<Home>Qualifying Exams><Analysis>Spring 2016 - Problem 6

Spring 2016 - Problem 6

Hilbert spaces

Suppose that $\set{\phi_n}_n$ is an orthonormal system of continuous functions in $L^2\p{\br{0,1}}$ and let $S$ be the closure of the span of $\set{\phi_n}_n$ . If $\sup_{f \in S \setminus \set{0}} \frac{\norm{f}_{L^\infty}}{\norm{f}_{L^2}}$ is finite, prove that $S$ is finite dimensional.

Solution.

Suppose $\set{f_1, \ldots, f_n}$ are linearly independent continuous functions in $S$ . Via Gram-Schmidt, we may assume without loss of generality that this set is orthonormal with respect to the $L^2\p{\br{0,1}}$ inner product.

Let $C = \sup_{f \in S \setminus \set{0}}\frac{\norm{f}_{L^\infty}}{\norm{f}_{L^2}}$ so that $\norm{f}_{L^\infty} \leq C\norm{f}_{L^2}$ For $a_1, \ldots, a_n \in \C$ , observe that for any $x \in \br{0, 1}$ , we have

\abs{\sum_{i=1}^n a_if_i\p{x}} \leq \norm{\sum_{i=1}^n a_if_i}_{L^\infty} \leq C \p{\sum_{i=1}^n \abs{a_i}^2}^{1/2}

by assumption. Thus, applying the inequality for each $x$ with $a_i\p{x} = \conj{f_i\p{x}}$ , we see

\sum_{i=1}^n \abs{f_i\p{x}}^2 \leq C\p{\sum_{i=1}^n \abs{f_i\p{x}}^2}^{1/2} \implies \sum_{i=1}^n \abs{f_i\p{x}}^2 \leq C^2.

Integrating both sides and using the fact that the $f_i$ are orthonormal, we get $n \leq C^2$ . Thus, the number of linearly independent elements in $S$ is bounded by $C^2$ , which is finite, so $S$ must be finite dimensional.