|t|)$ for $t$ near zero which is clearly integrable as $\log (1/|t|) << t^{-\epsilon}$ so choosing $\epsilon$ small enough so $p\epsilon <1$ shows it