# Prove that there is no self-adjoint extension using deficiency indices

Consider an operator $$P =-i\frac{d}{dx} : dom(P) \to L^2(\mathbb{R}^+)$$ where $$dom(P) = \{ f \in \mathcal{D}(\mathbb{R}^+) : f(0)=0\}$$ where $$\mathcal{D}(\mathbb{R}^+)$$ - smooth compactly supported functions (test-functions).

I need to prove that this operator has no self-adjoint extension by calculating deficiency indices.

But I'm stuck calculating those indices by their definition:

$$n_+ (P) = dim(im(P+i)^\perp) \\ n_-(P)=dim(im(P-i)^\perp)$$

So I have: $$im(P+i)^\perp = \{ \phi \in L^2(\mathbb{R}^+) | <\phi, -i\frac{d}{dx}f+if>=0, \forall f \in dom(P)\}$$ That is, a set of such $$\phi \in L^2(\mathbb{R}^+)$$ such that: $$\int^\infty_0 \bar{\phi}(x)(-i\frac{df}{dx}+if)dx=0$$

How do I proceed from here? I would try to do integration by parts, but general $$\phi$$ does not have to be differentiable (only square integrable).

I need to show somehow that one of the deficiency indices is not equal to zero.

The closure $$\overline{P}$$ of $$P$$ in $$L^2[0,\infty)$$ has a domain $$\mathcal{D}(\overline{P})$$ consisting of every $$f\in L^2[0,\infty)$$ that is equal a.e. to an absolutely continuous function $$\tilde{f}\in L^2[0,\infty)$$ such that $$\tilde{f}'\in L^2[0,\infty)$$ and $$\tilde{f}(0)=0$$. The ajdoint $$P^*$$ has the same action and domain except that $$\tilde{f}(0)$$ is unconstrained.
Because of the homogeneous endpoint condition $$\overline{P}$$ has no non-trivial eigenfunctions. However $$P^* e^{-x}=-ie^{-x}$$ does hold, and $$e^{-x}\in L^2[0,\infty)$$. $$P^* f = if$$ has no non-trivial solutions $$f\in\mathcal{D}(P^*)$$ because $$e^{x}\notin L^2[0,\infty)$$.
Summarizing, $$\mathcal{R}(P-iI)^{\perp}= \mathcal{N}(P^*+iI)=[\{ e^{-x}\}] \\ \mathcal{R}(P+iI)^{\perp}= \mathcal{N}(P^*-iI)=[\{0\}].$$
• Thanks! How do I prove that $P^*$ has the same action and domain? – Sergey Dylda Jan 14 at 10:36
• @SergeyDylda Start with the adjoint relation, meaning $g\in\mathcal{D}(P^*)$ iff $\int_0^{\infty}(-if'(t))\overline{g(t)}dt = \int_0^{\infty}f(t)\overline{L^*g}dt$ holds for all $f\in\mathcal{D}(L)$. Use limits of functions $f$ in the domain to replace $f$ with piecewise linear functions that approximate a step function, and take a limit of both sides to conclude $ig(s)-ig(r)=\int_r^s \overline{L^*g} dt$ a.e. for $g\in\mathcal{D}(L^*)$. Conclude that $g$ is equal a.e. to an absolutely continuous function, and $-ig'=L^*g$. Finish by showing all such functions work in the adjoint relation. – DisintegratingByParts Jan 14 at 17:24