If $V\subset L^\infty[0,1]$ with $\|f\|_\infty \leq c\|f\|_2$, then $V$ is finite dimensional

Question

If $V$ is a linear subspace of $L^\infty[0,1]$ with $\|f\|_\infty \leq c\|f\|_2$ for all $f\in V$, then $V$ is finite dimensional. The proof is an explicit calculation:

Since $L^\infty[0,1] \subset L^2[0,1]$, take $e_1,\cdots , e_n$ to be $L^2$-orthonormal vectors in $V$. Fix some $x$ in $[0,1]$. We have, for all $y\in[0,1]$, $$\left|\sum e_i(x)e_i(y)\right| \leq \left\|\sum e_i(x) e_i(\cdot) \right\|_\infty \leq c\left\|\sum e_i(x)e_i(\cdot)\right\|_2 = c \sqrt{ \sum e_i^2(x)},$$ take $y = x$ this implies $$\sum e_i^2(x) \leq c^2.$$ Integrate both side and we get $$ n=\int_0^1 \sum e_i^2(x) \leq c^2.$$ This proof is simple but it is not really intuitive for me. Could you guys help me with a more functional analysis argument of this?

Since for functions in $L^\infty[0,1]$ we always have $\|f\|_2 \leq \|f\|_\infty$ , paired with $\|f\|_\infty \leq c \|f\|_2$, this means that the identity map is a continuous bijection between $(V, \|\cdot \|_2)$ and $(V, \|\cdot \|_\infty)$. Also $\|\cdot\|_2$ and $\|\cdot\|_\infty$ are equivalent on $V$. I know any two norms are equivalent in a finite dimensional space, but I dont know if there is anything special about the $L^2$ and $L^\infty$ norm to make the converse of that true as well.

Answer 1

The following functional-analytic proof is an application of the Riesz lemma: a normed space is finite-dimensional if and only if its unit ball is compact, or equivalently, if the identity operator is compact.

The following statement is what we want to prove.

Proposition. Let $V$ be a vector space of bounded functions on $[0, 1]$ such that there is $c>0$ satisfying $\lVert f\rVert_{\infty}\le c\lVert f\rVert_2$ for all $f\in V$. Then $V$ is finite-dimensional.

Remark. Notice how we are avoiding the mention of $L^\infty([0, 1])$. The reason is that $V$ is a space of functions, whereas $L^\infty$ is typically defined as a space of equivalence classes of functions, up to almost everywhere equivalence.

Proof. For $x\in [0, 1]$ and $f\in V$ define the evaluation functional $\mathrm{ev}_x(f):=f(x).$ (Note: this would not be well-defined on $L^\infty([0, 1])$, as mentioned in the remark). The assumption implies that this is a bounded linear functional in the $L^2$ topology. So the Riesz representation theorem yields the existence of $K_x\in L^2([0, 1])$ such that $$ \mathrm{ev}_x(f)=\langle K_x | f \rangle_{L^2([0,1])}=\int_0^1 K(x, y) f(y)\, dy, $$ where we have let $K(x, \cdot):=K_x$.

By assumption, $$ \lvert \langle K_x | f \rangle_{L^2([0,1])} \rvert \le c \lVert f\rVert_{L^2([0,1])}$$ for arbitrary $f$. Letting $f=K_x$ we see that $\lVert K_x\rVert_2\le c$. The constant on the right-hand side is independent on $x$, so integrating we get $\int_{[0, 1]^2} K(x, y)^2\, dxdy\le c^2$ and we conclude that, in particular, $K\in L^2([0, 1]^2)$.

Now consider $T_K(g):=\int_0^1 K(\cdot, y)g(y)\, dy$, seen as a linear operator of $L^2([0, 1])$ into itself. A consequence of $K\in L^2([0, 1]^2)$ is that this operator is compact. We infer that the identity operator on $V$ is compact in the $L^2$ topology. By the Riesz lemma we conclude that $V$ is finite-dimensional. $\Box$

Concluding remarks.

1. The proof of Grothendieck, given in the main question, is superior to this one for two reasons. First, it is more elementary, and second, it gives an explicit bound on the dimension of $V$ in terms of $c$.
2. The function $K$ is known as a reproducing kernel. A posteriori, given a basis $e_1, \ldots, e_n$ of $V$ that is $L^2$-orthonormal, we have that $K(x, y)=\sum_j e_j(x)e_j(y)$. The function on the right is exactly the one appearing in Grothendieck's proof.