Does a symetric complex function $k(t,s)$ verify $\overline{k(t,s)}=k(t,s)$? I am trying to figure out why an integral operator is self-adjoint.
The operator is:
$$K(f)=\int_{0}^{1} k(t,s)f(s)ds$$
From $L^2([0,1])$ to $L^2([0,1])$ and $0, \leq t,s \leq 1$
So I did a bit of algebra and obtained that for the adjoint operator $K^*$ to be equal to $K$ then $k$ must satisfy  $\overline{k(t,s)}=k(t,s)$. The problem is that the only thing I know is that  $k(t,s)=k(s,t)$ for all $t,s$ which satisfy $0, \leq t,s \leq 1$.
So, Can I conclude it? 

$k(t,s)=k(s,t) \Rightarrow \overline{k(t,s)}=k(t,s)$


The actual condition is $k(t,s) = \overline{k(s,t)}$. This is very similar to the matrix case: if you think of ${k(s,t)}_{s,t\in[0,1]}$ as being a matrix that has uncountably many rows and columns, then swapping $t$ and $s$ is akin to taking a transpose and the complex conjugate is self-explanatory.
\begin{align}
\langle g,Kf\rangle &= \int_0^1 g(t)\overline{Kf(t)}\,dt \\
&= \int_0^1 g(t)\overline{\int_0^1 k(t,s)f(s)\,ds}\,dt \\
&= \int_0^1 g(t)\int_0^1 \overline{k(t,s)f(s)}\,ds\,dt \\
&= \int_0^1 \left(\int_0^1 \overline{k(t,s)}g(t)\,dt\right)\overline{f(s)}\,ds.
\end{align}
Evaluating $\langle Kg,f\rangle$, we get
\begin{align}
\int_0^1 Kg(t)\overline{f(t)}\,dt &= \int_0^1\left(\int_0^1 k(t,s)g(s)\,ds\right)\overline{f(t)}\,dt.
\end{align}
Swapping the roles of $t$ and $s$ in the last expression, we have
$$\langle Kg,f\rangle = \int_0^1\left(\int_0^1 k(s,t)g(t)\,dt\right)\overline{f(s)}\,ds.$$
Hopefully this helps.
