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Can someone please explain to me how I can show the following operator is compact? $$ Af(x)=\int_0^xf(t)dt$$ where $A:L_2[0,1]\rightarrow L_2[0,1]$.

There are two definitions of compactness that I've encountered:

Def1. An operator $L$ is compact if for a bounded sequence $\{ x_n\}$, there exists a convergent subsequence $\{ x_{n_k}\}_k$.

Def2. An operator $L$ is compact if it's continuous and maps a bounded set to a sequentially compact set.

What is the easier/standard method to prove compactness for integral operators?

I also know that "If $L$ is a bounded linear operator on a separable Hilbert space, with orthonormal basis $\{e_n\}$, and $\sum_n\| L e_n\|^2<\infty$, then $L$ is compact. "

Teodorism
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  • Also, https://math.stackexchange.com/questions/151425/spectrum-of-indefinite-integral-operators – cmk Dec 08 '19 at 01:34
  • @cmk Thank you. There's an easier way I think though. $L$ is a linear operator and satisfies $\sum_n| L e_n|<\infty$. If I show it's bounded I can use my very last argument in the question. How can I prove it's bounded? – Teodorism Dec 08 '19 at 01:40

2 Answers2

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To show that $A$ is bounded, Cauchy-Schwarz works well $$ \|Af\|^2 = \int_0^1\left|\int_0^x f(t)dt\right|^2dx \\ \le \int_0^1 \int_0^x|f(t)|^2dt\int_0^xdt\,dx \\ = \int_0^1 \int_0^x|f(t)|^2dt xdx \\ \le \int_0^1 \int_0^1 |f(t)|^2dt \, 1 dx = \|f\|^2. $$ Then apply your theorem using the orthonormal basis $\{ e_n(t)= e^{2\pi int} \}_{n=-\infty}^{\infty}$: $$ Ae_n = \int_0^x e_n(t)dt = \frac{1}{2\pi in}(e_n(t)-1), \; n\ne 0, \\ \|Ae_n\|\le \frac{1}{\pi|n|},\;\; n\ne 0. $$

Disintegrating By Parts
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$A$ is an integral operator with kernel $K(x,t) = \mathsf1_{(0,x)}(t)$ and $$ \iint_{[0,1]^2} |K(x,t)|^2\ \mathsf d(x\times t) = \int_0^1\int_0^x\ \mathsf dt \ \mathsf dx = \frac 12<\infty, $$ so $K\in L^2([0,1]^2)$. It follows that $A$ is a Hilbert-Schmidt operator and therefore is compact.

Math1000
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