Is the following statement true?
Let $T:L^p[0,1] \to L^p[0,1]$ be a bounded operator for $1 < p < \infty$ and suppose that $\operatorname{Im}(T) \subset C[0,1]$ consists of continuous functions. Then $T$ is compact.
I have tried to prove it by using the reflexivity of $X=L^p[0,1]$: Given $f_n \in X$ a bounded sequence, $Tf_n$ is also bounded, and thus by weak compactness, there exists $g \in L^p[0,1]$ such that $Tf_n \overset{w}{\to}g$ (denoting the subsequence again by $Tf_n$). One can show that actually $g = Tf$ for some $f \in X$ (by the fact that $TB_X$ is closed and convex and thus weakly closed) and thus $T(f_n – f) \overset{w}{\to}0$. I am stuck here and can't seem to understand how by continuity of $T(f_n – f)$ we can conclude strong convergence.
Maybe this approach is not fruitful, or the statement is just false. Any leads are appreciated.
Best Answer
EDIT: the edge cases $p = 1$ and $p = \infty$ are treated in Shalop's answer.
Here is a relatively elementary answer combining ideas presented in the comments of David Mitra, Davide Giraudo and Shalop.
We first prove the following case: If $T:L^p[0,1] \to C[0,1]$ is bounded (the range is with the supremum norm), then it is compact when the range is considered with the $L^p$ norm.
Denote $B$ the closed unit ball of $L^p$, by reflexivity it's weakly compact. Now $T:L^p[0,1] \to C[0,1]$ is also continuous when both space are equipped with the suitable weak topology (this holds in generally for bounded operators between normed spaces). Thus $T(B)$ is weakly compact in $C[0,1]$. This means that for any sequence $f_n \in T(B)$ there is a subsequence $f_{n_k}$ and $f \in C[0,1]$ such that for all $\varphi \in C[0,1]^*$ we have $\varphi(f_{n_k}) \to \varphi(f)$. For $x \in [0,1]$ take $\varphi$ to be the evaluation functional at $x$ (it is bounded) and get that $$ \forall x \in [0,1] \; f_{n_k}(x) \underset{k \to \infty}{\rightarrow} f(x)$$ Thus we have pointwise convergence. Furthermore, $T(B)$ is bounded in the $\Vert \cdot \Vert_\infty$ norm, so by the bounded convergence theorem on $\vert f - f_{n_k} \vert^p$ (from measure theory): $f_{n_k} \overset{L_p}{\to}f$. Thus $T(B)$ is compact in the $L^p$ norm $\blacksquare$.
For the General case we use Davide Giraudo's argument, simplified by Shalop: If $T:L^p[0,1] \to L^p[0,1]$ bounded with $\operatorname{Im}(T) \subset C[0,1]$. $T$ has a closed graph in $L^p[0,1] \times L^p[0,1]$, but if $x_n \overset{L^p}{\to} x$ and $Tx_n \overset{\Vert \cdot \Vert_\ \infty}{\to} y \in C[0,1]$ then in particular $Tx_n \overset{L^p}{\to} y$, so $Tx=y$. So $T$ has a closed graph as a function from $L^p$ to $(C[0,1], \Vert \cdot \Vert_\infty$). By the closed graph theorem it's continuous between these spaces, and this is exactly the case we proved $\blacksquare$.
Any corrections are welcome.