3
$\begingroup$

This is a problem I have encountered in my studies and work in unbounded differential operators

Let us define the operator $T = -\frac{d^2}{dx^2}$ as an operator on $L^2(\mathbb{R})$ with domain $C_0 ^{\infty} (\mathbb{R})$ where the domain is the set of infinitely differentiable complex-valued functions on $\mathbb{R}$ with compact support. We are asked to compute the adjoint of this operator along with its domain (which extend the original $T$). Also, we are asked if this operator is essentially self-adjoint.

To be honest, I am new to this area of functional analysis and operator theory and so I find myself struggling, I do not know how to compute the adjoint and its domain. I know a basic criterion of essential self-adjointness is to check if the closure $\bar{T}$ is self-adjoint. I also know $T$ is symmetric (please see below) via integration by parts and using the boundary conditions arising from compact support. I have no idea how to do these things here or in practice in general. I thank all persons who can help with both parts of the problem.

***** Note: please let me clarify a few definitions. If $T$ is a densely defined linear operator on a Hilbert space $H$, the domain $D(T^*)$ is the set of $\phi \in H$ for which there is a $\eta \in H$ with $$ \langle T\psi,\phi \rangle = \langle \psi,\eta \rangle $$ for all $\psi \in D(T)$. For each such $\phi \in D(T^*)$ we define $T^* \phi = \eta$, and $T^*$ is called the adjoint of $T$. A densely-defined operator is said to be symmetric if $\langle T\phi,\psi \rangle = \langle \phi,T\psi \rangle$ for all $\phi,\psi \in D(T)$, and in this case $D(T) \subseteq D(T^*)$ and $T=T^*$ on $D(T)$ and $T^*$ is said to extend $T$. A symmetric operator is self-adjoint iff $D(T)=D(T^*)$ and thus $T=T^*$. An operator $T$ is said to be essentially self-adjoint if its closure $\bar{T}$ is self-adjoint, which is equivalent to $\ker(T^* \pm i) = \{0\}$ or $\text{Ran}(T \pm i)$ are dense in $H$.

$\endgroup$
0

2 Answers 2

4
$\begingroup$

The adjoint $T^*$ is defined as the set of $g\in L^2(\mathbb{R})$ for which there exists a constant $C_{g}$ such that $$ |\langle Tf,g\rangle_{L^2}| \le C_g\|f\|_{L^2},\;\;\; \forall f\in \mathcal{D}(T). $$ This inequality holds iff there is a unique $T^*g\in L^2$ such that $$ \langle Tf,g\rangle = \langle f,T^*g\rangle,\;\;\; \forall f\in\mathcal{D}(T). $$ ($T^*g$ is unique if it exists because $\mathcal{D}(T)$ is dense in $L^2(\mathbb{R})$.) The Fourier transform $\mathcal{F}$ on $L^2$ can be brought to bear on $|\langle Tf,g\rangle| \le C_g\|f\|_{L^2}$: $$ \langle \widehat{Tf},\widehat{g}\rangle=\langle \widehat{f},\widehat{T^*g}\rangle \\ \langle -\xi^2\widehat{f},\widehat{g}\rangle=\langle \widehat{f},\widehat{T^*g}\rangle \\ \langle\widehat{f},-\xi^2\widehat{g}\rangle=\langle \widehat{f},\widehat{T^*g}\rangle \\ \implies \widehat{T^*g}=-\xi^2\widehat{g} \in L^2 \\ T^*g = -\mathcal{F}^{-1}\xi^2\mathcal{F}g $$ So the adjoint $T^*$ is fully characterized in terms of the Fourier transform: it is unitarily equivalent to multiplication by $-\xi^2$ in the Fourier domain. Multiplication operators on $L^2(\mathbb{R})$ are self-adjoint. $$ T^*=-\mathcal{F}^{-1}\xi^2\mathcal{F} \\ \implies T^c = (T^*)^*=-\mathcal{F}^{-1}\xi^2\mathcal{F}, $$ where $T^c$ is the closure of $T$. $T^c$ is self-adjoint because it is unitarily equivalent to a multiplication operator.

