Fall 2015 - Problem 4

Solution.

1. Let $x \in \mathcal{H}$ be such that $\inner{x, e_n} = 0$ for all $n \geq 1$. Since $\set{f_n}_n$ is complete, we can write $x = \sum_{n=1}^\infty \inner{x, f_n}f_n$. To make use of the condition given, set $y = \sum_{n=1}^\infty \inner{x, f_n}e_n$, which gives

   $$\begin{aligned} \norm{x - y}^2 &= \norm{\sum_{n=1}^\infty \inner{x, e_n}\p{e_n - f_n}}^2 \\ &\leq \p{\sum_{n=1}^\infty \abs{\inner{x, f_n}}\norm{e_n - f_n}}^2 \\ &\leq \p{\sum_{n=1}^\infty \abs{\inner{x, f_n}}^2} \p{\sum_{n=1}^\infty \norm{e_n - f_n}^2} && \p{\text{Cauchy-Schwarz}} \\ &\leq \norm{x}^2. \end{aligned}$$

   But by assumption, $\inner{x, e_n} = 0$ for all $n \geq 1$, which implies that $\inner{x, y} = 0$. Hence, by the Pythagorean theorem,

   $$\norm{x}^2 + \norm{y}^2 \leq \norm{x}^2 \implies y = 0,$$

   and so $\inner{x, f_n} = 0$ for all $n \geq 1$ as well. Since $\set{f_n}_n$ was complete, this implies that $x = 0$, so $\set{x_n}_n$ is complete as well.

2. For $N \in \N$, let $E_N = \cl{\span{\set{e_N, e_{N+1}, \ldots}}}$ and $F_N = \cl{\span{\set{f_N, f_{N+1}, \ldots}}}$. Since these are closed subspaces, we have orthogonal projections $\pi_{E_N}$ and $\pi_{V_N}$, respectively. Then for any $x \in \mathcal{H}$,

   $$\begin{aligned} \norm{\pi_{E_N}\p{x} - \pi_{V_N}\p{x}} &= \norm{\sum_{n=N}^\infty \inner{x, e_n}e_n - \sum_{n=N}^\infty \inner{x, f_n}f_n} \\ &\leq \norm{\sum_{n=N}^\infty \inner{x, e_n}\p{e_n - f_n}} + \norm{\sum_{n=N}^\infty \inner{x, e_n - f_n}f_n} \\ &\leq \sum_{n=N}^\infty \abs{\inner{x, e_n}} \norm{e_n - f_n} + \p{\sum_{n=N}^\infty \abs{\inner{x, e_n - f_n}}^2}^{1/2} \\ &\leq \p{\sum_{n=N}^\infty \abs{\inner{x, e_n}}^2}^{1/2} \p{\sum_{n=N}^\infty \norm{e_n - f_n}^2}^{1/2} + \p{\sum_{n=N}^\infty \norm{x}^2 \norm{e_n - f_n}^2}^{1/2} && \p{\text{Cauchy-Schwarz}} \\ &\leq 2\norm{x} \p{\sum_{n=N}^\infty \norm{e_n - f_n}^2}^{1/2}. && \p{\text{Bessel's inequality}} \end{aligned}$$

   We will show that for large values of $N$, $E_N$ and $F_N$ intuitively capture the same amount of information. More precisely, we will show that because $F_N^\perp$ is finite dimensional ($\set{f_n}_n$ is a complete orthonormal system), this forces $E_N^\perp$ to be finite dimensional as well.

   Hence, $\norm{\pi_{E_N} - \pi_{F_N}} \to 0$ as $N \to \infty$. Notice also that $\mathcal{H} = E_N \oplus E_N^\perp = F_N \oplus F_N^\perp$, which means $\pi_{E_N^\perp} = \id - \pi_{E_N}$ and $\pi_{F_N^\perp} = \id - \pi_{F_N}$. Thus,

   $$\norm{\pi_{E_N^\perp} - \pi_{F_N^\perp}} = \norm{\p{\id - \pi_{E_N}} - \p{\id - \pi_{F_N}}} = \norm{\pi_{E_N} - \pi_{F_N}}.$$

   Let $N$ be large enough so that $\norm{\pi_{E_N^\perp} - \pi_{F_N^\perp}} < \frac{1}{2}$. Since $F_N^\perp$ has dimension $N - 1$, any set $\set{x_1, \ldots, x_n}$ of $N$ elements must be linearly dependent after projecting to $F_N^\perp$. In particular, if $x_1, \ldots, x_n \in E_N^\perp$, then there exist $c_1, \ldots, c_n$ such that

   $$0 = \sum_{n=1}^N c_n\pi_{F_N}\p{x_n} = \pi_{F_N}\p{\sum_{n=1}^N c_nx_n}.$$

   Then by the triangle inequality,

   $$\begin{aligned} \norm{\sum_{n=1}^N c_nx_n} &= \norm{\pi_{E_N}\p{\sum_{n=1}^N c_nx_n}} \\ &\leq \norm{\p{\pi_{E_N} - \pi_{F_N}}\p{\sum_{n=1}^N c_nx_n}} + \norm{\pi_{F_N}\p{\sum_{n=1}^N c_nx_n}} \\ &\leq \norm{\pi_{E_N} - \pi_{F_N}} \norm{\sum_{n=1}^N c_nx_n} \\ &\leq \frac{1}{2}\norm{\sum_{n=1}^N c_nx_n}, \end{aligned}$$

   which implies that $\sum_{n=1}^N c_nx_n = 0$, i.e., $\set{x_1, \ldots, x_n}$ is linearly dependent in $E_N^\perp$. Since these vectors were arbitrary, it follows that $\dim{E_N^\perp} \leq N - 1$. But $E_N^\perp$ contain the orthonormal set $\set{e_1, \ldots, e_{N-1}}$, so $\dim{E_N^\perp} = N - 1$, i.e., $\set{e_1, \ldots, e_{N-1}}$ spans $E_N^\perp$, and so $\set{e_n}_n$ is a complete orthonormal system.