Power series on the unit disc Consider a power series $P(z) = 1 + a_1 z^2 + a_2 z^4+ a_3z^6 . . .$. It is given that $P$ converges in the open unit disc centered at the origin. Also, for any $\omega$ such that $|\omega|=1$, we have that $\lim_{z \to \omega} |P(z)| = 1$. Can we conclude that $a_i=0$ for all $i\geq 1$?
Intuitively this seems to be true but I can't find a proof or a counterexample.
 A: Let $D$ be the open unit disc. Choose $z_n\in D$ such that $|P(z_n)| \to \sup_{D} |P|.$ Then there is a subsequence $z_{n_k}$ converging to some $w\in \overline D.$ If $w \in D,$ then $|P|$ has an absolute maximum at $w.$ By the maximum modulus theorem, $P$ is constant in $D,$ giving $a_n=0, n>0.$ If $|w|=1,$ then by hypothesis, $|P(z_{n_k})| \to 1.$ But $P(0) = 1.$ Again by the MMT, $P$ is constant and we have $a_n=0, n>0.$ (I'm not sure why we assume the power series has only even powers; I didn't use it.)
A: Another (cruder) proof is to observe that $v_r(\varphi) = u(re^{i\varphi})=\Re \log P(re^{i\varphi})$ converges uniformly to $0$ as $r\to 1$. But as $v_r$ is subharmonic this leads to $v_r < 0$. This implies that if $P(z) \ne 1$ we would have $a_1z^2 + a_2z_4 + a4_z6 + ...$ in the left half plane and by continuity this applies for all $z$ in an neighbourhood. But this can't be because it would mean that the average would be left of the origin which would imply that it should have a constant term.
