# 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.

• thank you very much for the great answer so what is the domain of the adjoint, based on what you wrote? Is the domain the Sobolev space $H^2$? – kroner Oct 28 at 3:00
• or is $H^2$ only contained in the domain? – kroner Oct 28 at 3:17
• @kroner : Yes, the domain of $T^*$ is $H^2(\mathbb{R})$, and the closure of $T$ is also this operator. – Disintegrating By Parts Oct 28 at 3:36
• thank you so much – kroner Oct 28 at 3:37
• @kroner : You're welcome. I had to rethink these things again. It's a good question. – Disintegrating By Parts Oct 28 at 3:38

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

• thank you but this just shows the operator is symmetric. The adjoint extends the operator on a possibly larger domain. – kroner Oct 26 at 16:33
• the operator and its adjoint agree on the common domain but the domain of the adjoint can extend that of the original. We need to find the adjoint and its larger domain. There is a difference between symmetric and self adjoint for unbounded operators – kroner Oct 26 at 16:36
• Of course, You are right. I edited my answer accordingly. – Peter Melech Oct 26 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 $T$ – Peter Melech Oct 26 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. – Peter Melech Oct 26 at 17:06