Given a real valued function $V\in C_0^\infty (\mathbb{R}^n)$, let $$A_\lambda=-\Delta+V(x)+\lambda $$ denote an unbounded operator on $L^2(\mathbb{R}^n)$ with domain $D(A)=C_0^\infty (\mathbb{R}^n)$, and for $u\in D(A)$, let $$a_\lambda(u)=\langle A_\lambda u,u\rangle_{L^2}$$ One part of the problem is to show that $A_\lambda$ is essentially self-adjoint.

I proved, using Lax-Milgram, that there exists a bounded, self adjoint operator $B_\lambda(u): H^1\rightarrow H^1$ such that $$\langle B_\lambda u, u \rangle_{H^1}=a_\lambda(u)$$ and also $a_\lambda$ can be extended to $H^1$.

My guess is that $B_\lambda=\overline{A_\lambda}$, but I do not know how to show this. Any idea/hints?

  • $\begingroup$ Thanks, I will look at @ChristophKehle $\endgroup$ – nerd May 16 '15 at 2:26

Let $X$ be a Hilbert space, and let $A : \mathcal{D}(A) \subseteq X\rightarrow X$ be a densely-defined and closable linear operator with closure $A^{c}$. If $B$ is a bounded linear operator on $X$, then $(A+B)^{c}=A^{c}+B$ and $(A+B)^{\star}=A^{\star}+B^{\star}$. The operator $u\mapsto Vu$ is bounded and selfadjoint. Therefore $$ (-\Delta +V)^{\star} = (-\Delta)^{\star}+V = (-\Delta)^{c}+V = (-\Delta+V)^{c}. $$ (I'm assuming you know that $-\Delta$ is essentially selfadjoint.)


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