$(A_K^*f)(x) = \int_0^1 K^*(x,y) f(y)\, dy$ for all $f\in L^2[0,1]$ - Fubini's theorem?

Let $K$ be a square-integrable kernel on $[0,1] \times [0,1]$, i.e. $$\int_0^1 \int_0^1 |K(x,y)|^2\, dx\, dy < \infty$$ and let $A_K$ be the integral operator induced by it on $L^2[0,1]$, i.e. $$(A_K f)(x) = \int_0^1 K(x,y) f(y)\, dy$$ for all $f\in L^2[0,1]$. Define $K^*(x,y) = \overline{K(y,x)}$. Show that the adjoint operator $A_K^*$ is the integral operator induced by the kernel $K^*$.

Hint: Fubini's theorem.

My thoughts:

In mathy language, we want to prove that $$(A_K^*f)(x) = \int_0^1 K^*(x,y) f(y)\, dy = \int_0^1 \overline{K(y,x)} f(y)\, dy$$ By definition of the adjoint, we have $\langle A_K f,g\rangle = \langle f, A_K^*g\rangle$ for all $f,g\in L^2[0,1]$. The usual inner product on $L^2[0,1]$ is defined by $\langle f,g\rangle = \int_0^1 f(x) \overline {g(x)}\, dx$ for all $f,g\in L^2[0,1]$. Thus, we have $$\langle A_K^* g, f \rangle = \int _0^1 g(x) \overline{\int_0^1 K(x,y) f(y)\, dy}\, dx = \int_0^1 \int_0^1 g(x) \overline{K(x,y)} \overline{f(y)}\, dy \, dx$$ Swapping $x$ and $y$ in the last (definite) integral, i.e. via a change of variables, we have $$\int_0^1 \int_0^1 g(x) \overline{K(x,y)} \overline{f(y)}\, dy \, dx = \int_0^1 \int_0^1 g(y) K^*(x,y) \overline{f(x)}\, dx \, dy$$ Also, $$\langle A_K^* g, f \rangle = \int_0^1 (A_K^*g)(x) \overline{f(x)}\, dx$$ Certainly, the constant function equal to $1$ on $[0,1]$ is in $L^2[0,1]$, so putting $f = 1$ we get $$\int_0^1 (A_K^*g)(x) \, dx = \int_0^1 \int_0^1 g(y) K^*(x,y)\, dx \, dy$$ for all $g\in L^2[0,1]$.

Thanks!

$$\langle g,A_Kf\rangle=\int_0^1\int_0^1 g(x)\overline{K(x,y)}\overline{f(y)}dy dx=\int_0^1\int_0^1 g(x)\overline{K(x,y)}\overline{f(y)}dx dy$$

by Fubini's theorem as the function $\vert g(x)K(x,y)f(y)\vert$ is integrable on $[0,1]\times[0,1]$. To see this use Cauchy-Schwarz inequality twice

$$\begin{align}\int_0^1 \int_0^1 \vert g(x)\vert \vert K(x,y)\vert \vert f(y)\vert dx dy &=\int_0^1 \vert f(y)\vert\left[\int_0^1\vert g(x)\vert \vert K(x,y)\vert dx\right] dy\\ &\leq \int_0^1 \vert f(y)\vert\left(\int_0^1 \vert g(x)\vert^2 dx\right)^{1/2} \left(\int_0^1\vert K(x,y)\vert^2dx \right)^{1/2} dy\\ &=\|g\|_2\int_0^1\vert f(y)\vert \left(\int_0^1\vert K(x,y)\vert^2dx \right)^{1/2} dy\\ &\leq\|g\|_2\left(\int_0^1\vert f(y)\vert^2 dy\right)^{1/2}\left(\int_0^1\int_0^1\vert K(x,y)\vert^2 dxdy\right)^{1/2}\\ &=\|g\|_2\|f\|_2\|K\|_2\end{align}$$

Then we continue

$$\begin{align}\int_0^1\int_0^1 g(x)\overline{K(x,y)}\overline{f(y)}dx dy&=\int_0^1\overline{f(y)}\left[\int_0^1 g(x)\overline{K(x,y)}dx\right] dy\\ &=\int_0^1 \overline{f(y)}(A_{K^*}g)(y)dy\\ &=\int_0^1 (A_{K^*}g)(y)\overline{f(y)}dy\\ &=\langle A_{K^*}g,f\rangle\end{align}$$

So this shows $A_{K^*} = A_{K}^*$.