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Consider the Hilbert space $L^2(\mathbb{R})$, and unbounded operator $Au:=iu’$ with domain $$D(A)= \{u \in L^2(\mathbb{R}) | u \text{ is absolutely continuous and } u’ \in L^2(\mathbb{R})\} $$ I’m showing that it’s self-adjoint. It’s not hard to show that it’s symmetric, and suffice to show that $D(A^*) \subseteq D(A)$, which I don’t know how to deal with. Any help will be appreciated.

milky sausage
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3 Answers3

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Let $u\in D(A^\ast)$. For all $\phi\in C_c^\infty(\mathbb{R})$ we have $$ \langle A^\ast u,\phi\rangle=\langle u,A\phi\rangle=-i\int u\phi'=i\langle u',\phi\rangle, $$ where $u'$ is the distributional derivative of $u$.

Thus $$ |\langle u',\phi\rangle|=|\langle A^\ast u,\phi\rangle|\leq \|A^\ast u\|_2\|\phi\|_2. $$ Hence $u'\in L^2(\mathbb{R})$ by the Riesz representation theorem.

Now let $$ \tilde u(x)=\int_0^x u'(y)\,dy. $$ This function is absolutely continuous and has weak derivative $u'$. It follows that there exists a constant $C\in\mathbb{R}$ such that $u=\tilde u+C$ a.e. as explained here.

MaoWao
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  • I have one more question, $u=u^~+C$ doesn’t mean $u$ is square integrable if $C$ is arbitrary, does it? – milky sausage Dec 19 '20 at 06:26
  • In this part I only showed that if the weak derivative of $u$ is square-integrable, then $u$ is absolutely continuous. But the domain of $A^\ast$ is by definition contained in $L^2$, so $u$ must be square-integrable. – MaoWao Dec 19 '20 at 09:06
  • Got it! Thanks you! – milky sausage Dec 19 '20 at 09:09
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Suppose $v\in\mathcal{D}(A^*)$. Then there exists $w\in L^2(\mathbb{R})$ such that, for all $u\in\mathcal{D}(A)$, one has $$ \int_{-\infty}^{\infty} (Au)vdx=\int_{-\infty}^{\infty}uwdx. $$ Information about $w$ is obtained by choosing specific $u \in \mathcal{D}(A)$. For example, let $$ u_{a,b,\epsilon,\delta}=\int_{-\infty}^{x}\frac{1}{\epsilon}\chi_{[a-\epsilon,a]}(t)-\frac{1}{\epsilon}\chi_{[b,b+\delta]}(t)dt. $$ This function is $0$ for $x < a-\epsilon$ and $0$ for $x > b+\delta$. It is $1$ in $[a,b]$, and it is linear on $[a-\epsilon,a]$ and on $[b,b+\delta]$. $u_{a,b,\epsilon,\delta}\in\mathcal{D}(A)$. So, the first equation applied with $u=u_{a,b,\epsilon}$ gives $$ \frac{1}{\epsilon}\int_{a-\epsilon}^{a}vdx-\frac{1}{\epsilon}\int_{b}^{b+\delta}vdx=\int_{-\infty}^{\infty}u_{a,b,\epsilon,\delta}wdx $$ The limit as $\epsilon\downarrow 0$ exists, because it exists on the right. Similarly the limit as $\delta\downarrow 0$ exists. It follows that the following holds for almost all $a,b\in\mathbb{R}$: $$ v(a)-v(b)=\int_{a}^{b}w(x)dx $$ Therefore, $v$ is equal a.e. to an absolutely continuous function that is in $L^2$ with a derivative $v'=w\in L^2$. So $v\in\mathcal{D}(A)$. In other words $\mathcal{D}(A^*)\subseteq\mathcal{D}(A)$. The opposite inclusion follows directly using the symmetry of $A$.

Disintegrating By Parts
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On the Hermite Function basis, $A$ is a tri-diagonal matrix $$const \times \begin{bmatrix} 0 & -i\sqrt{1} & 0 & 0 & 0 & \cdots \\ i\sqrt{1} & 0 & -i\sqrt{2} & 0 & 0 & \cdots \\ 0 & i\sqrt{2} & 0 & -i\sqrt{3} & 0 & \cdots \\ 0 & 0 & i\sqrt{3} & 0 & -i\sqrt{4} & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{bmatrix}$$ (Wikipedia). Since the off-diagonal elements are $\mathscr O(\sqrt n)$, by Carleman's test $A$ is essentially self-adjoint on the span of the Hermite Functions. Since the Hermite Functions are complete, $A$ is essentially self-adjoint on the larger domain.

Keith McClary
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