What assumptions are needed for compactness and self-adjointness? I am studying functional analysis and got stuck with a textbook exercise. I greatly appreciate some hints/help or a push in the right direction!
Define for $g\in C^0[-1,1]$ the integral operator $T_g:L^2\to L^2$, by
$$T_g(f)(s):=\int_0^1g(s-t)f(t)dt$$
Now I need to oppose conditions on $g$ such that $T_g$ becomes a compact self-adjoint operator.
My progress: First for we look at when $T_g$ is self-adjoint. So we want some conditions such that: $\langle T_g(f)(s),h(s)\rangle=\langle f(s), T_g(h)(s)\rangle$. For the left hand side we get the following: $$\langle T_g(f)(s),h(s)\rangle=\int_0^1T_g(f)(s)\overline{h(s)}ds=\int_0^1\int_0^1g(s-t)f(t)\overline{h(s)}dtds$$
And for the right hand side we get the following:
$$\langle f(s), T_g(h)(s)\rangle=\int_0^1f(s)\overline{T_g(h)(s)}ds=\int_0^1 \int_0^1f(s)\overline{g(s-t)} \overline{h(t)}dsdt$$
Well now I am not to sure I thought that $\overline g=g$ is the only condition for $T_g$ to be self-adjoint.
Now for compactness:
I wanted to pick a bounded subset $M \subset L^2$, and show that $T_g(M) $ is compact.
For that I want to show that $T_g(M)$ is closed, bounded and equicontiuous.
I have no good idea/intuition on how to show these properties with the given information I have.
Can someone help me solve this exercise I think it is a exercise from which I can learn a lot and get a better grasp! :)
 A: For $T_g$ to be selfadjoint, you need
$$\tag1
\int_0^1\int_0^1g(s-t)f(t)\overline{h(s)}\,dt\,ds=\int_0^1\int_0^1\overline{g(s-t)}f(s)\overline{h(t)}\,dt\,ds,
$$
for all $f,h\in L^2[0,1]$. Changing the names of $s$ and $t$,
$$\tag2
\int_0^1\int_0^1\big[g(s-t)-\overline{g(t-s)}\big]f(s)\overline{h(t)}\,dt\,ds=0
$$
for all $f,h\in L^2[0,1]$. Given any rectangle $R=E\times F$, we have $1_R(s,t)=1_{E}(s)\,1_{F}(t)$. So
$$\tag3
\iint_R\big[g(s-t)-\overline{g(t-s)}\big]\,dt\,ds=0
$$
for every measurable rectangle $R=E\times F$. By Lebesgue Differentiation (or see another argument  here), $g(s-t)-\overline{g(t-s)}$ a.e. As $g$ is continuous, this is the same as
$$\tag4
g(t)=\overline{g(-t)},\qquad t\in[-1,1].
$$
As for compactness, the continuity of $g$ guarantees that $T$ is always compact. In fact, continuity is not really necessary, as a sufficient condition is that $\tilde g\in L^2[0,1]^2$, where $\tilde g(s,t)=g(s-t)$ (see, for instance Proposition II.4.7 in Conway's A Course in Functional Analysis). In the case where $g$ is continuous, a more or less direct proof can be made. Indeed, suppose first that $g(t)=t^k$. Then
$$\tag5
(T_gf)(s)=\int_0^1(s-t)^k\,f(t)\,dt=\sum_{j=0}^k(-1)^j\binom kj s^j\int_0^1t^{k-j}\,f(t)\,dt.
$$
The right-hand-side is a polynomial on $s$. So the rank of $T_g$ is contained in the (finite-dimensional) set of polynomials of degree at most $k$. When $g$ is a polynomial, $T_g$ is a linear combination of operators of the form $(5)$; thus finite-rank, as a linear combination of finite-rank is finite-rank.
Now for arbitrary continuous $g$, there exists a sequence $\{g_n\}$ of polynomials with $\|g-g_n\|_\infty\to0$. Then
\begin{align}
\|T_gf-T_{g_n}f\|_2^2
&=\int_0^1\Big|T_gf(s)-T_{g_n}f(s)\Big|^2\,ds\\[0.3cm]
&=\int_0^1\Big|\int_0^1 \big[g(s-t)-g_n(s-t)\big]\,f(t)\,dt\Big|^2\,ds\\[0.3cm]
&\leq\int_0^1\Big(\int_0^1 \big|g(s-t)-g_n(s-t)\big|\,|f(t)|\,dt\Big)^2\,ds\\[0.3cm]
&\leq\|g-g_n\|_\infty^2\,\int_0^1\Big(\int_0^1 |f(t)|\,dt\Big)^2\,ds\\[0.3cm]
&=\|g-g_n\|_\infty^2\,\Big(\int_0^1 |f(t)|\,dt\Big)^2\\[0.3cm]
&\leq\|g-g_n\|_\infty^2\,\int_0^1 |f(t)|^2\,dt\\[0.3cm]
&=\|g-g_n\|_\infty^2\,\|f\|_2^2.
\end{align}
It follows that $T_g=\lim_nT_{g_n}$, and so $T_g$ is compact as it is a limit of finite-rank operators.
