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.

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    $\begingroup$ If we assume $T$ is also bounded as an operator into $C[0,1]$, then it must be compact (into $L_p[0,1]$). This follows since $C[0,1]$ has the Dunford-Pettis property (weakly compact operators map weakly compact sets to norm compact sets) and the identity from $C[0,1]$ to $L_p[0,1]$ is weakly compact. I'm not sure what happens in the general case. $\endgroup$ – David Mitra May 15 at 7:27
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    $\begingroup$ Maybe an application of Baire theorem combined with David Mitra's argument can help. Define $F_n:= \{f\in L^p, \left\lvert Tf(x)\right\rvert\leqslant n a.e.\}$. By using the fact that each convergent sequence in $L^p$ has an almost everywhere convergent subsequence and continuity of $T$, it follows that $F_n$ is closed. Moreover, for all $f$, $Tf$ is bounded on $[0,1]$ hence $L^p=\bigcup_n F_n$. One of the $F_n$ has a non-empty interior and one can show that $\left\lVert Tf\right\rVert_\infty\leqslant C\left\lVert f\right\rVert_p$. $\endgroup$ – Davide Giraudo May 15 at 8:06
  • $\begingroup$ @DavideGiraudo Fantastic argument for reduction! I think it works. $\endgroup$ – pitariver May 15 at 17:16
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    $\begingroup$ This is fascinating! I wonder if it’s true when we replace $C[0,1]$ by $L^q[0,1]$ for some $q>p$. Also in Davide’s argument it seems that one may alternatively use the closed graph theorem. $\endgroup$ – Shalop May 16 at 4:46
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    $\begingroup$ @DavideGiraudo I took the liberty of doing so. $\endgroup$ – David Mitra May 16 at 16:40

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.

  • $\begingroup$ I think so! Very nice. $\endgroup$ – David Mitra May 16 at 17:43

The answer is yes, and the proof follows easily from the following three observations:

1) If either of $X$ or $Y$ is reflexive, then any bounded operator from $X$ to $Y$ is weakly compact. This follows from the fact that a Banach space is reflexive if and only if its closed unit ball is weakly compact.

2) If $T:L_p[0,1]\rightarrow L_p[0,1]$, $1<p<\infty$, is bounded and if $T$ maps into $C[0,1]$ (note some care in interpretation is needed here since elements of $L_p$ are equivalence classes of functions), then $T$ is bounded when regarded as a map into $C[0,1]$. This follows from Davide Giraudo's argument in the comments, or from an easy application of the Closed Graph Theorem, as suggested by Shalop.

3) $C[0,1]$ has the Dunford-Pettis property: any weakly compact operator from $C[0,1]$ into any Banach space maps weakly convergent sequences to norm convergent sequences. A proof can be found in, e.g., VI.7.4 of Linear Operators, vol. 1, Dunford and Schwartz, or in chapter 5 of Albiac and Kalton's Topics in Banach Space Theory.

So, write your operator as $T=I\circ T_C$ where $T_C:L_p[0,1]\rightarrow C[0,1]$ is defined by $T_C f=Tf$ and $I$ is the "identity" from $C[0,1]$ to $L_p[0,1]$. Both $T_C$ and $I$ are linear, $T_C$ is bounded by 2), and $I$ is bounded since $\Vert\cdot\Vert_p\le\Vert\cdot\Vert_\infty$.

Take a bounded sequence $(x_i)$ in $L_p[0,1]$. It has a weakly convergent subsequence $(y_i)$. Since bounded operators are weak-weak continuous, the image of $(y_i)$ under $T_C$ is weakly convergent. Now, according to 3), $I$, being weakly compact by 1), maps $(T_C y_i)$ to a norm convergent sequence $(Ty_i)$, as desired.

$$\ $$

Remark: The Dunford Pettis result is a big gun; I wonder if a more elementary argument can be made.

  • $\begingroup$ Very nice answer! I added an elementary spinoff that avoids Dunford Pettis. $\endgroup$ – pitariver May 16 at 17:50

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.

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    $\begingroup$ Thanks a lot for expanding the usefulness and scope of this thread! $\endgroup$ – pitariver May 22 at 4:56

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