# Lecture 6: Spectral Theorem for Compact Operators ###### tags: 224a 2020 ## Compact Operators Rather than trying to understand all self-adjoint operators, we begin by focusing on the a simpler subclass. The simplest operators in $L(H)$ are the *finite rank* operators: $$A = \sum_{n\le N} (\phi_n, \cdot)\psi_n$$ for some $\phi_n,\psi_n\in H$. These operators always have closed range (since finite dimensional subspaces are closed), and can be understood using the methods of linear algebra. The compact operators are a larger class which inherit many of the nice properties of finite rank operators. **Definition.** An operator $T\in L(H)$ is *compact* if for every bounded set $B\subset H$, the closure of $T(B)$ is compact. Compactness may seem like a weird condition, but it comes in handy because it allows one to mimick finite dimensional arguments where one "optimizes" over the unit norm ball in $\mathbb{R}^n$ to produce e.g. an eigenvector. This is not directly possible in $H$ because the norm ball $B=\{x\in H:\|x\|\le 1\}$ is not compact: consider the infinitely many basis vectors $e_1,e_2,\ldots$, which have pairwise distance $\sqrt{2}$. A more intuitive understanding of this class is provided by the following theorem. **Theorem.** An operator $T\in L(H)$ is compact iff it is the norm limit of finite rank operators. Note that the content of the above theorem is that it is a *norm* limit. It is easy to see that every bounded operator is a limit of finite rank operators in the strong topology (i.e., there are finite rank $T_n$ such that for every $x$, $\|T_nx-Tx\|\rightarrow 0$ where the rate of convergence may depend on $x$.) The most obvious non-example of a compact operator is the identity, since it maps the unit ball to itself. A large class of examples is provided by integral kernel operators. We begin by proving one direction of this theorem (the other direction is similar and not hard). *Proof.* Assume $T\in K(H)$, let $B\subset H$ be the unit ball, and let $K=\overline{T(B)}$. Let $n$ be an integer. The set $$\cup_{z\in K} \{ y: \|z-y\|<1/n\}$$ is an open cover of $K$ so it must have a finite subcover; call this $S_n$. Let $P_n$ be the orthogonal projection onto $span(S_n)$. Note that $P_nT$ is a finite rank operator since $S_n$ is finite. Consider a point $x\in H$. Choose $z\in S_n$ such that $\|Tx-z\|<1/n$. Since $P_nTx$ is the closest point to $Tx$ in $span(S_n)$, we must have $$\|P_nTx-Tx\|\le \|z-Tx\|<1/n,$$ whence $\|P_nT-T\|<1/n$, so that $\|P_nT-T\|\rightarrow 0$ as $n\rightarrow \infty.$. $\square$ Two interesting families of compact operators are the following. **Integral Kernel Operators.** Suppose $K\in L^2(0,1)^2$. A Cauchy-Schwartz exercise shows that $\|T_K\|\le \|K\|_{L^2(0,1)^2}$, where $$T_Kf:=\int_0^1 K(x,y)f(y)dy.$$ Let $\{\phi_i\}$ be an ONB of $L^2(0,1)$. It is easy to check that the family of bivariate functions $\{\phi_i(x)\phi_j(y)\}_{i,j}$ is an ONB of $L^2(0,1)^2$, so we may expand $$K(x,y)=\sum_{ij} c_{ij} \phi_i(x)\phi_j(y).$$ Let $K_n(x,y)=\sum_{i,j\le n} c_{ij}\phi_i(x)\phi_j(y).$ As the difference $T_{K_n}-T_K$ is an integral kernel operator of the same type, we have $$\|T_K-T_{K_n}\|\le \|K-K_n\|_{L^2}=\sum_{i,j>n}|c_{ij}|^2\rightarrow 0$$ as $n\rightarrow \infty$, showing that $T_k$ is compact by the previous theorem. **Diagonal Operators.** If $\alpha_n$ is any sequence with $|\alpha_n|\rightarrow 0$, then the diagonal multiplication operator $T(e_i)=\alpha_i e_i$ is compact, since it is approximated by $T_n(e_i)=\alpha_i e_i \{i\le n\}$. Note that diagonal multiplication operators in $L^2(0,1)$, $T_gf = g(x)f(x)$ for $g\in L^\infty(0,1)$ are *not* compact, as seen by considering $g$ to be the indicator of any set of positive Lebesgue measure, since then the image $\overline{T_g(B)}$ contains infinitely many vectors pairwise separated by a constant distance. Finally, we mention that the set $K(H)$ of compact operators on $H$ is a 2-sided $*-$ideal in $L(H)$, which means that if $T\in K(H)$ then $T^*, BT, TB\in K(H)$ for any $B\in L(H)$. These properties are easy to verify from the characterization as of $K(H)$ as norm limits of finite rank operators. ## The Spectral Theorem for Compact Operators We will now show that the diagonal operators above are in a sense the *only* examples of compact operators, up to isomorphism. **Definition.** If $Tf=\lambda f$ for some $f\in H\setminus\{0\}$ then $f$ is called an eigenvector of $T$ and $\lambda$ is called an eigenvalue. **Theorem.** Suppose $T=T^*\in K(H)$. Then: 1. The eigenvalues of $T$ are real and may be ordered $\lambda_1,\lambda_2,\ldots \rightarrow 0$. 