domain of the adjoint of the momentum operator.

Question

I was reading the book by Roman where he discusses the linear momentum operator on the half line: $pf= -if’$, this operator is densely defined and in the book he finds the adjoint by using the inner product on $L^2(0,\infty)$, one wishes to find the class of $h$ such that: $$ \int _0 ^\infty h^*(-ig’) dx = \int _0 ^\infty \psi^* g \: dx $$ for some $\psi$ and all $g$ in the domain of $p$. So clearly this is easy to do if $h$ is differentiable and with square integrable derivative by integration by parts. The question is: could there be non differentiable $h$ that satisfy this? I have been trying to show that there aren’t any such functions but can’t get anywhere and the book assumes this.

Answer 1

Suppose your equation holds for all $g\in\mathcal{D}(p)$. Let $g_y$ be the function that is $1$ on $[0,y]$, is linear on $[y,y+\epsilon]$ and is $0$ at $y+\epsilon$, as well as for all $x > y+\epsilon$. Then your adjoint equation would require $$ \int_0^{\infty}h^* (-ig_y')dx=\int_0^{\infty}\psi^* g_ydx \\ -i\frac{1}{\epsilon}\int_{y}^{y+\epsilon}h^*dx=\int_0^{\infty}\psi^*g_ydx $$ Now, letting $\epsilon\downarrow 0$ gives $$ -ih^*(y)=\int_0^y\psi^*dx $$ Hence, $h$ must be differentiable a.e., $h(0)=0$, and $$ h'(y)=-i\psi(y). $$