Spectral Theory of Compact Operators

Kreyszig Functional Analysis

Let $X$ and $Y$ be normed spaces. A linear operator $T:X\to Y$ is called compact for every bounded subset $M$ of $X$, $T(M)$ is relatively compact in $Y$; that is, the closure $\overline{T(M)}$ in $Y$ is compact under the strong (norm) topology.

An equivalent formulation of operator compactness is that a linear operator $T:X\to Y$ between normed spaces is compact if and only if for every bounded sequence $(x_n)$ in $X$ the corresponding sequence $(T{x}_n)$ in $Y$ has a convergent subsequence.

We state the following elementary facts without proof:

• If $T:X\to Y$ is a bounded linear operator between normed spaces and $dimT(X)<\infty$, the operator $T$ is compact.
• A subset $S\subset X$ is relatively compact in $X$ if and only if every sequence $(s_n)$ in $S$ has a subsequence convergent in the ambient space $X$.
• Relative compactness implies total boundedness, and total boundedness implies separability.
• If $B$ is a totally bounded subset of a complete metric space $X$, then $B$ is relatively compact.
• If $T:X\to X$ is a compact linear operator and $S:X\to X$ is a bounded linear operator on a normed space $X$, $TS$ and $ST$ are both compact.

Theorem 1: Let $X$ be a normed space and $T:X\to X$ be a compact linear operator. If $dimX=\infty$, $0\in \sigma (T)$.

Proof) Assume for the sake of contradiction that $0\in \rho (T)$. That is, $T$ has an inverse $R_0(T)$ that is a bounded linear operator defined on a dense subspace of $X$. For any $x,y\in X$, $\Vert x-y\Vert =\Vert R_0(T)(T(x)-T(y))\Vert \le \Vert R_0(T)\Vert \Vert T(x)-T(y)\Vert \Rightarrow \Vert T(x)-T(y)\Vert \ge \Vert x-y\Vert /\Vert R_0(T)\Vert$. If $dimX=\infty$, by Riesz’s Lemma there exists an infinite sequence of elements $(x_n)$ such that $\Vert x_n\Vert =1$ and $\Vert x_n-x_m\Vert >1/2$ for any $n\ne m$. Applying $T$ forces $\Vert T{x}_n-T{x}_m\Vert >1/(2\Vert R_0(T)\Vert )$. Thus, $(T{x}_n)$ cannot have a convergent subsequence, contradicting compactness.

Theorem 2: The set of eigenvalues of a compact linear operator $T:X\to X$ on a normed space $X$ is at most countable, and the only possible accumulation point is $\lambda =0$.

Proof) It suffices to show for any real $k>0$ the set of all $\lambda \in {\sigma }_p(T)$ with $|\lambda |\ge k$ is finite. FTSOC, assume such a $k>0$ exists such that there is an infinite sequence $({\lambda }_n)$ of distinct eigenvalues such that $|{\lambda }_n|\ge k$. If we associate each ${\lambda }_n$ with an eigenvector $x_n$ such that $T_{{\lambda }_n}{x}_n=0$, clearly the vectors $x_n$ are all linearly independent. Thus, taking $M_n=Span\{x_1,\dots ,x_n\}$ supplies us with an infinite dimensional space spanned by eigenvectors scaled $\ge k$. We claim this contradicts compactness, as, intuitively, bounded subsets must map to clustered, relatively compact subsets and cannot be uniformly spread out.

For any $n$, by Riesz’s Lemma we can take $v_n$ such that $\Vert v_n\Vert =1$, $v_n\in M_n$, and $d(v_n,x)>1/2$ for any $x\in M_{n-1}$. As $T_{{\lambda }_n}{x}_n=0$, $T_{{\lambda }_n}{v}_n\in M_{n-1}$. For any $m<n$, $Tm=T_{{\lambda }_n}m+\lambda m\in M_m\subset M_{n-1}$, and therefore,

$$T{v}_n-T{v}_m={\lambda }_n{v}_n+T_{{\lambda }_n}{v}_n+Tm={\lambda }_n{v}_n+\text{(element of }{M}_{n-1})\Rightarrow \Vert T{v}_n-T{v}_m\Vert >{\lambda }_n/2\ge k/2$$

Therefore, $(v_n)$ is bounded but $(T{v}_n)$ doesn’t have a convergent subsequence, contradicting compactness.

Theorem 3: Let $T:X\to X$ be a compact linear operator on a normed space $X$. For any complex number $\lambda \ne 0$, the null space $\mathcal{N}(T_{\lambda })$ of $T_{\lambda }=T-\lambda I$ is finite-dimensional (and thus complete and closed).

Proof) It suffices to show the closed unit ball of $\mathcal{N}(T_{\lambda })$, $M=\{x\in \mathcal{N}(T_{\lambda })|\Vert x\Vert \le 1\}$, is compact. Take any sequence $(x_n)$ in $M$. By the compactness of $T$, $(T{x}_n)=(\lambda x_n)$ must have a convergent subsequence, and thus as $\lambda \ne 0$, $(x_n)$ must also have a convergent subsequence. We are done.

Theorem 4: Let $T:X\to X$ be a compact linear operator on a normed space $X$. For any complex number $\lambda \ne 0$, the range $T_{\lambda }(X)$ is closed.

