TITLE: If $T:L^p[0,1] \to L^p[0,1]$ bounded for $1 < p < \infty$ with continuous image, then it's compact
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: I just wanted to address the side cases $p=1$ and $p=\infty$ even though the question doesn't ask about it.
For $p=1$ it is still true that $T$ is weak-to-norm continuous on $L^1$ (a property equivalent to compactness for reflexive spaces, but weaker in general). Indeed, if the image of $T$ is contained in $L^{\infty}$ then the closed graph theorem tells us that $T$ is bounded from $L^1$ to $L^{\infty}$, and it is actually true that every bounded operator from $L^1 \to L^{\infty}$ is given by a kernel. More specifically, there necessarily exists $k \in L^{\infty}([0,1]^2)$ such that $(Tf)(x) = \int_0^1 k(x,y)f(y)dy$ (note that once we know this, the weak-to-norm continuity follows easily by the bounded convergence theorem). This can be proved by exploting nice properties of projective tensor products of Banach spaces and then using that $(L^1)^*=L^{\infty}$, see page 1 of this paper for instance. A more elementary proof of this fact can also be obtained by using the Radon-Nikodym theorem. Specifically, if $T:L^1\to L^{\infty}$ is bounded then we can define a pre-measure $\mu$ on $[0,1]^2$ by sending product sets $A \times B \mapsto \int_B (T1_A)(x)dx$. Then it is trivial that $\mu(A \times B) \leq \|T\|_{L^1 \to L^{\infty}}\cdot m(A \times B)$, where $m$ is just 2d Lebesgue measure. Using this fact together with the Caratheodory extension theorem, one can show that $\mu$ extends to a Borel measure on $[0,1]^2$ with the property that $\mu(E)\leq \|T\|m(E)$ for all Borel $E \subset [0,1]^2$. Then $\mu$ is trivially absolutely continuous, so it has a density $k$ with respect to $m$, which defines our kernel.
However, just because $T$ is weak-to-norm continuous does not imply that $T$ is compact on $L^1$ (because $L^1$ is not reflexive). Indeed, if we define our kernel by $k(x,y) = \cos(x \log y) +\sin(x \log y)$, then one easily verifies that $\int_0^u k(x,y)dy = u\sin(x \log u)$ (differentiate both sides in $u$ to see this). In particular, if we let $f_n = n\cdot 1_{[0,n^{-1}]}$, then we see that $Tf_n(x) = \sin(-x \log n)$. Thus $f_n$ is a bounded sequence in $L^1$ and $Tf_n$ converges weakly but not strongly to zero (the sinusiodal frequencies become larger and larger as $n \to \infty$). Hence $T$ is not compact.
For $p=\infty$, an operator $T$ mapping $L^p$ boundedly into $C[0,1]$ need not be compact or even weak-to-norm continuous. That's because $C[0,1]$ is a universal separable Banach space so any separable Banach space embeds isomorphically into it. In particular let $J:L^2[0,1] \to C[0,1]$ be such an embedding. Since $L^{\infty} \stackrel{i}{\to} L^2$ is not a compact or even weak-to-norm continuous embedding (consider $f_n(x) = \sin(nx)$), the composition $L^{\infty} \stackrel{i}{\to} L^2 \stackrel{J}{\to} C[0,1]$ will not be either.