Finding the adjoint of a compact operator?

Question

Let $H = L^2([0,1])$ and $A:H \to H$ be defined as

$$ Af(t) = \int_0^t f(s) ds, \quad f \in H $$ By the Ascoli-Arzela theorem, $A$ is compact. Now the adjoint of $A$ is given as

$$ A^\ast g(s) = \int_s^1 g(t) dt $$

Questions

1. How does the Ascoli-Arzela theorem show that $A$ is compact?
2. More importantly, how was the adjoint $A^\ast$ determined? How do we get from the expression for $A$ to the expression for $A^\ast$?

Answer 1

For determining the adjoint :

You have that

$$\langle Af, g \rangle = \int_0^1 g(t) \int_0^t f(s) ds dt $$

And this is equal (by Fubini) to

$$ = \int_0^1 f(t) \int_t^1 g(s) ds dt = \langle f, Bg \rangle$$

with $Bg(t) = \int_t^1 g(s) ds$

And as this is true for every $f$ and $g$, you have that $A^* = B$

Answer 2

For Questions 1, take $\{f_n\}$ be uniformly bounded in $L^2$, consider the class $\{Af_n\}$.

check 1.$\{Af_n\}$ is uniformly bounded in $L^2$.

check 2.$\{DAf_n\}$ is well-define and uniformly bounded in $L^2$, where D is Differential operator.

pf of check 1:

This follows by Holder,

$|\int_0^t f(s) ds| \leq \sqrt{t }\|f_n\|_2\leq \|f_n\|_2$, so $\|Af_n\|_2\leq \|Af_n\|_\infty \leq \|f_n\|_2$

pf of check 2:

Since the form $\int_0^t f(s) ds$ is absolutely continuous, so $DAf_n=f_n$ is well-define and they are uniformly bounded in $L^2$ follows by assumption.

So by the Ascoli-Arzela theorem, there's a subsequence of $\{Af_n\}$ converge to a continuous function $f$ pointwise. On the other hand, since domain is finite and $\{Af_n\}$ are uniformly bounded in $L^2$, you can check by DCT that they converge in $L^2$. That means $A$ is compact.

Answer 3

For $f\in L^2$ and $0 \le r \le s \le 1$, \begin{align} |(Af)(r)-(Af)(s)| & \le \int_{r}^{s}|f(x)|dx \\ & \le \left(\int_{r}^{s}|f(x)|^2dx\right)^{1/2}\left(\int_{r}^{s}dx\right)^{1/2} \\ & = \|f\|\sqrt{s-r}. \end{align} Therefore the imagine of a bounded subset of $L^2$ is uniformly bounded and equicontinuous, which proves that $A$ is compact.

The adjoint is determined using integration by parts: \begin{align} (Af,g) & = \int_{0}^{1}\int_{0}^{x}f(t)dt\overline{g(x)}dx \\ & = -\int_{0}^{1}\left(\int_{0}^{x}f(t)dt\right)\left(\frac{d}{dx}\int_{x}^{1}\overline{g(t)}dt\right)dx \\ & = -\left.\int_{0}^{x}f(t)dt\int_{x}^{1}\overline{g(t)}dt\right|_{x=0}^{1} + \int_{0}^{1}f(x)\left(\int_{x}^{1}\overline{g(t)}dt\right)dx \\ & =\int_{0}^{1}f(x)\left(\overline{\int_{x}^{1}g(t)dt}\right)dx = (f,A^*g) \end{align}