# What is the definition of a self-adjoint operator for any operator?

What is the definition of a self-adjoint operator for any operator? In the books they always mention the self-adjoint condition on Hilbert or Banach spaces. But what is the general definition?

I ask this then, I thought that by proving that if $$A:X\to Y$$ operator and $$A*:Y'\to X'$$ adjoint, with $$(x,Ay)=(x,A^*y)$$ was sufficient to conclude that the operator is self-adjoint, which is false, this only proves that it is symmetric.

I also ask this because in the link Computing the adjoint operator of Laplacian operator and applying it to the Gaussian function., I came to the conclusion that the Laplacian is self-adjoint, but I only proved that it is symmetric. Should I prove that $$D(\Delta)=D(\Delta^*)$$?

If $$\Delta:D(\Delta)\to L^2$$ with $$D(\Delta)=\left\{\varphi\in L^2: \Delta \varphi\in L^2\right\}$$. By definition, the adjoint of $$\Delta$$ is $$\Delta^{*}:D(\Delta)^{*}\subset (L^2)'\to (D(\Delta))'$$ where $$D(\Delta^{*})=\left\{y'\in (L^2)':\exists x'\in (D(\Delta))',\, y'(\Delta \varphi)=x'(\varphi) \forall \varphi\in D(\Delta)\right\}$$ (definition by Wong).

Affirmation $$D(\Delta)\subset D(\Delta^{*})$$.

Proof. Let $$\psi\in D(\Delta)$$ then $$\psi\in L^2$$ and $$\Delta \psi\in L^2$$. Now, $$|(\Delta \varphi,\psi)|=|(\varphi,\Delta \psi)|\leq C\left\|\varphi\right\|_{2}$$ for all $$\varphi\in D(\Delta)$$, where $$C=\left\|\Delta\psi\right\|_{2}<\infty$$.

Therefore, $$\psi\in D(\Delta^{*})$$

Question 1. Why $$D(\Delta^*)\subset D(\Delta)$$?

My attempt: Let $$u\in D(\Delta^*)$$. By exercise 13.2, Wong, $$u\in L^2$$ (because $$(L^{2'}=L^2)$$ and exists a function $$f\in L^2$$ such that $$(u,\Delta v)=(f,v),\quad v\in D(A)$$. Therefore, $$\int u(x)\Delta v(x)dx=\int f(x)v(x)dx$$. Because $$\Delta$$ is symmetric, $$\int \Delta u(x)v(x)dx=\int f(x)v(x)dx$$ then (Why? By uniqueness?) $$\Delta u=f$$. Therefore $$\Delta u\in L^2$$, i.e. $$u\in D(\Delta)$$. Therefore $$D(\Delta^{*})\subset D(\Delta)$$.

This is correct? Is true that $$\Delta u=f$$? (By uniqueness of Riesz representation)

• For an operator to be self-adjoint (as in equal to its adjoint), the adjoint $A^*$ needs to be defined on the same domain and codomain as $A$, so that the equation $A = A^*$ makes sense. This means than the operator needs to be an operator on the same vector space, so $A \colon X \to X$ rather than $A \colon X \to Y$, and also some extra data about the vector spaces has to be used so that the adjoint of an operator is defined on the original vector spaces rather than on their duals. A common piece of extra data to use (eg in Hilbert spaces) is a fixed inner product on each space. May 23, 2021 at 5:17

## 2 Answers

It's somewhat technical. Let $$T:D(T)\subset X\rightarrow X$$ be a densely-defined, unbounded operator on a Hilbert space $$X$$. $$T$$ is said to be symmetric if $$D(T)\subset D(T^*)$$ and $$T=T^*$$ on $$D(T)$$, which is equivalent to the condition that $$(Tx,y)=(x,Ty)$$ for all $$x,y\in D(T).$$ That is, $$T^*$$ is an extension of $$T$$. We say that $$T$$ is self-adjoint if $$T=T^*$$ (so, $$D(T)=D(T^*)$$). Now, there is also the notion of essential self-adjointness. $$T$$ is essentially self-adjoint if there exists a unique extension of $$T$$ which is self-adjoint (or, equivalently, if its closure is self-adjoint). In this case, we often conflate the operator with its self-adjoint extension. In your prior post, you only showed symmetry because you were working in the Schwartz space. However, the Laplacian is essentially self-adjoint on this dense domain (called a core), and the domain of its self-adjoint extension is the Sobolev space $$H^2$$.

You need the Hilbert space structure to define self-adjointness since you need an inner product. As stated in the comments, you need the domain and codomain to be the same to make statements about the operator equaling its adjoint.

Suppose $$L : \mathcal{D}(L)\subseteq\mathcal{H}\rightarrow\mathcal{H}$$ is a symmetric linear operator on a dense linear subspace $$\mathcal{D}(L)$$ of a complex Hilbert space $$\mathcal{H}$$. Then the graph $$\mathcal{G}(L)$$ is a closed subspace of the graph $$\mathcal{G}(L^*)$$ in $$\mathcal{H}\times\mathcal{H}$$. Furthermore, one has the following orthogonal decomposition in $$\mathcal{H}\times\mathcal{H}$$: $$\mathcal{G}(L^*)=\mathcal{G}(L)\oplus\mathcal{N}_{-}\oplus\mathcal{N}_{+},$$ where $$\mathcal{N}_{+}=\{ \phi\in\mathcal{D}(L^*) : L^*\phi = i \phi \}$$ and $$\mathcal{N}_{-}=\{ \phi\in\mathcal{D}(L^*) : L^*\phi = -i\phi \}$$. These spaces are the so-called deficiency spaces of $$L^*$$. So, in order to verify that the closed operator $$L$$ is self-adjoint, it is necessary and sufficient to show that the so-called deficiency spaces $$\mathcal{N}(L^*\pm iI)$$ are $$\{0\}$$.