The Poisson kernel is a boundary approximate identity

Let $\epsilon >0$. By uniform continuity of $\phi$ on the compact unit circle, choose $\delta \in (0,\pi ]$ such that $\mathrm{\mid }\phi (e^{it})-\phi (e^{i\alpha })\mathrm{\mid }<\epsilon \mathrm{/}2$ whenever the circular distance from $t$ to $\alpha$ is less than $\delta$.

Writing $z=r{e}^{i\alpha }$ and subtracting $\phi (e^{i\alpha })$ inside the Poisson integral, positivity and total mass one from [L1] give $\mathrm{\mid }P[\phi ](z)-\phi (e^{i\alpha })\mathrm{\mid }\le \frac{1}{2\pi }{\int }_{\mathrm{\mid }t-\alpha \mathrm{\mid }<\delta }{P}_r(\alpha -t)\frac{\epsilon }{2}dt+\frac{1}{2\pi }{\int }_{\mathrm{\mid }t-\alpha \mathrm{\mid }\ge \delta }{P}_r(\alpha -t)2\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}dt\mathrm{.}$

The first integral in step 2.1 is at most $\epsilon \mathrm{/}2$ because the kernel mass is $1$, and the second is at most $2\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}$ times the far-arc mass from [L1], which is $<\epsilon \mathrm{/}2$ for all $\alpha$ once $r$ is close enough to $1$. Therefore $\mathrm{\mid }P[\phi ](r{e}^{i\alpha })-\phi (e^{i\alpha })\mathrm{\mid }<\epsilon$ uniformly in $\alpha$.

Since $\epsilon$ was arbitrary, the convergence is uniform as $r\to 1^{-}$.