Let $X = L^2(\Omega)$, $\Omega \subset \mathbb{R}^N$ be an open set (or a $\sigma$-finite measure space), $B \in L^2( \Omega \times \Omega)$. Then the Hilbert-Schmidt operator $T \in \mathcal L(X)$ defined by $Tf(x) = \int_\Omega B(x,y) f(y) \mbox{d}y$ is known to be compact. I would like to find its transpose $T'$, $T' \in \mathcal L(X')$, and show that it is compact too.
The last is easy: $T'$ is compact by Schauder' theorem, since $T$ is compact. The first part is more difficult. I know that $L^2$ is a Hilbert space and moreover $L^2$ is isomorphic to its dual $(L^2)'$ as a special case of the Riesz' theorem, then $T'$ coincide with the adjoint $T^*$, but at this point of the lecture notes Hilbert spaces framework has not been introduced yet. Furthermore, I know the solution (for example, Yosida pp 197-198), i.e. $(T^*g)(y) = \int_\Omega \overline{B(y,x)} g(x) \, \mbox{d}x$. This is my reasoning:
By definition of dual operator adapted to this case, $\langle T' F, f \rangle_{X' X} = \langle F, Tf \rangle_{X' X} $, so $T' F (f) = F(T_f)$. $F \in (L^2)'$ is represented via Riesz' theorem as
\begin{equation} F_g (f) = \int_\Omega g(x) f(x) \, \mbox{d}x, \end{equation}
for all $f \in L^2$ and where $g \in L^2$. Hence
\begin{equation} F_g (Tf) = \int_\Omega g(x) \int_\Omega B(x,y)f(y) \, \mbox{d}y \, \mbox{d}x . \end{equation}
Since $g(x)$ does not depend on $y$, with an application of Fubini-Tonelli theorem we can write
\begin{equation} F_g (Tf) = \int_\Omega \int_\Omega g(x) B(x,y)f(y)\,\mbox{d}x \, \mbox{d}y = T' F (f). \end{equation}
Now, I'm stuck. How to achieve the conclusion? And what is wrong? (maybe would be easier "what is correct?")
Notation. $F \in X'$ is a functional. I dropped dependencies on variables when not necessary, e.g $F_g( Tf )$ is a shorthand for $F_g( Tf (x) )$.