This is from the handbook of spectral theory...
Take $H = L^2(\mathbb{R})$. Consider $Dom(T) = H^1(\mathbb{R})$ and $T = i \partial_x $. What is $(Dom(T^*), T^*)$? And if we choose $Dom(T) = C_0^{\infty}(\mathbb{R})$?
Solution: Let $f \in L^2(\mathbb{R})$. We find \begin{equation} \Phi_x : (Dom(T), || \cdot ||_{L^2}) \longrightarrow (\mathbb{C}, |\cdot |) \end{equation} \begin{equation} g \longrightarrow -i \int_{\mathbb{R}} g'(x) \bar{f}(x) dx \end{equation} which is continuous if and only if $f \in H^1(\mathbb{R})$. It follows that $Dom(T^*) = H^1(\mathbb{R})$. Moreover, knowing that $f \in \mathbb{H}^1(\mathbb{R})$ and integration by parts gives \begin{equation} \Phi_f(g) = -i \int_{\mathbb{R}} g'(x) \bar{f}(x) dx = \int_\mathbb{R} g(x) \bar{-i f'(x) } dx \end{equation} which means that $T^* = T$. Now, if we choose $Dom(T) = \mathcal{C}_0^{\infty}(\mathbb{R})$, for the same reasons as before, we still have $Dom(T*) = H^1(\mathbb{R})$. But $T$ is not self-adjoint, (we do not have $T = T*$).
I am confused about the line continuous if and only if $f \in H^1$, is there a theorem on continuous linear functionals that I am missing? Moreover, how does this work regarding an example where the operator is second order?
Best wishes,
Catherine