Analysis Problem of the Day 38

Today’s problem appeared as Problem 6 on the UCLA Fall 2010 Analysis Qual:

Problem 6: Consider the Hilbert space $H$ of functions $f: \overline{\mathbb{D}} \to \mathbb{C}$ given by $f(z) = \sum_{k=0}^\infty \widehat{f}(k) z^k$ with $\|f\|^2:= \sum_{k=0}^\infty (1+|k|)^2|\widehat{f}(k)|^2<\infty.$

a) Prove that the linear functional $L: f \to f(1)$ is bounded.

b) Find the element $g$ such that $L(f)=\langle f,g \rangle.$

c) Show that $f \to \text{Re }L(f)$ achieves a unique maximum on $\{\|f\| \leq 1, f(0)=0\}$ and find its value.

Solution: a) By Cauchy-Schwarz,

$|f(1)| = |\sum \widehat{f}(k)(1+|k|^2)^{\frac12}(1+|k|^2)^{-\frac12}|\leq \sqrt{\sum (1+|k|^2)|\widehat{f}(k)|^2} \sqrt{\sum (1+|k|^2)^{-1}} \leq C\|f\|,$

with $C \leq \frac{\pi}{\sqrt{6}}.$ Thus, $L$ is bounded.

b) Note that $\langle f,g \rangle = \sum \langle k \rangle \widehat{f}(k) \overline{\langle k \rangle \widehat{g}(k)},$ where the Japanese bracket is defined by $\langle k \rangle :=(1+|k|^2)^{\frac12}.$ Thus, by the Riesz Representation Theorem, if $\langle f,g \rangle = L(f) = f(1) = \sum \widehat{f}(k),$ then $(1+|k|^2) \widehat{g}(k)=1,$ i.e. $g(z) = \sum_{k=0}^\infty (1+|k|^2)^{-1} z^k$ (a more explicit formula is likely impossible).

c) Note that by the Cauchy-Schwarz argument above, $|\text{Re }f(1)| \leq |f(1)| \leq \sum_{k=1}^\infty (1+|k|^2)^{-1}$ when $f(0) = \widehat{f}(0)=0.$ It follows that equality occurs if and only if $\widehat{f}(k) = (1+|k|^2)^{-1},$ i.e. the maximum is uniquely achieved and equals to $\sum_{k=1}^\infty (1+|k|^2)^{-1}.$

Remark: Interpreting these as holomorphic functions on the unit disc, the Hilbert space in question is the space of holomorphic functions on the unit disc that admit an extension $f$ to $S^1$ with $\|f\|_2^2+\|f'\|_2^2 < \infty,$ i.e. $f \in H^1(S^1).$

Leave a Reply Cancel reply