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

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.

• 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. – David Mitra May 15 at 7:27
• 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 $\left\lVert Tf\right\rVert_\infty\leqslant C\left\lVert f\right\rVert_p$. – Davide Giraudo May 15 at 8:06
• @DavideGiraudo Fantastic argument for reduction! I think it works. – pitariver May 15 at 17:16
• 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. – Shalop May 16 at 4:46
• @DavideGiraudo I took the liberty of doing so. – 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.

• I think so! Very nice. – 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 $$T:L_p[0,1]\rightarrow L_p[0,1]$$, $$1, is bounded and if $$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.

• Very nice answer! I added an elementary spinoff that avoids Dunford Pettis. – 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.

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