Closed subspace of $L^{p}( [a,b])$ consisting of continuous functions I'm wondering if a closed subspace of $L^p [a,b](1\leq p<\infty)$ consisting of continuous functions must be finite-dimensional?
I already know it's true for $L^2([a,b]): $ this question
 A: The identity map $Id:(X ,\|\cdot\|_\infty) \to (X,\|\cdot\|_p)$ is linear, continuous, and bijective, hence invertible. This implies there is $c>0$ such that $\|x\|_\infty \le c \|x\|_p$ for all $x\in X$.
But this implies $\|x\|_2 \le \sqrt{b-a} \|x\|_\infty \le c \sqrt{b-a} \|x\|_p$ for all $x\in X$. Hence, $X$ is a closed subspace of $L^2$, and hence finite-dimensional.
A: The answer to the OP's question is yes, a closed subspace $M$ of $L_p([a,b])$ consisting of continuous functions is finite dimensional. Since $M\subset\mathcal{C}([a,b])$, then $M\subset L_\infty$ and the result follows from a Theorem by Grothendieck (see, Rudin, W., Functional Analysis, McGraw Hill, Chapter 5.)

Theorem: Suppose $(X,\mathscr{B},\mu)$ is a finite measure space, $M$ closed in $L_p(\mu)$ for some $0< p<\infty$ and $M\subset L_\infty(\mu)$. Then
$M$ is finite dimensional.

The theorem above  follows from the $L_2(\mu)$ case as we show below.
Sketch of the proof $1\leq p<\infty$:  Without loss of generality assume $\mu(X)=1$. The case $p=2$ is the subject of this posting.
Notice that for all $1\leq r<\infty$, $L_\infty(\mu)\subset L_r(\mu)$  and that the inclusion map $\iota_r:f\mapsto f$ is injective and a continuous operator since $\|f\|_r=0$ iff $f=0$ $\mu$-almost surely, and
$$\|\iota_r (f)\|_r=\|f\|_r\leq \|f\|_\infty$$
If $M$ closed in $L_p(\mu)$ for some $1\leq p<\infty$, then $M$ is also closed in $L_\infty(\mu)$: suppose $(f_n:n\in\mathbb{N})\subset M$ converges to some $f\in L_\infty(\mu)$, i.e. $\|f_n-f\|_\infty\xrightarrow{n\rightarrow\infty}0$, then
$$\|f_n-f\|_p\leq\|f_n-f\|_\infty\xrightarrow{n\rightarrow\infty}0$$
Since $M$ is closed in $L_p(\mu)$, it follows that $f\in M$.
An application of the  open map theorem yields that $\iota^{-1}_p:(M,\|\;\|_p)\rightarrow (M,\|\;\|_\infty)$ is also bounded. Thus, there is a constant $K>0$ such that
$$\|f\|_p\leq \|f\|_\infty\leq K\|f\|_p,\qquad f\in M$$

*

*If $1\leq p\leq 2$, then for any $f\in M$
$$\|f\|_2\leq\|f\|_\infty\leq K\|f\|_p\leq
K\|f\|_2$$
Thus $M$ is also closed in $L_2(\mu)$, and then the result in the posting cited above can be used.


*If $2<p<\infty$ then
\begin{align}
\|f\|_p\leq\|f\|^{1-2/p}_\infty\|f\|^{p/2}_2
\end{align}
It follows that
$$\|f\|_\infty\leq K\|f\|_p\leq K\|f\|^{1-2/p}_\infty\|f\|^{p/2}_2$$
and so,
$$\|f\|_2\leq \|f\|_\infty\leq K^{p/2}\|f\|_2$$
Consequently $M$ is closed in $L_2(\mu)$.


*If $0<p<1$, the space $L_p(\mu)$  equipped with translation invariant metric $d_p(f, g)=\|f-g\|^p_p=\int|f-g|^p\,d\mu$ is complete metrizable linear space (an $F$-space). It is easy to prove that  a set in $B\subset (L_p,d_p)$  is bounded iff $\sup\{\|f\|_p: f\in B\}<\infty$. It follows that a linear operator $P:(L_\infty(\mu),\|\;\|_\infty)\rightarrow (L_p(\mu),d_p)$ is bounded iff there is a constant $k>0$ such that $\|Pf\|_p\leq k\|f\|_\infty$. Conversely, a linear operator $R:(L_p(\mu),d_p)\rightarrow(L_\infty(\mu),\|\;\|_\infty)$ is bounded iff there is a contant $K>0$ such that $\|Rf\|_\infty\leq K\|f\|_p$. The proof of the Theorem for this case as we can see, becomes similar to that of $1\leq p<2$.
