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Let $T: X \to Y$ be the following integral operator, where $f$ is just assumed to be integrable (not neccessarilly continuous):

$$(Tf)(x)=\int_0^x f(t)dt$$

In an exercise I have to check whether this operator is compact, depending on what the spaces $X$ and $Y$ are. I have already shown that it is compact in the cases where $X=Y=C[0,1]$, and when $X=L^p[0,1]$ and $Y=C[0,1]$. To do it I used the Arzela-Ascoli theorem, and checked that the image under $T$ of the unit ball of $X$ is bounded and equicontinuous.

I ran into travel however in the last part of my exercise, when $X=Y=L^p[0,1]$. This is since here, of course, I can not use the Arzela Ascoli criterion to check compactness of $T$. Hence my question is: How would you prove/disprove compactness of this operator under such spaces? What can be a convenient formulation of compactness of $T$ to check? Thanks!

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  • $\begingroup$ Do you only need the case $X=Y=L^p$ or $X=L^p$ and $Y=L^{p'}$ for possibly distinct $p,p'$? $\endgroup$ Jan 23 at 19:34
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    $\begingroup$ Perhaps write $(Tf)(x) = \int K(x,t) f(t)dt$, where $K(x,t) = 1_{[t \le x]}(x,t)$? $\endgroup$
    – copper.hat
    Jan 23 at 19:37
  • $\begingroup$ @JustDroppedIn I need it just for when $p=p'$. $\endgroup$
    – user770533
    Jan 23 at 19:41
  • $\begingroup$ @copper.hat Sorry, but how would that be useful? $\endgroup$
    – user770533
    Jan 23 at 19:42
  • $\begingroup$ Perhaps you could approximate $K$ in some way to show $T$ is the limit of finite rank operators. Just an idea, have not thought it through. $\endgroup$
    – copper.hat
    Jan 23 at 19:50

1 Answer 1

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Lemma: Let $X,Y$ be Banach spaces. If $T_n:X\to Y$ are bounded operators with finite dimensional range and $T_n\to T\in B(X,Y)$ in operator norm, then $T$ is a compact operator.

For the case $X=Y=L^p$, $1< p\le\infty$, we will write $T$ as a norm limit of finite rank operators and deduce compactness of $T$ by the above lemma.

Let $n>1$ be an integer and partition the interval $[0,1]$ in $n$ subintervals of equal length, namely $I_{j,n}:=[\frac{j-1}{n},\frac{j}{n})$, $j=1,\dots,n$. Let $\phi_{j,n}$ denote the indicator function of $I_{j,n}$. Set $$T_n:L^p[0,1]\to L^p[0,1], \;\;\;T_n(f)=\sum_{j=1}^n\bigg(\int_0^{\frac{j}{n}}f(t)dt\bigg)\cdot\phi_{j,n}$$ Note that the operator $T_n$ has automatically finite dimensional range, since $T_n(f)$ lies in the linear span of the functions $\{\phi_{j,n}:j=1,\dots,n\}\subset L^p[0,1]$. Also,

for $x\in[0,1]$ there exists a unique $j_x\in\{1,\dots,n\}$ such that $x\in I_{j_x,n}$ so $\phi_{i,n}(x)=1$ if and only if $i=j_x$. So

$$|T_n(f)(x)|^p=\bigg|\sum_{j=1}^n\bigg(\int_0^{\frac{j}{n}}f(t)dt\bigg)\cdot\phi_{j,n}(x)\bigg|^p=\bigg|\int_0^{\frac{j_x}{n}}f(t)dt\bigg|^p\le\bigg(\int_0^1|f(t)|dt\bigg)^p\le\|f\|_p^p$$

where in the last inequality we have used the fact that $\|f\|_1\le\|f\|_p$ for $f\in L^p[0,1]$. Thus $$\|T_n(f)\|_p^p=\int_0^1|T_n(f)(x)|^pdx\le\int_0^1\|f\|_p^pdx=\|f\|_p^p$$ and thus $\|T_n\|\le1$, so $T_n$ are indeed bounded operators with finite rank.

We now show that $\|T_n-T\|_p\to0$. Indeed, we have $$|T_n(f)(x)-T(f)(x)|^p=\bigg|\sum_{j=1}^n\int_0^{\frac{j}{n}}f(t)dt\cdot \phi_{j,n}(x)-\int_0^xf(t)dt\bigg|^p=$$ $$=\bigg|\int_0^\frac{j_x}{n}f(t)dt-\int_0^xf(t)dt\bigg|^p\;\;(\star)$$ where $j_x\in\{1,\dots,n\}$ is the unique integer such that $x\in I_{j_x,n}$ (and the above equality occurs because $\phi_{j_x,n}(x)=1$ and $\phi_{i,n}(x)=0$ for $i\ne j_x$). Continuing from $(\star)$, if we denote by $q$ the conjugate exponent $(1/p+1/q=1)$, we have $$(\star)=\bigg|\int_x^{\frac{j_x}{n}}f(t)dt\bigg|^p\le\bigg|\int_0^1\chi_{I_{j_x,n}}(t)f(t)dt\bigg|^p\le$$ $$\le\bigg(\int_0^1\chi_{I_{j_x,n}}(t)\cdot|f(t)|dt\bigg)^p\le\bigg(\mu(I_{j_x,n})^{1/q}\cdot\|f\|_p\bigg)^p=\frac{1}{n^{p/q}}\cdot\|f\|_p^p$$ where we used Holder's inequality. Therefore, $$\|T_n(f)-T(f)\|_p^p=\int_0^1|T_n(f)(x)-T(f)(x)|^pdx\le\int_0^1\frac{1}{n^{p/q}}\cdot\|f\|_p^pdt=\frac{1}{n^{p/q}}\cdot\|f\|_p^p$$ and thus $\|T_n-T\|_p\le\frac{1}{n^{1/q}}$. Letting $n\to\infty$ gives $T_n\to T$.

P.S: Why it is reasonable to define the operators $T_n$ the way we did? First, we need them to have finite dimensional range. Second, we look at $T(f)(x)=\int_0^xf(t)dt$. This is a number very close to $\int_0^{j/n}f(t)dt$ for some suitable $j,n$. So it feels natural to partition the unit interval in small intervals of length $1/n$ and define $T_n(f)$ by the rule "take a $x$, determine in which small interval it lies (i.e. find the proper $j_x$), then assign the value $\int_0^{j_x/n}f(t)dx$. Implicitly we have been multiplying with $\phi_{j,n}$ and adding up, to make sure we evetually obtained the correct $j_x$. I hope this helps you understand the reasoning here.

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  • $\begingroup$ Thank you very much for your answer. I would like to ask you two questions: 1) Why next to (star) you write an integral with inferior limit $\frac{j_{x-1}}{n}$ and not just $\frac{j_x}{n}$? 2) How about the case when $p=\infty$? It seems that your exact same argument does the job. $\endgroup$
    – user770533
    Jan 23 at 21:26
  • $\begingroup$ @Mathias Sorry, I had a mistake there, I edited it. I am unsure about the case $p=\infty$ because then the norm is the essential supremum and not an integral, but I think you can check the same argument, doing the necessary modifications :) $\endgroup$ Jan 23 at 21:32
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    $\begingroup$ @Mathias Apparently, the $p=\infty$ case probably works as you said. But there is a problem with the $p=1$ case, since then $q=\infty$ and this causes a problem. There are some answers for the $p=1$ case in this post math.stackexchange.com/questions/4366135/… $\endgroup$ Jan 26 at 13:16

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