Fall 2010 - Problem 6

Solution.

1. By Cauchy-Schwarz,

   $$\begin{aligned} \abs{f\p{1}} &\leq \sum_{k=0}^\infty \abs{\hat{f}\p{k}}\sqrt{1 + k^2} \cdot \frac{1}{\sqrt{1 + k^2}} \\ &\leq \p{\sum_{k=0}^\infty \p{1 + k^2}\abs{\hat{f}\p{k}}^2}^{1/2} \p{\sum_{k=0}^\infty \frac{1}{1 + k^2}}^{1/2} \\ &= \p{\sum_{k=0}^\infty \frac{1}{1 + k^2}}^{1/2} \norm{f}, \end{aligned}$$

   and because the series converges, $L$ is bounded.

2. We assume that the inner product is given by

   $$\inner{f, g} = \sum_{k=0}^\infty \hat{f}\p{k}\conj{\hat{g}\p{k}} \p{1 + k^2}.$$

   If $g$ represents $L$, then it must satisfy

   $$\inner{f, g} = f\p{1} \iff \sum_{k=0}^\infty \hat{f}\p{k}\conj{\hat{g}\p{k}} \p{1 + k^2} = \sum_{k=0}^\infty \hat{f}\p{k}$$

   for any $f \in \mathcal{H}$. Thus, if we set

   $$g\p{z} = \sum_{k=0}^\infty \frac{z^k}{1 + k^2},$$

   then the identity is satisfied. This definition makes sense as the series converges on $\cl{\D}$, so it remains to show that $g \in \mathcal{H}$ by showing that $\norm{g} < \infty$. But this is clear:

   $$\norm{g}^2 = \sum_{k=0}^\infty \frac{1}{1 + k^2} < \infty,$$

   so $g \in \mathcal{H}$ and we are done.

3. Observe that if $0 < \norm{f} < 1$, then

   $$\Re{L\p{f}} < \frac{\Re{L\p{f}}}{\norm{f}} = \Re{L\p{\frac{f}{\norm{f}}}}.$$

   Thus, if the maximum is attained, it must occur with $\norm{f} = 1$. Also, notice that for $z \in \C$, $\Re{z} \leq \abs{z}$, and the maximum is attained if and only if $z$ is real. Thus, if there exists $k \geq 0$ such that $\hat{f}\p{k}$ is not real, then

   $$\Re{f\p{1}} = \sum_{k=0}^\infty \Re\hat{f}\p{k} < \sum_{k=0}^\infty \abs{\hat{f}\p{k}}.$$

   Thus, if we replace $\hat{f}\p{k}$ with $\abs{\hat{f}\p{k}}$, then the resulting function is still in $\mathcal{H}$ and gives a strictly larger value. Thus, the maximum must be attained by a function $f$ with $\hat{f}\p{k}$ real. Similarly, if we remove negative values of $\hat{f}\p{k}$, we get a strictly larger value, so the maximum also has non-negative coefficients. Hence, from now, we only need to consider $f$ with $\hat{f}\p{k} \geq 0$ for all $k \geq 0$ and $\norm{f} = 1$.

   Observe that with this restriction, $\Re{L\p{f}} = f\p{1}$. By the same argument as in (2), we also see that on $\mathcal{B}$, this functional is represented by

   $$g\p{z} = \sum_{k=1}^\infty \frac{z^k}{1 + k^2},$$

   i.e., the same $g$ as in (2), but with the constant term removed. Thus, by Cauchy-Schwarz,

   $$\Re{f\p{1}} = f\p{1} = \inner{f, g} \leq \norm{f}\norm{g} = \norm{g},$$

   since $\norm{f} = 1$. From here, we see that the maximum is attained uniquely by $f = \frac{g}{\norm{g}}$, so it remains to show that this choice of $f$ is in $\mathcal{B}$.

   It's clear that $\norm{f} = 1$, and

   $$f\p{0} = \frac{g\p{0}}{\norm{g}} = 0$$

   by construction, so $f \in \mathcal{H}$. Finally, the maximal value is given by

   $$\frac{g\p{1}}{\norm{g}} = \p{\sum_{k=1}^\infty \frac{1}{1 + k^2}}\p{\sum_{k=1}^\infty \frac{1}{1 + k^2}}^{-1/2} = \p{\sum_{k=1}^\infty \frac{1}{1 + k^2}}^{1/2},$$

   which completes the proof.