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.



Subspace of $L^2$ that lies in $L^\infty$
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 $n \leq C^2$.


