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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.

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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, $$ \mathcal{R}(P-iI)^{\perp}= \mathcal{N}(P^*+iI)=[\{ e^{-x}\}] \\ \mathcal{R}(P+iI)^{\perp}= \mathcal{N}(P^*-iI)=[\{0\}]. $$

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  • $\begingroup$ Thanks! How do I prove that $P^*$ has the same action and domain? $\endgroup$ – Sergey Dylda Jan 14 at 10:36
  • $\begingroup$ @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. $\endgroup$ – DisintegratingByParts Jan 14 at 17:24

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