# Examples of self adjoint compact operators on Hilbert spaces

As the titles says, I'm looking for examples of self adjoint compact operators on Hilbert spaces.

So far I know of the diagonal operator on $$\ell^2$$,

$$(Tx)_i = \alpha_ix_i$$

for some sequence $$\alpha_i \to 0$$; and the Hilbert Schmidt integral operator in $$L^2(\Omega)$$,

$$Kg = \int_{\Omega}K(x,y)g(y)dy$$

with $$K \in L^2(\Omega^2)$$ a symmetric Hilbert-Schmidt kernel.

I would also like to know of some applications that use Hilbert Schmidt integral operators.

Edit: I'd be really grateful to know about the behaviour of the norm of the operators as well.

• @cmk awesome, thank you! – Guido A. Jul 24 at 1:44

An important example is the inverse of the Laplace operator: $$T=(-\Delta)^{-1}$$ as mapping from $$H^1_0(\Omega)^*$$ to $$H^1_0(\Omega)$$. Here, $$\Omega$$ is an open and bounded domain in $$\mathbb R^n$$. That is, given $$z\in H^1_0(\Omega)$$, $$u:=Tz$$ is the weak solution of $$-\Delta u = z,$$ which is equivalent to $$\int_\Omega \nabla u \cdot\nabla v \ dx = \int_\Omega zv \ dx \quad\forall v\in H^1_0(\Omega).$$

There are tons of examples, but I'll give a few from the context of PDEs.

@daw already gave a very good example. Let me provide a similar one, in the form of the heat propagator. You may know that the study of the heat equation is fundamental in many fields of PDEs. To that end, let $$M$$ be a compact, Riemannian manifold without boundary with metric $$g$$. If $$p$$ denotes the fundamental solution to the heat equation on $$M\times (0,\infty),$$ then for $$t>0,$$ we define the heat propagator as the map $$e^{-t\Delta_g}:L^2(M)\rightarrow L^2(M)$$ via $$e^{-t\Delta_g} f(x)=\int\limits_M p(x,y,t) f(y)\, \omega_g(y),$$ where $$\omega_g$$ is the volume form. It's not too hard to show that $$e^{-t\Delta_g}$$ is a self-adjoint, compact operator on $$L^2$$.

Here is another example, which is of significant importance in the realm of semi-classical analysis. Note: I'm going to use $$\mathcal{S}$$ to denote the Schwartz space, and I'm going to use it for both $$\mathcal{S}(\mathbb{R}^n)$$ and $$\mathcal{S}(\mathbb{R}^{2n})$$. Symbols will always be in the latter, and everything else in the former.

Let $$a\in\mathcal{S},$$ and $$h>0.$$ We define the Weyl quantization to be the operator $$a^w(x,hD)$$ on $$\mathcal{S}$$ so that $$a^w(x,hD)u(x)=\frac{1}{(2\pi h)^n}\int\limits_{\mathbb{R}^{n}}\int\limits_{\mathbb{R}^{n}}e^{\frac{i}{h}(x-y)\cdot\xi} a\left(\frac{x+y}{2},\xi\right)u(y)\, dyd\xi.$$ This is related to applying the conjugation of $$a$$ by the semi-classical Fourier transform to $$u$$, but the convex combination of $$x$$ and $$y$$ stops this from being quite a semi-classical Fourier multiplier (unless $$a$$ is independent of $$x$$). This is called a pseudifferential operator (which can also be defined on manifolds, but I'd rather not make that digression). This operator can be defined as a map $$a^w(x,hD):\mathcal{S}'\rightarrow\mathcal{S},$$ and it is formally self-adjoint here, provided $$a$$ is real-valued. In fact, in a proof similar to proving that the Hilbert-Schmidt kernel operator is bounded in $$L^2$$, you get that the Weyl quantization sends $$L^2\rightarrow L^2$$, and it's bounded. In a proof similar to that of the Arzela-Ascoli theorem, one can show this is compact, as well.

