I want to show that the adjoint of the identity $i: H_0^s(\Omega)\rightarrow L^2(\Omega), x\mapsto x$ where $s>0$ is arbitrary is the fractional Laplacian i.e. $(-\Delta)^{s}$ where $(-\Delta)^s$ is defined by measurable functional calculus. We assume $\Omega$ to be open and bounded. Any ideas to support me?
What I know is that if $\Omega=\mathbb{R}^d$ then we could use the definition of $H_0^s(\Omega)$ via Fourrier transform.
But such an definition is not available if $U$ is assumed to be open and bounded.
However I think we one might identify elements in $H_0^s(\Omega)$ via zero extention as elements of $H_0^s(\mathbb{R})$.
It depends what you mean by 'adjoint'.
First possibility is to be the adjoint in the sense of dual spaces. Then $i^*:(L^2)^* \to (H^s)^*$. If one identifies $L^2=(L^2)^*$, then $i^*(f)$ applied to $u\in H^s$ is $i^*(f)(u) = \int_\Omega f\cdot u$.
Second possibility is the Hilbert space adjoint. The definition of $i^*$ is $\langle i(u),v\rangle_{L^2} = \langle u,i^*(v) \rangle_{H^s}$ for all $u\in H^s$, $v\in L^2$. In order to compute $v$, one has to solve the variational problem $$ \langle u,i^*(v) \rangle_{H^s} = \langle i(u),v\rangle_{L^2} = \int_\Omega uv\ dx \quad \forall u\in H^s. $$ The left-hand side is the inner product in $H^s$, induced by the fractional Laplacian. In this sense, $i^*(v) = (-\Delta^s)^{-1}v$.