I'm reading a book on composition operators, and the author makes the following claim:
Given a self-map of the unit disc, and a $H^2$ function $f$, where $H^2$ is the Hardy space, if we fix a radius $r<1$ and define $f_m$ to be the $m$th partial sums in the power series expansion of $f$, then $$ \frac{1}{2 \pi} \int_0^{2\pi} |f(\phi(re^{i\theta}))|^2d\theta = \lim_{m \to \infty} \frac{1}{2 \pi} \int_0^{2\pi} |f_m(\phi(re^{i\theta}))|^2d\theta, $$ which follows from the fact that $f_m \circ \phi$ converges uniformly on compact subsets of the unit disc to $f \circ \phi$. And this is all well and good, but when the function becomes revalued by using the continuity of the modulus, that is, when we write $$ |f(\phi(re^{i\theta}))|^2 = \lim_{m \to \infty}|f_m(\phi(re^{i\theta}))|^2, $$ how are we guaranteed that the convergence is still uniform?
Any help is appreciated.