# Question concerning the proof of the Stein-Weiss interpolation theorem

Currently I am going through the proof of the Stein-Weiss interpolation theorem for a seminar paper and in particular through the proof of Lemma 1 which begins at page 320 (I will use a particular function for $\Phi$). First I will establish the setting:

Let $S := \{z \in \mathbb{C} : 0 < \text{Re }z < 1\}$ and $F$ be holomorphic in $S$ and continuous on $\overline{S}$. Further define for $z$ in the closed unit circle with $z \neq \pm 1$ $$h(z) := \frac{1}{\pi i}\log \left( i\frac{1 + z}{1 - z}\right)$$ Then $h$ is holomorphic, $\log|F(h(z))|$ is subharmonic in the open unit disc and upper semicontinuous on the closed unit circle with $z \neq \pm 1$.

Now on p. 320, bottom (eq. 2.4), it is written that by the subharmonic character of $\log|F(h(z))|$ we have for $0 \leq \rho < R <1$ $$\log|F(h(\rho e^{i\theta}))| \leq \frac{1}{2\pi}\int_{-\pi}^\pi \log\vert F(h(Re^{it}))\vert\frac{R^2 - \rho^2}{R^2 - 2R\rho\cos\left( \theta - t \right) + \rho^2}dt$$ I wanted to know exactly why this is true. So my idea was to define a function \begin{aligned} H(\rho e^{i\theta}) := \begin{cases} \displaystyle \log\vert F(h(Re^{i\theta}))\vert & \rho = R\\ \displaystyle \frac{1}{2\pi}\int_{-\pi}^\pi \log\vert F(h(Re^{it}))\vert\frac{R^2 - \rho^2}{R^2 - 2R\rho\cos\left( \theta - t \right) + \rho^2}dt & 0 \leq \rho < R \end{cases} \end{aligned} and using the maximum modulus principle for subharmonic functions (see here p. 336) since $H$ is harmonic and continuous. But my problem is now, that I am not sure if $\log|F(h(z))|$ is continuous when $|z| = R$ (and this would make my arguing useless). So my question:

• Is $\log|F(h(z))|$ continuous for $|z| = R$?
• Is there any other approach to show the inequality above?

Thanks.