You also asked about applications of Hilbert-Schmidt operators. First, here's a classical result:

If $$T:L^2(X_1,\mu_1)\rightarrow L^2(X_2,\mu_2)$$ is HS, then there exists $$K\in L^2(X_1\times X_2,\mu_1\times \mu_2)$$ such that $$(Tu,v)_{L^2}=\iint K(x_1,x_2)u(x_1)\overline{v(x_2)}\, d\mu_1(x_1)d\mu_2(x_2).$$ Note that $$\|T\|_{\text{HS}}=\|K\|_{L^2}.$$ Conversely, if $$K\in L^2(X_1\times X_2,\mu_1\times \mu_2)$$, then the above integral defines a Hilbert-Schmidt operator $$T$$, with the same norm equality.

If a bounded operator on a separable Hilbert space can be written as the product of two HS operators, then we say that this operator is trace-class. One characterization of such operators is that a bounded operator $$C$$ is trace-class if and only if $$C$$ is compact and $$(C^*C)^{1/2}$$ has summable eigenvalues. The norm on trace-class operators is given by the infimum over all decompositions of the operator (in the way described by the definition of trace-class) of the product of their respective Hilbert-Schmidt norms. This forms the space into a Banach space. A nice thing about trace-class operators is that we can define their determinants in an appropriate sense (although I won't go through this construction). Trace-class operators get their name because we can take their trace, in the sense that if $$C$$ is trace-class, then we can define its trace as $$\text{Tr}\ C=\sum c_{jj},$$ where $$c_{jj}=(Cu_j,u_j)$$, and $$\{u_j\}$$ is the orthonormal basis for the given Hilbert space. Alternatively, if we write $$C=AB^*$$, then $$\text{Tr}\ C=(A,B)_{\text{HS}}=\sum_{j,k} a_{jk}\overline{b}_{jk},$$ where $$a_{jk}=(Au_j,u_k).$$ We can talk about the trace of an integral operator as follows: Let $$A,B$$ be Hilbert-Schmidt on $$L^2(X,\mu)$$ with integral kernels $$K_A,K_B\in L^2(X\times X,\mu\times \mu).$$ If $$C=AB$$, then we can write $$Cu(x)=\iint K_A(x,z)K_B(z,y) u(y)\, d\mu(y)d\mu(z),$$ so we get the trace as $$\text{Tr}\ C=\iint K_A(x,z)K_B(z,y) \, d\mu(y)d\mu(z).$$ The integral kernel of $$C$$ is given by $$K_C(x,y)=\int K_A (x,z)K_B(z,y)\, d\mu(z).$$ So, we might want the trace to be $$\text{Tr}\ C=\int K_C(x,x)\, d\mu(x).$$ The problem is that the diagonal might have measure zero, so this is a bit of a technical point. Essentially, we can define $$K_C(x,x)$$ by restriction a.e., and for nice spaces (those commonly looked at), we can get from the first expression of the trace to the second by continuity.

One more thing related to Hilbert-Schmidt operators. Earlier, I mentioned something that is sometimes referred to as the Hilbert-Schmidt kernel theorem. There is a similar result for distributions which is utilized constantly in microlocal analysis and the study of pseduodifferential operators, called the Schwartz kernel theorem:

Let $$A:\mathcal{S}\rightarrow\mathcal{S}'$$ be a continuous linear map. Then, there exists $$K_A\in\mathcal{S}'(\mathbb{R}^n\times\mathbb{R}^n)$$ so that for all $$u,v\in\mathcal{S},$$ $$\langle Au,v\rangle=\langle K_A, u\otimes v\rangle,$$ where $$(u\otimes v)(x,y)=u(x)v(y)\in \mathcal{S}(\mathbb{R}^n\times\mathbb{R}^n).$$

We sometimes abuse notation and write this as $$Au(x)=\int K_A(x,y) u(y)\, dy,$$ so that $$\langle Au,v\rangle=\iint K_A(x,y) v(x)u(y)\, dydx.$$ Seriously, this is used all the time.

References-

Michael Taylor: Partial Differential Equations I

Maciej Zworski: Semiclassical Analysis