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Fall 2013 - Problem 11

Banach spaces

Consider the space $C\p{\br{0,1}}$ of real-valued continuous functions on the unit interval $\br{0, 1}$ . We denote by $\norm{f}_\infty = \sup_{x \in \br{0,1}} \abs{f\p{x}}$ the supremum norm and by $\norm{f}_2 = \p{\int_0^1 \abs{f\p{x}}^2 \,\diff{x}}^{1/2}$ the $L^2$ -norm of a function $f \in C\p{\br{0,1}}$ .

Let $S$ be a subspace of $C\p{\br{0,1}}$ . Show that if there exists a constant $K \geq 0$ such that $\norm{f}_\infty \leq K \norm{f}_2$ for all $f \in S$ , then $S$ is finite dimensional.

Solution.

Viewing $C\p{\br{0, 1}} \subseteq L^2\p{\br{0, 1}}$ , we have an inner product $\inner{f, g} = \int_0^1 f\p{x} g\p{x} \,\diff{x}$ for $f, g \in C\p{\br{0, 1}}$ .

Let $\set{f_1, \ldots, f_n} \subseteq C\p{\br{0, 1}}$ be linearly independent, and via Gram-Schmidt, we may assume without loss of generality that this set is orthonormal. For $a \in \R^n$ , define $\func{\phi_a}{\br{0,1}}{\R}$ ,

\phi_a\p{x} = \sum_{k=1}^n a_kf_k\p{x}.

Thus, by the Pythagorean theorem, $\norm{\phi_a}_{L^2}^2 = \norm{a}^2$ , and we also have $\phi_a \in S$ as a linear combination of the $f_k$ . By assumption,

\abs{\sum_{k=1}^n a_kf_k\p{x}} \leq \norm{\phi_a}_\infty \leq K\norm{\phi_a}_{L^2} = K\norm{a}.

If we fix $x \in \br{0, 1}$ and set $a_k = f_k\p{x}$ , we get

\sum_{k=1}^n f_k^2\p{x} \leq K\p{\sum_{k=1}^n f_k^2\p{x}}^{1/2} \implies \sum_{k=1}^n f_k^2\p{x} \leq K^2.

Integrating over $\br{0, 1}$ , we have

n \leq K^2,

so any linearly independent set can have at most $K^2$ elements, which completes the proof.