Prove $i\frac{d}{dx}$ is not essentially self-adjoint.

Question

Consider $T=i\frac{d}{dx}$ on $C_c^{\infty}(0,\infty)\subset L^2(0,\infty)$, the infinitely differentiable functions with compact support away from the origin. Prove that $T$ is not essentially self-adjoint.

My guess is to show that $(T^*-i)$ or $(T^*+i)$ has non trivial kernel, but I dont know how to do this. For instance, I don't know what $T^*$ is in the first place.

Answer 1

By the definition of the adjoint, for any $f \in C^\infty_c(0,\infty)$ and $g \in L^2(0,\infty)$, we must have $$ \langle Tf, g \rangle = \langle f, T^*g \rangle. $$ Whenever $g$ is differentiable and within the domain of $T^*$, integration by parts yields the following: $$ \begin{align} \langle Tf,g \rangle &= \int_0^\infty i \frac d{dt}f(t) \bar g(t)\,dt \\ & = \left.if(t)\bar g(t)\right|_0^\infty - \int_0^\infty f(t) \cdot \overline{-i\frac{d}{dt}g(t)}\,dt = \int_0^\infty f(t) \cdot \overline{i\frac{d}{dt}g(t)}\,dt \end{align} $$ so that $T^*g = i \frac{d}{dx} g$. With this is established, consider the function $g(x) = e^{-x}$. It suffices to note that

1. $g \in \mathrm{dom}(T^*)$ (i.e. $f \mapsto \langle Tf,g \rangle$ is continuous over $\mathrm{dom}(T)$),
2. $g \in \ker(T^* + i)$.