# Check whether integral operator is compact

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!

• Do you only need the case $X=Y=L^p$ or $X=L^p$ and $Y=L^{p'}$ for possibly distinct $p,p'$? Jan 23 at 19:34
• Perhaps write $(Tf)(x) = \int K(x,t) f(t)dt$, where $K(x,t) = 1_{[t \le x]}(x,t)$? Jan 23 at 19:37
• @JustDroppedIn I need it just for when $p=p'$.
– user770533
Jan 23 at 19:41
• @copper.hat Sorry, but how would that be useful?
– user770533
Jan 23 at 19:42
• 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. Jan 23 at 19:50

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.

• 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.
– user770533
Jan 23 at 21:26
• @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 :) Jan 23 at 21:32
• @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/… Jan 26 at 13:16