Analysis Problem of the Day 47

Today’s problem is Problem 11 from UCLA’s Fall 2001 Analysis Qual:

Problem 11: Let $F \in H(\mathbb{D}) \cap C(\overline{\mathbb{D}}).$

a) Prove $\lim_{\lambda \to 1} F(\lambda z) = F(z)$ uniformly on $\overline{\mathbb{D}}.$

b) $F(z) = \sum_{n=0}^\infty a_n z^n$ on $\mathbb{D}.$ Prove that $\lim_{n \to \infty} a_n = 0.$

Solution: a) Since $F$ is continuous on a compact set, it is uniformly continuous, i.e. for any $\epsilon>0$ there exists a $\delta>0$ such that $|z-w|<\delta \implies |F(z)-F(w)|<\epsilon.$ Now, since $|\lambda z-z|<\delta$ when $|\lambda-1|<\delta,$ it follows that $|F(\lambda z)-F(z)|<\epsilon$ whenever $|\lambda-1|<\delta,$ i.e. $\lim_{\lambda \to 1} F(\lambda z) = F(z)$ uniformly on $\overline{\mathbb{D}}.$

b) Setting $z=r e^{i\theta},$ we get by convergence of holomorphic functions on their disc of convergence that for every $0 < r < 1,$

$g_r(e^{i\theta}):=F(re^{i\theta}) = \sum_{n=0}^\infty r^n a_n e^{i n \theta},$

which is a continuous $2\pi$ -periodic function that therefore admits a Fourier series. By uniqueness of Fourier coefficients, it follows $\widehat{g_r}(n)= r^n a_n.$ Let $g_1(e^{i\theta}):=F(e^{i\theta}).$ By a), $g_r \to g_1$ uniformly, so since the Fourier transform is a bounded operator from $L^1$ to $l^\infty,$ it follows that $g_r \to g_1$ in $L^1,$ i.e. $\widehat{g_r} \to \widehat{g_1}$ in $l^\infty.$ But $\widehat{g_r}(n) \to 0$ as $n \to \infty,$ so $\widehat{g_1}(n) \to 0$ as $n \to \infty.$ But $\widehat{g_r}(n) \to a_n$ pointwise as $r \to 1,$ so by uniqueness of limits, it follows that $\widehat{g_1}(n)=a_n \to 0,$ i.e. $\lim_{n \to \infty} a_n=0.$

Leave a Reply Cancel reply