2. If $\lambda\neq 0$ is an eigenvalue of $T$, the eigenspace $V_\lambda:=\ker(\lambda-T)$ is finite dimensional. 3. There is an orthonormal basis of $H$ consisting of eigenvectors of $T$. The last property implies that we have the expansion $$T=\sum_n \lambda_n \phi_n \phi_n^*$$ where $\phi_n^*$ denotes the linear functional $(\phi_n,x)$ dual to $\phi_n$. Alternately, this can be expressed as $$T = UDU^*$$ where $U(e_n)=\phi_n$ is unitary and $D$ is a diagonal multiplication operator. *Proof of Theorem.* We begin by observing that for any eigenvectors $Tv=\lambda v$ and $Tw=\mu w$: $$\mu(v,w) = (v,Tw) = (T^*v,w)=(Tv,w)=\overline{\lambda}(v,w).$$ Plugging in $v=w$ shows that $\lambda=\overline{\lambda}$ so $\lambda$ must be real. For $\mu\neq \lambda$ we then must have $(v,w)=0$, so eigenvectors from distinct eigenvalues must be orthogonal. **Lemma 1.** For every $\epsilon>0$, the subspace $$S_\epsilon := span\{x: Tx=\lambda x, |\lambda|\ge \epsilon\}$$ is finite dimensional. *Proof of Lemma.* We first show that for every nonzero eigenvalue $\lambda$, $V_\lambda$ is finite dimensional. Assume not, i.e., there is an infinite sequence of orthonormal vectors $\{x_n\}$ such that $Tx_n=\lambda x_n$. Then $Tx_n\in K=\overline{T(B)}$ and we have $$\|Tx_n-Tx_m\|=|\lambda|\|x_n-x_m\|=\sqrt{2}|\lambda|$$ whenever $n\neq m$. But compactness implies that every sequence in $K$ must have a convergent subsequence, so this is impossible. The orthogonality of distinct eigenspaces (which are closed since they are kernels) implies that $$S_\epsilon = \bigoplus_{|\lambda|\ge \epsilon} V_\lambda.$$ Assume for contradiction that there are infinitely many direct summands, and choose one unit eigenvector $x_n$ from each eigenspace. By the same argument above, we have $$\|Tx_n-Tx_m\|=\|\lambda_nx_n-\lambda_mx_m\|\ge \epsilon$$ whenever $n\neq m$, since the hypotenuse of a right triangle is longer than its shorter side. This is impossible by compactness. $\square$ Lemma 1 implies properties (1) and (2) and that there are at most countably many eigenvalues. We now show that there are enough to form an orthonormal basis; the key is to show that we can always find one eigenvector, and the rest will follow by induction. **Lemma 2.** Either $\|T\|$ or $-\|T\|$ is an eigenvalue of $T$. *Proof of Lemma.* Let $c=\sup_{\|x\|=1} \|Tx\|$. If $c=0$ then we are done since $T=0$ and any unit vector in the kernel will do. Otherwise assume $c>0$ and let $\{x_n\}\subset B$ be a sequence such that $\|Tx_n\|\rightarrow c$. By compactness of $K$ we may pass to a subsequence such that $Tx_n$ converges to some vector $y\neq 0$. Let $A=c^2-T^2$, and observe that $(x,Ax)\ge 0$ for every $x\in H$, since $(x,T^2x) = \|Tx\|^2\le c^2$ for every $x$. Observe that $$(x_n,Ax_n)=c^2-\|Tx_n\|^2\rightarrow 0.$$ Since $A$ is positive it has a square root $\sqrt{A}$, so we have $$(x_n,\sqrt{A}^2x_n)=\|\sqrt{A}x_n\|^2\rightarrow 0.$$ Since $T\sqrt{A}$ is a bounded operator and $TA=AT$, this implies $$\|TAx_n\|=\|ATx_n\|\rightarrow 0,$$ which by continuity of $A$ implies $Ay=0$. Thus we have $$(c^2-T^2)y = (c-T)(c+T)y=0.$$ If $(c+T)y=0$ then $-c$ is an eigenvalue; otherwise $c$ is an eigenvalue with eigenvector $(c+T)y\neq 0$.$\square$ To finish the proof of the theorem, let $\gamma_n$ denote the dimension of $V_{\lambda_n}$ and let $\{\phi_{nj}\}_{n=1,j=1}^{\infty, \gamma_n}$ denote a sequence of orthonormal eigenvectors for all of the (countably many by Lemma 1) *nonzero* eigenvalues $\lambda_n$. Set $M=span\{\phi_{nj}\}_{n,j}$ (meaning the set of finite linear combinations). Observe that $T(M)\subset M$ by construction, and since $T$ is continuous this implies $T(\overline{M})\subset \overline{M}.$ On the other hand, if $y\in M^\perp$ and $x\in M$ we have $$(Ty,x) = (y,Tx)=0$$ since $Tx\in M$. Thus, both $\overline{M}$ and $M^\perp$ are closed invariant subspaces of $T$. Observe that if $P=proj(M^\perp)$, the restricted operator $PTP$ is also compact. Since it has no nonzero eigenvectors (by construction), Lemma 2 implies that it must satisfy $\|PTP\|=0$, which means $Tx=0$ whenever $x\in M^\perp$. Thus, every vector in $M^\perp$ is an eigenvector of $T$ with eigenvalue $0$. To complete the proof, take $\{\psi_k\}$ to be any orthonormal basis of $M^\perp$. Since $H=\overline{M}\oplus M^\perp$ the union $\{\phi_{nj}\}_{n,j}\cup \{\psi_k\}$ is an ONB of $H$ consisting of eigenvectors of $T$, as desired.$\square$ **Remark.** One can also prove Lemma 2 by showing that $\|y/c-x_n\|^2\rightarrow 0$, which can be seen by expanding the left hand side in inner products. This proof has the advantage of not using a square root.