Question

Let $E$ be a closed subspace of $L^2[0,1]$. Suppose that $E\subset{}L^\infty[0,1]$. Is it true that $E$ is finite dimensional?

PS. This is actually a question from the real analysis qualifier. I came across it as I was teaching qualifier preparation course, and was solving problems from old qualifiers. So, though it might follow from some advanced theory of Banach spaces, I am most interested in the 'elementary' solution, using only methods from standard real analysis course. Note: if $E\subset{}C[0,1]$, then it is a problem from Folland, and there is a solution there. However, it does not work for $L^\infty$, not without some trick.

Answer 1

Another solution: as Mikael wrote, $||f||_{\infty} \leq C ||f||_2$ for every $f \in E$. Let $f_1,\ldots,f_n$ be an orthonormal family in your subspace. Then for every $x \in [0,1]$, $f_1(x)^2+\ldots+f_n(x)^2 \leq ||f_1(x)f_1+\ldots+f_n(x)f_n||_{\infty} \leq C \|f_1(x)f_1+\ldots+f_n(x)f_n\|_2$ $$=C \sqrt{f_1(x)^2+\ldots+f_n(x)^2},$$

and by squaring we get $f_1(x)^2+\ldots+f_n(x)^2 \leq C^2$, and integrating gives $n \leq C^2$.

Answer 2

Here's one solution. There may be cleaner ones.

Let $E$ be as supposed. The natural inclusion $T : L^\infty([0,1]) \hookrightarrow L^2([0,1])$ is bounded, so $E = T^{-1}(E)$ is therefore also closed in $L^\infty$. By the open mapping theorem, it follows that $T^{-1}$ is bounded on $E$, so there exists $C$ such that for all $f \in E$, $||f||_\infty \le C ||f||_2$. Now if $||f||_2 = 1$, we have $||f||_\infty \le C$, and so by noting $$1 = \int |f|^2 \le C^2 m(|f| > \epsilon) + \epsilon^2$$ and taking, say, $\epsilon = 1/2$, we have $m(|f| > 1/2) \ge 1/4C^2$.

Now suppose $E$ is infinite dimensional; then it contains an $L^2$-orthonormal sequence $\{f_n\}$. By replacing $f_n$ by $-f_n$ as necessary we may assume that for each $f_n$, $m(f_n > 1/2) \ge 1/8C^2$. By a pigeonhole argument there is a set $A$ of positive measure where $f_{n_k} > 1/2$ for infinitely many $n_k$ (edit: actually I am not sure about this step). Now $f = \sum_k k^{-1} f_{n_k}$ converges in $L^2$ and so $f \in E$; since the $L^2$ and $L^\infty$ norms are equivalent on $E$, the sum also converges uniformly to $f$ on a set of full measure. But this implies that $f = +\infty$ a.e. on $A$, which is absurd.

Answer 3

First note that by the closed graph theorem, there is a $C$ such that $\|f\|_\infty \leq C\|f\|_2$ for any $f$ iin $E$. (this can also be checked directly if you consider that the closed graph theorem is advanced Banach space theory).

Assume $E$ is infinite dimensional, and take $(f_n)_{n\ge 1}$ an orthonormal basis of $E$. Let $A_n$ be the subset of $[0,1]$ on which the real part of $f_n$ is greater than $1/10$. Replacing if necessary $f_n$ by $-f_n$ or $i f_n$ or $-if_n$, we can assume that $\|Re(f_n)^+\|_2 \ge 1/2$, so that the measure of $A_n$ is greater than some constant $\delta$ depending on $C$ only ($\delta = (1/2^2-1/10^2)/C^2$ works for example).

Since the integral of $1_{A_1}+1_{A_2}+...+1_{A_n}$ is greater than $n\delta$, there exist $i_1,...i_k$ with $k$ being the integer part of $n\delta$, such that $A_{i_1},...,A_{i_k}$ have non-trivial intersection. Then the sum $1/\sqrt k (f_{A_{i_1}}+...f_{A_{i_k}})$ has $L^2$ norm $1$, but $L^\infty$ norm greater than $\sqrt k/10$ (because its real part is greater than $\sqrt k /10$ on a non-trivial subset). A contradiction.