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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.

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  • $\begingroup$ Sorry, I left a reference (to Rudin's book) as an answer without properly reading your question, and in particular the fact that you wanted a proof avoiding functional-analytic tricks $\endgroup$
    – Yemon Choi
    Commented Jan 19, 2011 at 19:12
  • $\begingroup$ Since it seems that using the open mapping theorem is allowed, perhaps the reference to Rudin might still be of interest? It's Theorem 5.2 in Functional Analysis (2nd ed) $\endgroup$
    – Yemon Choi
    Commented Jan 20, 2011 at 1:46
  • $\begingroup$ Also check problem 349 of Bernard Gelbaum's Problems in Analysis (Springer 1982) , chapter on Hilbert spaces, where it is also shown by elementary inequalities: "If $\|f\|_\infty\le C\|f\|_2$ for all $f\in M$, $M$ a subspace of $L^2([0,1])$, then $\rm{dim}(M)\le C^2$". $\endgroup$
    – Pietro Majer
    Commented Jul 10, 2017 at 19:56
  • $\begingroup$ Also see Borel, Automorphic forms on SL2(R), Lemma 8.3, where it is stated that this result is due to Godement. $\endgroup$
    – Bart Michels
    Commented Jul 9, 2018 at 16:33

3 Answers 3

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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 $n \leq C^2$.

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    $\begingroup$ I like this!!!! $\endgroup$
    – Nate Eldredge
    Commented Jan 19, 2011 at 17:06
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    $\begingroup$ Many thanks! This is actually the trick to extend the solution from Folland from $C[0,1]$ to $L^\infty$. $\endgroup$
    – Rostyslav Kravchenko
    Commented Jan 19, 2011 at 18:44
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    $\begingroup$ This is the "right" elementary solution and is what the makers of the qualifying exam had in mind. At a deeper conceptual level, the problem is obvious (the 2-summing norm of the identity operator on any $n$-dimensional space is $\sqrt {n}$). $\endgroup$
    – Bill Johnson
    Commented Jan 19, 2011 at 22:53
  • $\begingroup$ [Part 1] I seem to be stuck at a technical difficulty. The inequality $a_{1} f_{1}( x) \cdots +a_{n} f_{n}( x) \leq C\sqrt{|a_{1} |^{2} +\cdots +|a_{n} |^{2}}$ holds for all $x\in E_a$, where $E_a$ is some full measure set. The important point is that $E_a$ depends on $\displaystyle a_{i}$'s. Now when we put $a_{i} =\overline{f_{i}( y)}$ we have $\overline{f_{1}( y)} f_{1}( x) +\cdots +\overline{f_{n}( y)} f_{n}( x) \leq C\sqrt{|f_{1}( y) |^{2} +\cdots +|f_{n}( y) |^{2}}$. This holds for all $\displaystyle x\in E_{y}$ where $\displaystyle E_{y}$ has full measure. $\endgroup$
    – Mohith Nagaraju
    Commented Nov 1, 2022 at 14:07
  • $\begingroup$ [Part 2] From this we want to conclude that $\overline{f_{1}( x)} f_{1}( x) +\cdots +\overline{f_{n}( x)} f_{n}( x) \leq C\sqrt{|f_{1}( x) |^{2} +\cdots +|f_{n}( x) |^{2}} \text{ for almost all } x.$ This might actually be completely false. For instance, it can be the case that $y\notin E_{y}$. $\endgroup$
    – Mohith Nagaraju
    Commented Nov 1, 2022 at 14:07
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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.