$\endgroup$
4
  • 2
    $\begingroup$ @kroner : Yes, the domain of $T^*$ is $H^2(\mathbb{R})$, and the closure of $T$ is also this operator. $\endgroup$ Oct 28, 2020 at 3:36
  • 1
    $\begingroup$ @kroner : You're welcome. I had to rethink these things again. It's a good question. $\endgroup$ Oct 28, 2020 at 3:38
  • $\begingroup$ how did you prove that domain of T* is $H^2(\mathbb{R})$ $\endgroup$
    – Biplab
    Mar 4, 2023 at 4:43
  • $\begingroup$ Oh I get that . That’s because of Riesz Representation theorem if $f\to \langle Tf,g \rangle$ Is bounded there exists $g_1$ such that $\langle Tf,g \rangle=\langle f, g_1\rangle $ hence by the definition of $H^2(\mathbb{R})$ ,$g\in H^2(\mathbb{R})$. $\endgroup$
    – Biplab
    Mar 4, 2023 at 5:34
2
$\begingroup$

Let $T:L^2(\mathbb{R})\rightarrow L^2(\mathbb{R})$ be the unbounded operator defined on the dense set $C_0^{\infty}(\mathbb{R})$ by $T=-\frac{d^2}{dx^2}$ then for $f,g\in C_0^{\infty}(\mathbb{R})$ we have by integration by parts: $$\langle Tf,g\rangle=\int_{-\infty}^{\infty}-\frac{d^2}{dx^2}f(x)\overline{g(x)}dx=\int_{-\infty}^{\infty}\frac{d}{dx}f(x)\overline{\frac{d}{dx}g(x)}dx=-\int_{-\infty}^{\infty}f(x)\overline{\frac{d^2}{dx^2}g(x)}dx=\langle f,Tg\rangle$$ where there are no boundary values since $f,g$ have bounded support. This shows $T$ is symmetric on its domain.

$\textbf{Edit:}$ The question remains, for which $g\in L^2(\mathbb{R})$ is $$f\mapsto\int_{-\infty}^{\infty}-\frac{d^2}{dx^2}f(x)\overline{g(x)}dx$$ a continuous functional on $C_0^\infty(\mathbb{R})$.

$\textbf{Edit:}$ By the Cauchy-Schwarz inequality and the above we have $$|\langle Tf,g\rangle|\leq ||f||_{L^2(\mathbb{R})}\left(\int_{-\infty}^{\infty}|\frac{d^2}{dx^2}g(x)|^2dx\right)^{\frac{1}{2}}$$ which shows that the Sobolev space of twice weakly differentiable functions $H^2(\mathbb{R})$ is contained in the domain of the adjoint: $H^2(\mathbb{R})\subseteq D(T^*).$

$\endgroup$
9
  • 1
    $\begingroup$ Of course, You are right. I edited my answer accordingly. $\endgroup$ Oct 26, 2020 at 16:45
  • 1
    $\begingroup$ Can you think of a function $g\in L^2(\mathbb{R})\backslash C_0^\infty(\mathbb{R})$ so that the latter mapping is a continuous linear functional on $C_0^\infty(\mathbb{R})$?In this case it would be in the domain of the adjoint, which would thus be larger than that of $T$ $\endgroup$ Oct 26, 2020 at 16:56
  • 1
    $\begingroup$ Honestly I am not sure if such a function exists. Since we are actually talking about classes of functions one would have to find one with no smooth compactly-supported representative. I would guess it does not exist, but of course this would have to be proven. $\endgroup$ Oct 26, 2020 at 17:06
  • 1
    $\begingroup$ In this case my guess was false and there was a unique extension of $T$ being its adjoint. I promise to think about it. $\endgroup$ Oct 26, 2020 at 17:14
  • 1
    $\begingroup$ You´re welcome! $\endgroup$ Oct 26, 2020 at 17:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .