Let $p \in \mathbb{N}$, $a, b \in \mathbb{R}$. We define on $L_p (a,b)$ the operator
$$D = \frac{1}{i} \frac{d}{dx}$$ with the domain $$\operatorname{dom}D = \left\{ u \in W_p^1 (a,b) \ | \ u (a) = u(b) = 0 \right\}$$
Show, that $D^* = \frac{1}{i} \frac{d}{dx}$ on $L_{p'} (a,b)$, with its domain being equal to $W_{p'}^1 (a,b)$ (where $p'$ is the conjugate exponent of $p$)
I honestly don't know how to do this. My approach goes like this:
$$\langle Du, f \rangle = \frac{1}{i} \int_a^b u'(x) \overline{f (x)} dx = \frac{1}{i}u(x) \overline{f(x)} \ |_a^b - \frac{1}{i}\int_a^b u(x) \frac{d \overline{f}}{dx} (x) dx = \int_a^b u(x) \overline{\frac{1}{i}} \frac{d \overline{f}}{dx}(x) dx = \langle u, D f \rangle $$
But I think this is just proving it for a special case, right?
Because where are we proving the fact that $D^*$ is on the $L_{p'} (a,b)$ space, with the domain $W_{p'}^1 (a,b)$?
Should I try using a different inner product? Which one? With the integrand taken to the p-th power?
You want to prove that $D^* f \in L_{p'}$ for all $f \in W_{p'}^1$.
This follows almost from the definitions:
$$ W_{p'}^1(a,b) = \left \{ f: [a,b] \to \mathbb R: \frac{df}{dx} \text{ exists and } \frac{df}{dx} \in L_{p'}(a,b) \right\}$$
Since $D^* f = \frac{1}{i} \frac{df}{dx}$ it should be easy to see $D^* f \in L_{p'}$