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Let $p \in [1, \infty]$. I want to prove that this integral operator is compact: $$ T_p: L^p([0, 1]) \to L^p([0, 1]), \quad T(f(x)) := \int_0^x f(t)dt $$

I can prove it for $L_1$ case and I can prove the following facts:

  1. $L_p([0, 1]) \subset L_1([0, 1])$ for any p (including $\infty$)

  2. Jensen's inequality (for convex $\phi(t) = t^p$) gives me $\lVert x \rVert_1^p \leq \lVert x \rVert_p^p$ for all $x \in B_{1, L_p([0, 1])}$ hence $B_{1, L_p} \subset B_{1, L_1}$

  3. $\forall x \in L_p([0, 1])$, $T_p(x) = T_1(x)$ (in sence of inclusion in (1)), hence we can say that $T_p(B_{1, L_p}) \subset T_1(B_{1, L_1})$

What I want to say:

Compact operator send unit ball to totally bounded set. $\forall p \in [1, \infty)$, $T_p(B_{1, L_p}) \subset T_1(B_{1, L_1})$ and last one - totally bounded. Hence - the first one set is also totally bounded

Is it a correct way to proof compactness of integral operator $T_p$?

Here are 3 open questions I can't fix right now:

  1. $T_1(B_{1, L_1})$ totally bounded, but for fix $\epsilon > 0$ we can't take epsilon-net from $T_1(B_{1, L_1})$ and say it is also epsilon-net for $T_p(B_{1, L_p})$, since first epsilon-net consists from element from $L_1$ and last one should consists from element from $L_p$. Do we have some density conditions on $L_p$ inside $L_1$? I saw only this Why is $L^{1} \cap L^{\infty}$ dense is in $L^{p}$?
  2. First epsilon-net also use norm from $L_1$, so even if $L_p$ dense in $L_1$ (all spaces take place on interval $[0, 1]$), we need equivalence of norms in both spaces (I know, that norm in $L_p$ and $L_q$ doesn't equivalent untill $p \neq q$: Are all the norms of $L^p$ space equivalent? but thats true for functions on real line, in my example functions defined on interval $[0, 1]$, so maybe in that smaller space they equivalent?)
  3. I didn't fully "proof" case when $p = \infty$, I can't give an easy proof of balls inclusion from (2)

Can you give any hint how I can fix my proof of compactness? Or maybe some countreexamples/hints how I can proof main statement about compactness of $T_p$?

gt6989b
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brokoner12
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  • You need a characterization of compact sets in $L^{p}$. Most of your arguments cannot be completed without this. Hwo did you solve he case $p=1$? For a characterization of compact sets in $L^{p}$ see Dunford and Schwarz Vol I. As far as I can see $L^{P} \subseteq L^{1}$ is not useful. – Kavi Rama Murthy Apr 23 '21 at 11:34
  • @KaviRamaMurthy for $L^1$ for fix epsilon I divide interval [0, 1] on equal intervals (of length at most $\frac{\epsilon}{2}$ and take all functions, constant on each interval (with value at most 1, with steps $\frac{\epsilon}{2}$.

    thanks for link! I`ll take a look to characterization inside this book

     – brokoner12 Apr 23 '21 at 11:43

1 Answers1

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For your particular operator, you can proceed without a general compactness criterion in $L_p$:

In case $p>1$, you can use Hölder's inequality to show that the image of the unit ball is equicontinuous and bounded in $C([0,1])$ (actually, the image is bounded in some Hölder space). Thus, a simple compactness criterion in $C([0,1])$ (Arzela-Ascoli) is sufficient.

The difficult case is $p=1$. In that case, you probably have to show that the adjoint operator $T^*\colon L_\infty\to L_\infty$, $T^*g(y)=\int_y^1g(s)ds$ is compact. The latter can again be shown by Arzela-Ascoli similarly as above.

Martin Väth
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  • In the case $p=1$ are you sure, that the operator $T_p$ is compact? I think I found a counter example: Let $f_n(x) = n$ for $x\leq 1/n$ and $f_n(x)=0$ otherwise. The sequence $T(f_n)$ converges to $F(0)=1$ and $F(x)=0$ for $x>1$, so $T_1$ cannot be a compact operator from $L^1([0,1])\to C([0,1])$. – Michael Dec 28 '21 at 12:35
  • I have not claimed that $T$ is compact as an operator from $L_1$ into $C$; I have claimed that it is ompact as an operator from $L_1$ into $L_1$. – Martin Väth Dec 29 '21 at 07:58
  • To give some more details about the proof for the case $p=1$: This is using Schauder's theorem that $T$ is compact if (and only if) its adjoint $T^*$ is compact. Now you have to know how $L_1^=L_\infty$ are canonically identified and to apply Fubini's (and Fubini-Tonelli's) theorem to see that if $Tf(x)=\int_0^1k(x,t)f(t)dt$ with bounded measurable $k$ the adjoint operator is $T^g(y)=\int_0^1k(s,y)g(s)ds$. Now apply this with $k$ defined by $k(x,t)=1$ for $t\le x$ and $k(t,x)=0$ otherwise. – Martin Väth Dec 29 '21 at 08:24
  • BTW, using that $L_q^=L_{q'}$ with $\frac1q+\frac1{q'}=1$ for $1\le q<\infty$, the proof given above (and using Hölder's inequality to show that $T^(B)$ is uniformly continuous if $B$ is the unit ball of $L_{q'}$, because $q'>1$) shows that actually $T^$ is compact as an operator from $L_q^$ into $L_1^*$, that is, $T$ is compact as an operator from $L_1$ into $L_q$ if $1\le q<\infty$. Only the case $q=\infty$ has to be excluded as Michael's example shows. – Martin Väth Dec 29 '21 at 23:06