Let $\mathbb{D}$ denote the unit disk and $\mathbb{T}$ the unit circle , if $f\in H^1(\mathbb{D})$ is outer , then necessarily $\log|f|\in L^1(\mathbb{T})$ ?
We know from theory of Hardy spaces that a function $f\in H^1$ is outer if $f=\lambda[f]$ for some $\lambda\in\mathbb{C}$ with $|\lambda|=1$ and $[f]$ defined by $$[f](z)=\exp\left(\int_\mathbb{T}\frac{\xi+z}{\xi-z} \ \log|f(\xi)| \ dm(\xi)\right)$$ $dm$ being the normalized Lebesgue measure on $\mathbb{T}$ . Therefore we have $$\log|f(z)|=\int_\mathbb{T}\frac{1-|z|^2}{|\xi-z|^2} \ \log|f(\xi)| \ dm(\xi)$$ $$\begin{aligned}\implies\int_\mathbb{T}|\log|f(z)| \ | \ dm(z)\leq\int_\mathbb{T}\int_\mathbb{T}\frac{1-|z|^2}{|\xi-z|^2} \ |\log|f(\xi)| \ | \ dm(\xi) \ dm(z) \\ <\int_\mathbb{T}\int_\mathbb{T}\frac{1-|z|^2}{|\xi-z|^2} \ |f(\xi)| \ dm(\xi) \ dm(z)\qquad \ \ \end{aligned}$$ We need the RHS is finite , but I am unable to see why this would happen . If not , is there some suitable counterexample to this ? Any help is appreciated .
If $f$ is outer it is zero free so $\ln(f)$ is well defined and by applying the mean value property to its real part we get
$$\ln(|f(0)|)=\int_0^{2\pi} \ln(|f_r(e^{it})|)dm(t)$$
where $f_r(e^{it})=f(re^{it})$. $\max(\ln(|f|),0)=\ln^+(|f|)\le |f|\in L^1(\mathbb{T})$, so it suffices to prove that $\ln^-(|f|)=\min(\ln(|f|),0)$ is in $L^1(\mathbb{T})$. Since $f_r(e^{i\vartheta})\to f(e^{i\vartheta})$ almost everywhere with $r\to 1$, Fatou's lemma implies that
$$\ln(|f(0)|)-\|f\|_{H^1}\le \ln(|f(0)|)-\limsup_r\int_0^{2\pi}\ln^+|f_r(e^{it})|dm(t)\le\\ \le\lim_{r\to 1} \int_0^{2\pi} \ln^- |f_r|(e^{it})dm(t) \le \int_0^{2\pi} \ln^-|f|(e^{it})dm(t) $$
the claim follows.