It is known that for $L^2(\mathbb R)$ the operator $Tf(x) = if'(x)$ is unitary equivalent to $\hat T \hat f(\xi )= \xi \hat f(\xi) $. Where domain of T is $H^1(\mathbb R)$. Hence the Spectrum of T in this case is $EssRan (M_{\xi})=\mathbb R$
(EssRan $M_{\xi}$:= Essential range of multiplication by $h(\xi)=\xi$). Here $\hat T = M_{\xi}$
But in case of bounded domain (say $(0,1)$). If we consider $T'f(x) = f'(x)$, $D(T')=H^1(0,1)$.
The function $\phi (x)= e^{xz}$ satisfies $(T-zI)\phi=0$ for any $z \in \mathbb C$
Hence Spectrum of $T'$ is all $\mathbb C$.
It is clear that $T'$ is not unitary equivalent to multiplication in fourier space ($\{\hat f(n)\}_{n\in Z}\rightarrow \{n \hat f(n)\}$).
I need to know what is unitary equivalent operator of derivative in this case.(It is certainly not multiplication, then what it is?)
Let $T = i\frac{d}{dt}$ on $H^{1}(0,1)$. The restriction of $T$ to a linear subdomain is selfadjoint iff that subdomain is equal to the follow for some real $\alpha$: $$ \mathcal{D}(T_{\alpha})=\{ f\in H^{1} : f(0)+e^{i\alpha}f(1)=0 \}. $$ The only significant case (the others can be reduced to this case) is $f(0)=f(1)$, or $\alpha=0$. The restriction $T_{0}$ of $T$ to periodic functions in $H^{1}(0,1)$ is selfadjoint with an orthonormal basis of eigenvectors $\{ e^{-2\pi inx}\}_{n=-\infty}^{\infty}$. This operator is unitarily equivalent to multiplication on $\ell^{2}(\mathbb{Z})$ by $m(x)=2\pi x$ on $\mathbb{Z}$.
If you do not impose conditions on $T$, then the resulting operator has spectrum $\mathbb{C}$, and cannot be unitarily equivalent to a multiplication operator on any $L^{2}_{\mu}(\Omega)$ because $T$ is not normal.