Consider the vector space $V$ of complex-valued functions on $[0,1]$ with the inner product $\langle f,g \rangle = \int_{0}^{1} f(t)\overline{g(t)} dt$.
Let $K(s,t)$ be a continuous complex-valued function on $[0,1]\times[0,1]$. Define $T :V \rightarrow V $ by $T(f)(s) = \int_{0}^{1} K(s,t)f(t) dt$. What is the adjoint of $T$?
This is what I have done:
$\langle Tf,g \rangle = \int_{0}^{1}( \int_{0}^{1} K(s,t) f(t) dt )\overline{g(s)} ds$ this is the integral of $K(s,t)f(t)\overline{g(s)}$ over $[0,1]\times[0,1]$ so $T^*(g)$ ought to be $\int_{0}^{1} \overline{K(s,t)} g(t) dt$ because this would make $\langle f,T^*g\rangle$ also the integral of $K(s,t)f(t)\overline{g(s)}$ over $[0,1]\times[0,1]$.
Am I correct? I think there need to be extra conditions on $K(s,t)$, maybe something to do with Fubini's theorem.
Fubini conditions are surely met, because of continuity on a compact set assumptions over each function involved in your integrals.
Therefore you can swap iterated integrals to get:
$$\begin{split} \langle Tf, g\rangle &= \int_0^1 \left( \int_0^1 K(s,t) f(t)\ \text{d} t\right)\ \overline{g(s)}\ \text{d} s \\ &= \iint_{[0,1]\times [0,1]} K(s,t) f(t)\ \overline{g(s)} \text{d} t\ \text{d} s \\ &= \iint_{[0,1]\times [0,1]} \overline{\overline{K(s,t)}\ g(s)}\ f(t)\ \text{d} t\ \text{d} s \\ &= \int_0^1 f(t)\ \overline{\left( \int_0^1 \overline{K(s,t)}\ g(s)\ \text{d} s\right)}\ \text{d} t \end{split}$$
hence $T^* g(t) := \int_0^1 \overline{K(s,t)}\ g(s)\ \text{d} s$ as you’ve already found.