# Computing adjoint of differentiation operator

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 $$H$$.

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.

• @kroner : Yes, the domain of $T^*$ is $H^2(\mathbb{R})$, and the closure of $T$ is also this operator. Oct 28, 2020 at 3:36
• @kroner : You're welcome. I had to rethink these things again. It's a good question. Oct 28, 2020 at 3:38
• how did you prove that domain of T* is $H^2(\mathbb{R})$ Mar 4, 2023 at 4:43
• 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 $H^2(\mathbb{R})$ ,$g\in H^2(\mathbb{R})$. Mar 4, 2023 at 5:34

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: $$H^2(\mathbb{R})\subseteq D(T^*).$$

• Of course, You are right. I edited my answer accordingly. Oct 26, 2020 at 16:45
• 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 $T$ Oct 26, 2020 at 16:56
• 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. Oct 26, 2020 at 17:06
• In this case my guess was false and there was a unique extension of $T$ being its adjoint. I promise to think about it. Oct 26, 2020 at 17:14
• You´re welcome! Oct 26, 2020 at 17:15