The question (not homework) is
Let $f\in\mathcal{H}$ and suppose that $f(0) = 0$ and $\left|f(z)\right|\leq 1$ for $z\in\mathbb{U}$. Show that $$\left|f(z)\right|\leq \frac{2}{\pi} arg\left(\frac{1+ir}{1-ir}\right) = \frac{2}{\pi} \arctan\left(\frac{2r}{1-r^2}\right) = \frac{4}{\pi}\arctan r \leq \frac{4}{\pi}r$$ for $\left|z\right| \leq r < 1$.
Pretty sure I only need help with the first inequality, for which the following unhelpful hint is given:
Hint: $f\in \mathcal{H}_S$, where $S$ is the set of non-zero integers; hence
$$\left|f(re^{i\theta})\right|\leq \frac{1}{\pi}\int_{-\pi/2}^{\pi/2} \left|\mathcal{P}\left(re^{it}\right) - \alpha(r)\right|\:dt,$$ where $\alpha(r)$ is any function of $r$.
Where does the "any function of $r$" come in? I understand that I am supposed to perform a convolution with a harmonic function of the form $$g(re^{i\theta}) = \sum_{n\neq 0} a_nr^{|n|}e^{in\theta} = \mathcal{P}(re^{i\theta}) - h(re^{i\theta}),$$ but it would seem that $h(re^{i\theta})$ would be a constant (i.e., the $a_0$ term) rather than a general function of $r$. Help?
Let $h(r)$ some function of $0 \le r <1$ and fix $r<1$ and let $g_r(z)=P(z)-h(r)$ where $P$ is the Poisson Kernel and $z=\rho e^{i\theta}$; if $f$ is as in the hypothesis, $f*g_r=f$, so if $ z=r_1e^{i\theta}, r_1<r$ we get:
$2\pi|f(z)| =2\pi|f*g_r(z)|=|\int_0^{2\pi}f(\frac{r_1}{r}e^{i(\theta-t)})(P(re^{it})-h(r))dt| \le \int_0^{2\pi}|P(re^{it})-h(r))|dt$
Letting now $r_1 \to r$ gives the required inequality; so yes the convolution is with $g_r(z)=P(z)-c$ but $c=c(r)$ so in other words we do different convolutions (convolutions with different functions $g_r$ which are all harmonic in the unit disc) for different $r$'s if that makes things clearer