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I wish to show that any closed subspace $M$ of $L^{2}[0,1]$ consisting of continuous real-valued functions is finite-dimensional (which amazes me quite a bit, I must admit).

My approach proceeds as follows:

Step 0. Note that $L^{2}[0,1]$ is a separable Hilbert space with the usual scalar product on $L^2$. Moreover, since $M$ is closed, it is separable Hilbert, too, with inner product of $L^2([0,1])$ restricted to $M$.

Step 1. Let $I: (M, \| \cdot \|_{\infty}) \to (M, \| \cdot \|_{2})$ be the identity operator on $M$. Since $I$ is bijective, by the open mapping theorem, $I^{-1}$ exists in the space of bounded linear operators on $M$. Hence, for any $f \in M$, we have $$ \| f\|_{\infty} = \| I^{-1} (f)\|_{\infty} \leq C \|f \|_{2} \quad \forall f \in M,$$ where $C := \| I^{-1} \|.$

Step 2. Now, let $t \in [0,1]$ and consider the linear functional $\ell_{t}: M \to \mathbb{R}, f \mapsto f(t).$ As $M$ is Hilbert, we may apply Riesz representation theorem to deduce that $\ell_{t} = \langle \cdot, g_t \rangle =: \ell_{g_t}$ for some $g_t \in M$ and $\|\ell_{g_t}\| = \| g_t\|_{2}$. Moreover, $$ f(t) = l_t (f) = \langle f, g_t \rangle \quad \forall t \in [0,1].$$ Using step 1, we may further conclude that $\| g_t \|_{2} \leq C$.

Step 3. By seperability of $M$, we may suppose that $S = \{h_{n}: n \in \mathbb{N}\}$ is a orthonormal basis of $M$. Then, Parseval's identity yields

$$ \infty > \|g_t\|_{2} = \sum_{h \in S} | \langle h, g_t \rangle |^{2} = \sum_{h \in S} |h(t)|^{2} $$

I am stuck in the last line. I want to argue that this sum has to be finite. Can we bound $|h(t)|$ from below? I guess, this suffices to conclude, right? Does this argument work?

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  • $\begingroup$ I am not sure if this problem is formulated correctly. What does it mean that $M$ contains continuous functions? All continuous functions are dense in $L^{2}$ so we probably think about some subspace of continuous functions. Most likely, you want to prove that a subspace $M$ of $L^{2}$ where $||.||_{\infty}$ and $L^{2}$ norms are equivalent is finite dimensional. $\endgroup$
    – Salcio
    Feb 6, 2021 at 1:33
  • $\begingroup$ The conclusion even holds when $M\subset L^{\infty}([0,1])$. $\endgroup$ Feb 7, 2021 at 12:17
  • $\begingroup$ @Jonathan, if $M\subseteq L^\infty$, how would you define $l_t$? $\endgroup$
    – Ruy
    Feb 7, 2021 at 12:24
  • $\begingroup$ @Ruy, one has to do something different, see my answer $\endgroup$ Feb 7, 2021 at 14:13
  • $\begingroup$ related math.stackexchange.com/q/4553495/121671 $\endgroup$
    – Mittens
    Nov 1, 2022 at 15:19

2 Answers 2

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From the last line $$ C^2≥ \|g_t\|^2_{2} = \sum_{h \in S} | \langle h, g_t \rangle |^{2} = \sum_{h \in S} |h(t)|^{2} $$ so \begin{align} \infty &> \int_0^1 C^2 \, dx\\ &\geq \int_0^1 \sum_{h \in S} |h(t)|^{2} \, dx\\ &= \sum_{h \in S} \int_0^1 |h(t)|^{2} \, \\ &= \sum_{h \in S} \|h\|^2 = \#S. \end{align}


Here is an interesting consequence of the above fact:

It is known that for a bounded operator $T:H\to K$, between Hilbert spaces $H$ and $K$, the following are equivalent:

  1. $T$ is compact,

  2. the range of $T$ has no infinite dimensional closed subspaces.

While thinking about the present question I first tried to answer it by applying the above fact to the inclusion map $$ T:C([0,1])\to L^2([0,1]), $$ so it would be enough to show that $T$ satisfies (1), above. However I later noticed that $T$ is NOT compact, the reason being that the functions $$ e_n(x) = \sqrt2\sin(2\pi nx), \quad x\in [0,1], $$ form a bounded set in $C([0,1])$, but they do not admit a convergent sub-sequence in $L^2([0,1])$, as they form an orthonormal set in the latter space.

Of course I was wrong all along, as the equivalence between (1) and (2) is stated above only for Hilbert spaces!

While it is not hard to show that (1) implies (2) for all Banach spaces, the present question and answer provide a counter example for (2) $\Rightarrow$ (1) in the context of Banach spaces, as we now know that the range of $T$ contains no infinite dimensional closed subspaces, and yet $T$ is not compact!

Summarizing, the point I want to make is this:

There are non-compact bounded operators between Banach spaces whose range contains no infinite dimensional subspaces, even though no such operators exist between Hilbert spaces.