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    $\begingroup$ Could you give a hint on how you can check $\|f\|_{\infty} \leq C \|f\|_{2}$ without appealing to the open mapping theorem (or closed graph theorem)? $\endgroup$
    – Theo Buehler
    Commented Jan 19, 2011 at 16:45
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    $\begingroup$ In fact I was not very honest: the proof I had in mind was more or less a proof of the closed graph theorem (see arxiv.org/abs/1005.1585). Anyway, here it goes. A first attempt would be to take a sequence $g_n$ in the unit ball of $E$ such that $\|g_n\|_\infty > 4^n$, consider $f=\sum_n \omega_n 2^{-n} g_n$ with a good choice of $\omega_n\in \mathbb C$ with $|\omega_n|=1$, and hope that $f$ is not bounded. I do not believe that this works in general, but if the $g_n$ are chosen carefully this will work... $\endgroup$
    – Mikael de la Salle
    Commented Jan 19, 2011 at 22:05
  • $\begingroup$ ... In fact $\|g_n\|_\infty>4^n$ means that there is $A_n \subset [0,1]$ of positive measure such that the (modulus of the) average $E[g|A_n]$ of $g_n$ on $A_n$ is greater than $4^n$. Changing $g_n$ if necessary, we can assume that $g_n$ is "the worst" such function, ie that $E[g_n|A]= c_n > 4^n$, and that $|E[g|A]| \leq (c_n+1) \|g\|_2$ for any $g \in E$. So define $f_0=g_0$ and $f_{n+1} = f_n+ 2^{-n} e^{-i\theta_n} g_{n+1}$, where $\theta_{n}$ is an argument of $E[f_n|A_{n+1}]$, so that $|E[f_{n+1}|A_{n+1}]|\geq c_n/2^n$... $\endgroup$
    – Mikael de la Salle
    Commented Jan 19, 2011 at 22:15
  • $\begingroup$ Hum, I noticed that I already got it wrong: I should have replaced the $2^{-n}$ by $3^{-n}$ in the definition of $f_{n+1}$, and I would conclude that $|E[f_n|A_n]|\geq c_n 3^{-n+1}$. Then $f=\lim f_n$ belongs to $E$, and $\|f-f_n\|_2 \leq 3^{-n+1}/2$, so that $|E[f|A_n]| \ge |E[f_n|A_n]|-(c_n+1)\|f_f_n\|_2 \ge 3^{-n+1}(c_n -1)/2$. This goes to $\infty$, and proves that $f$ is not in $L^\infty$. $\endgroup$
    – Mikael de la Salle
    Commented Jan 19, 2011 at 22:38
  • $\begingroup$ Thank you very much, I think I see what you're saying and I'll have to have a closer look at Sokal's paper which looks really neat! While trying to answer the original question I came up with something similar as you're sketching now and I was afraid that a substantially simpler argument escaped me - I interpreted using the open mapping theorem as "cheating" from the PS in the question... $\endgroup$
    – Theo Buehler
    Commented Jan 19, 2011 at 23:11
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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.

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    $\begingroup$ Oops, Mikael beat me. $\endgroup$
    – Nate Eldredge
    Commented Jan 19, 2011 at 15:44
  • $\begingroup$ It is funny that we had exactly the same construction. I just do not follow your use of the pigeohole principle. $\endgroup$
    – Mikael de la Salle
    Commented Jan 19, 2011 at 16:29
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    $\begingroup$ Hm, in retrospect, neither do I. $\endgroup$
    – Nate Eldredge
    Commented Jan 19, 2011 at 17:07
  • $\begingroup$ @Nate, I think that you are taking $A = \limsup \{f_n > 1/2\}$ and observing that $m(A) \ge \limsup m(f_n > 1/2) \ge 1/8C^2$. $\endgroup$
    – LSpice
    Commented Jan 19, 2011 at 22:35
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    $\begingroup$ @L Spice: That's what I was thinking, but unfortunately, $A = \limsup \\{ f_n > 1/2\\}$ doesn't quite do the trick. It gives us that for each $x \in A$, there is some sequence $f_{n_k}$ with $f_{n_k}(x) > 1/2$ for all $n_k$; unfortunately, the particular sequence that works may vary with $x$, and it is not clear to me that there is a single sequence that works for all $x \in A$ (or even a positive-measure subset). $\endgroup$
    – Nate Eldredge
    Commented Jan 19, 2011 at 23:06

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