Spring 2009 - Problem 2

Solution.

1. Fix $x \in B$. If $x \in S$, then this is clear: set $x_n = x$ for all $n \geq 1$. So, from now on, we may assume that $\norm{x} < 1$.

   Consider the set $P$ of orthonormal linearly independent sets in $H$ containing $x/\norm{x}$ partially ordered by inclusion. This is non-empty since $\set{x/\norm{x}} \in P$. Moreover, given a chain $\mathcal{C}$, $\mathcal{B} \coloneqq \bigcup_{B \in \mathcal{C}} B \in P$ since $\mathcal{C}$ is totally ordered. Indeed, given a finite subcollection $v_1, \ldots, v_n \in \mathcal{B}$, then there exists $B \in \mathcal{C}$ such that $v_1, \ldots, v_n \in B$, which implies that any finite subcollection in the union is orthonormal. Moreover, every set contains $x/\norm{x}$, so $\mathcal{B} \in P$. By Zorn's lemma, there exists a maximal element $E = \set{x/\norm{x}, e_1, e_2, \ldots}$.

   Indeed, $E$ has infinitely many elements since $H$ is infinite dimensional. Otherwise, suppose that $E$ is finite with $n$ elements, which means that there exists $v \in H$ outside the closure of the span of $E$. Consider

   $$v' = v - \inner{v, \frac{x}{\norm{x}}}\frac{x}{\norm{x}} - \sum_{i=1}^n \inner{v, e_i}e_i,$$

   which is in $H$ since $H$ is a vector space. By orthogonality,

   $$\begin{aligned} \inner{v', e_k} &= \inner{v, e_k} - \inner{v, \frac{x}{\norm{x}}}\inner{\frac{x}{\norm{x}}, e_k} - \sum_{i=1}^n \inner{v, e_i}\inner{e_i, e_k} \\ &= \inner{v, e_k} - \inner{v, e_k} \\ &= 0. \end{aligned}$$

   Similarly, $\inner{v', x/\norm{x}} = 0$ as well, but this means that $E \cup \set{v'}$ is an orthonormal set containing $x/\norm{x}$ with a strictly larger span than $E$, which contradicts maximality of $E$. Hence, no $v$ could have existed to begin with, so $E$ must be infinite.

   Set $x_n = x + \sqrt{1 - \norm{x}^2}e_n$, and notice that by orthonormality,

   $$\norm{x_n}^2 = \norm{x}^2 + \p{1 - \norm{x}^2}\norm{e_n}^2 = 1 \implies x_n \in S.$$

   Moreover, given any $y \in H$, we have

   $$\inner{x_n - x, y} = \sqrt{1 - \norm{x}^2}\inner{e_n, y}.$$

   Observe that

   $$\begin{aligned} 0 &\leq \norm{y - \sum_{i=1}^n \inner{y, e_i}e_i}^2 \\ &= \norm{y}^2 - 2\sum_{i=1}^n \abs{\inner{y, e_i}}^2 + \sum_{i=1}^n \abs{\inner{y, e_i}}^2 \\ &= \norm{y}^2 - \sum_{i=1}^n \abs{\inner{y, e_i}}^2 \\ \implies \sum_{i=1}^n \abs{\inner{y, e_i}}^2 &\leq \norm{y}^2, \end{aligned}$$

   and by sending $n \to \infty$, we get

   $$\sum_{i=1}^\infty \abs{\inner{y, e_i}}^2 \leq \norm{y}^2 < \infty,$$

   so $\inner{y, e_n} \xrightarrow{n\to\infty} 0$ by the divergence test for series. Thus, $\inner{x_n, y} \xrightarrow{n\to\infty} \inner{x, y}$, which was what we wanted to show.

2. Inspired by the result of part (1), let $\set{e_1, e_2, \ldots}$ be an orthonormal set and define $\func{T_n}{H}{H}$ by $x \mapsto \inner{x, e_n}e_n$. Since inner products a bilinear, it follows that $T_n$ is linear. Moreover, by Cauchy-Schwarz,

   $$\norm{T_n\p{x}} = \norm{\inner{x, e_n}e_n} = \abs{\inner{x, e_n}} \leq \norm{x}^2$$

   and this maximum is achieved via $T_n\p{e_n} = \norm{e_n}^2e_n = e_n$. Thus, $\norm{T_n} = 1$, and by the same inequality as above,

   $$\sum_{i=1}^\infty \norm{T_n\p{x}}^2 = \sum_{i=1}^\infty \abs{\inner{x, e_i}}^2 \leq \norm{x}^2 < \infty,$$

   so $T_n\p{x} \xrightarrow{n\to\infty} 0$ for any $x \in H$.