# How to show operator is compact [duplicate]

Currently I'm self studying functional analysis, namely compact operators. In the text, the author gives the following example:

Example 1: Let $$C_1$$ and $$C_2$$ be positive constants and let $$M:=\left\{x(t)\in C[a,b]:|x(t)| Then $$M$$ is relatively compact.

I completely understand Example 1, it makes use of Arzela's theorem. Then the author gives the following example:

Example 2: The operator $$Ax:=\int_0^tx(\tau)d\tau.\tag{2}$$ on $$C[0,1]$$ is a compact operator (use the previous example).

My question is fairly straightforward: using Example 1, how is this operator compact on $$C[0,1]$$?

As of now I have the following two definitions of compact operators:

Definition 1: A linear operator $$A\colon X\to Y$$ is called a compact operator if and only if for every bounded sequence $$x_n\in X$$ the sequence $$Ax_n$$ has a Cauchy subsequence.

Definition 2: An linear operator $$A\colon X\to Y$$ is compact if and only if the image of the unit ball of $$X$$ is a relatively compact set in $$Y$$.

Just going off terminology, I would guess that Definition 2 should be used to solve Exercise 2, but I'm not seeing it through all the way. Any help is appreciated!

• Take $M$ to be the image of the unit ball and use Example 1 to show that $M$ is relatively compact. You are then done by Definition 2. Jun 22, 2021 at 20:19
• Can you find a constant $C>0$ so that for any $x\in C[0,1]$ satisfying $\|x\|_{C[0,1]}\le 1$, and every $t\in [0,1]$, 1. $|Ax(t)| < C$, and 2. $|(Ax)'(t)| < C$? If so, then you will have shown that the image of the unit ball in $C[0,1]$ is contained in the set $M$. Jun 22, 2021 at 20:19
• @AlexOrtiz: Yes, thanks I see it now! Jun 22, 2021 at 20:25

I think you just need to notice that $$|Ax|\leq \max_\limits{t\in[0,1]}x(t)$$ (which comes immediately since $$x$$ is continous on a compact and the integral function is monotonic). Such maximum actually controlls also the derivative of $$Ax$$, thus if you take a set $$M$$ of functions of the form $$Ax$$ where $$x\in B(0,1)$$ you obtain a precompact space by example $$1$$ and with the second definition. By $$B(0,1)$$ I denoted the unit ball of center $$0$$ in $$\mathcal{C}[0,1]$$.