What is the correct formula of the kernel of the A$^*$A

If the kernel of the linear integral operator A on L$^2$(0,1) where A is a linear bounded integral operator non self adjoint?

Thanks.

Solution 1:

It depends on how the order of multiplication is taken. For example, suppose $A$ is of the form $Af(x)=\int_{(0,1]}f(s)k(x,s)\,ds$. Then, by Fubini's theorem $$\begin{align} \langle A^*Af,g\rangle&=\langle Af, Ag\rangle=\int_{(0,1]}Af(x)\overline{Ag(x)}\,dx\\ &=\int_{(0,1]}\Big(\int_{(0,1]}f(s)k(x,s)\,ds\Big)\Big(\int_{(0,1]}\overline{g(t)k(x,t)}\,dt\Big)\,dx\\ &=\int_{(0,1]}\Big(\int_{(0,1]\times(0,1]}f(s)k(x,s)\overline{k(x,t)}\,ds\otimes dx\Big)\overline{g(t)}\,dt \end{align}$$ From this, it follows that $$\begin{align} A^*Af(t)&=\int_{(0,1]\times(0,1]} f(s)k(x,s)\overline{k(x,t)}\,dsdx\\ &=\int_{(0,1]}f(s)\Big(\int_{(0,1]}k(x,s)\overline{k(x,t)}\,dx\Big)\,ds \end{align}$$ Hence, $A^*A$ is of the for $A^*Af(t)=\int_{(0,1]}f(s)\,q(t, s)\,ds$ where $$q(t, s)=\int_{(0,1]}k(x,s)\overline{k(x,t)}\,dx, \qquad 0<s, t\leq 1$$

When $k(x,s)=\{s/x\}$, then $$q(t, s)=\int_{(0,1]}\{s/x\}\,\{t/x\}\,dx, \qquad 0<s, t\leq 1$$

When $k(x,s)=\{x/s\}$, then $$q(t, s)=\int_{(0,1]}\{x/s\}\,\{x/s\}\,dx, \qquad 0<s, t\leq 1$$

The computation of these integrals are left to the OP.