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  • $\begingroup$ Neat! It does not really make a difference for the conclusion but why does an inequality pop up when you exchange sum and integral. Is this not an equality by monotone convergence? $\endgroup$
    – ferhenk
    Feb 6, 2021 at 10:15
  • $\begingroup$ The inequality comes from the previous line. I've edited for clarification. $\endgroup$
    – Ruy
    Feb 6, 2021 at 11:25
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    $\begingroup$ Thanks for a very interesting question, and also for more than 90% of the answer! $\endgroup$
    – Ruy
    Feb 6, 2021 at 11:27
  • $\begingroup$ Oh, sure! That makes sense. $\endgroup$
    – ferhenk
    Feb 6, 2021 at 11:33
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    $\begingroup$ @StratosFair, by Riesz $$\|g_t\|_2=\|l_t\|=\|\delta_t\circ I^{-1}\|\leq$$$$\|\delta_t\|\|I^{-1}\|=C, $$ where $\delta_t$ is the evaluation functional on $C([0,1])$. $\endgroup$
    – Ruy
    Oct 18, 2022 at 23:48
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In fact, if $M$ is a closed subspace of $L^2(X)$ for some finite measure space $X$, and $M\subset L^{\infty}(X)$, then $M$ is finite dimensional.

First note that step $1$ doesn't change.: we still have $\|f\|_2\leq \mu(X)^{1/2}\|f\|_{\infty}$ for $f\in L^{\infty}(X)$ which 1) guarantees that $M$ is Banach with the $\|\cdot\|_\infty$ norm and 2) that the identity map $(M,\|\cdot\|_\infty)\rightarrow (M,\|\cdot\|_2)$ is continuous. So the bounded inverse theorem gives a constant $C$ such that $\|f\|_\infty\leq C\|f\|_2$, $f\in M$.

For the remainder of the proof we have to do somehting different. Take an orthonormal system $\{f_k\}_{k=1}^n$ in $M$. Then for any $c_1, \ldots, c_n \in \mathbb{C}$ the inequality $\|f\|_\infty\leq C\|f\|_2$ for the function $f=c_1 f_1+ c_2f_2+\ldots +c_n f_n$ means that $$|c_1f_1(x)+\ldots+c_nf_n(x)|^2\leq C^2(|c_1|^2+\ldots +|c_n|^2)\; \; \; \; \; \;(1)$$ for a.e. $x$. Choosing a countable dense subset $\Omega\subset \mathbb{C}^n$ we can find a measurable set $Y$ with $\mu(X\backslash Y)=0$ such that $(1)$ holds for all $x\in Y$ and $(c_1, \ldots, c_n)\in \Omega$. By continuity $(1)$ then holds for all $y\in Y$ and any scalars $c_1, \ldots, c_n \in \mathbb{C}$. In particular we can set $c_k=\overline{f_k(x)}$ to get $$|f_1(x)|^2+\ldots +|f_n(x)|^2\leq C^2$$ for all $x\in Y$. Now we simply integrate both sides over $X$ (or, equivalently, over $Y$) to get $n\leq C^2\mu(X)$. Hence we have a uniform bound on the size of any orthonormal system in $M$ which implies that $M$ is finite dimensional.

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  • $\begingroup$ Hi @Jonathan, I stopped reading your answer at the second paragraph because I realized I do not understand why is $M$ complete with $\|\cdot\|_\infty$. In fact I now realize that the same problem is also present in the OP's argument, perhaps putting everything at risk! Could you please clarify this point? $\endgroup$
    – Ruy
    Feb 7, 2021 at 14:28
  • $\begingroup$ Hi @Ruy, if we have a Cauchy sequence $\{f_n\}$ in $M$ with respect to the $\|\cdot\|_\infty$ norm then it converges to a function $f\in L^\infty(X)$ in $\|\cdot\|_\infty$ norm. Now the inequality $\|f-f_n\|_2\leq \mu(X)^{1/2}\|f-f_n\|_\infty$ shows that $f_n\rightarrow f$ also in $\|\cdot\|_2$ norm. Since $M$ is closed with respect to this norm we have $f\in M$. $\endgroup$ Feb 7, 2021 at 14:42
  • $\begingroup$ Another argument would be that any bounded orthogonal sequence of functions contains a sub sequence which is a Sidon set. Then, clearly "sup" norm on such subspace is $l_{1}$ and not $l_{2}$. $\endgroup$
    – Salcio
    Feb 7, 2021 at 14:57
  • $\begingroup$ Hy @Jonathan, thanks. I guess that was just a catnap! Sure, $M$ is its own inverse image under the continuous inclusion $L^\infty\to L^2$. $\endgroup$
    – Ruy
    Feb 7, 2021 at 15:18
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    $\begingroup$ Now, fully awake, I must say your argument is indeed very nice! Thanks! $\endgroup$
    – Ruy
    Feb 7, 2021 at 15:24

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