Proof) Assume some element $y\in \overline{T_{\lambda }(X)}-T_{\lambda }(X)$ exists and take an arbitrary sequence $(x_n)\in X$ such that $T_{\lambda }(x_n)\to y$. Note that $(x_n)$ cannot have a convergent subsequence as, if such a subsequence $(x_{n_k})\to w$ exists, by boundedness $T_{\lambda }w=y$, creating an immediate contradiction. As $T=T_{\lambda }+\lambda I⟺\lambda I=T-T_{\lambda }$, this implies $T{x}_n$ also cannot have a convergent subsequence. As $T$ is compact, this is only possible if $x_n$ has no bounded subsequence; that is, $\Vert x_n\Vert \to \infty$.

Note that this is true for ANY sequence $(x_n)\in X$ such that $T_{\lambda }(x_n)\to y$. This is important as it means, for any arbitrary sequence of vectors $z_n\in \mathcal{N}(T_{\lambda })$, $\Vert x_n+z_n\Vert \to \infty$ as $T_{\lambda }(x_n+z_n)=T_{\lambda }{x}_n\to y$. Intuitively, this seems impossible as it means the component of the vectors $x_n$ orthogonal to the null space of $T_{\lambda }$ (note “orthogonality” formally doesn’t exist here as $X$ is not an inner product space) go to infinity but the values $T_{\lambda }{x}_n$ are very neatly bounded and, in fact, convergent. We simply have to create a formal argument to express this incongruity.

We do so by first isolating the components of $x_n$ “orthogonal to $\mathcal{N}(T_{\lambda })$”. For any $x_n$, we can define

$${\delta }_n=\underset{v\in \mathcal{N}(T_{\lambda })}{\inf }\Vert x_n-v\Vert$$

and, for every $x_n$, take some $z_n\in \mathcal{N}(T_{\lambda })$ such that $\Vert x_n-z_n\Vert <2{\delta }_n$. The vector $x_n-z_n$ is about the best approximation of “row space component” we can get in non-inner product space. Clearly, $\Vert x_n-z_n\Vert \to \infty$ and $T_{\lambda }(x_n-z_n)\to y$. The key insight here is taking the unit vectors $w_n=(x_n-z_n)/\Vert x_n-z_n\Vert$. Clearly, $T_{\lambda }{w}_n\to 0$ and, as $w_n$ is bounded, there is a subsequence $w_{n_k}$ such that $T{w}_{n_k}\to l$ for some $l\in X$. As $\lambda I=T-T_{\lambda }$, this forces $w_{n_k}$ to be convergent in $x$.

We are close to our desired contradiction. Let $w_{n_k}\to w$. As $T_{\lambda }{w}_n\to 0$, this forces $T_{\lambda }w=0$ and thus $w\in \mathcal{N}(T_{\lambda })$. However, we obtained $w_n$ by isolating the “row space components” of $x_n$ and then normalizing them, so they clearly cannot converge to some point in the null space. This is shown formally as follows: for any $z\in \mathcal{N}(T_{\lambda })$,

$$\Vert w_n-z\Vert =\frac{1}{\Vert x_n-z_n\Vert }\Vert (x_n-z_n)-\Vert x_n-z_n\Vert z\Vert \ge \frac{{\delta }_n}{\Vert x_n-z_n\Vert }\ge \frac{1}{2}$$

Therefore, no subsequence of $(w_n)$ can possibly converge to an element of $\mathcal{N}(T_{\lambda })$. We are done.

Corollary of Theorems 3 & 4: For any compact operator $T:X\to X$ and nonnegative integer $n$, $T_{\lambda }^n=(T-\lambda I{)}^n=ST+(-\lambda {)}^{n}I$ where $S$ is some bounded operator. $ST$ is also compact, and thus by Theorems 3 and 4, $T_{\lambda }^n$ has a finite dimensional null space and closed range. The following are also trivially true:

$$\{0\}=\mathcal{N}(T_{\lambda }^0)\subset \mathcal{N}(T_{\lambda }^1)\subset \mathcal{N}(T_{\lambda }^2)\subset \dots ,X=T_{\lambda }^0(X)\supset T_{\lambda }^1(X)\supset \dots$$

Theorem 5: Let $T:X\to X$ be a compact linear operator on a normed space $X$, and let $\lambda \ne 0$ be an arbitrary complex number. There exists a smallest integer $r$ (dependent on $\lambda$) such that from $n=r$ onwards the null spaces $\mathcal{N}(T_{\lambda }^n)$ are all equal. Furthermore, for all $n<r$, the inclusions $\mathcal{N}(T_{\lambda }^n)\subset \mathcal{N}(T_{\lambda }^{n+1})$ are all proper.

Proof) It suffices to show an $r$ exists such that $\mathcal{N}(T_{\lambda }^r)=\mathcal{N}(T_{\lambda }^{r+1})$. Assume there is some $n>r$ such that $\mathcal{N}(T_{\lambda }^{n+1})-\mathcal{N}(T_{\lambda }^n)\ne \varnothing$. Then, there is some $x\in X$ such that $T_{\lambda }^{n+1}x=0$ and $T_{\lambda }^{n}x\ne 0$. $y=T_{\lambda }^{n-r}x$ must then satisfy $T_{\lambda }^{r+1}y=0$ and $T_{\lambda }^{r}y\ne 0$. This implies $\mathcal{N}(T_{\lambda }^r)\ne \mathcal{N}(T_{\lambda }^{r+1})$, a contradiction.

Assume, FTSOC, that $\mathcal{N}(T_{\lambda }^0)\subset \mathcal{N}(T_{\lambda }^1)\subset \dots$ is an infinite sequence of proper inclusions. By Riesz’s Lemma, we can take a sequence $(x_n)$ in $X$ such that $\Vert x_n\Vert =1$, $x_n\in \mathcal{N}(T_{\lambda }^n)$, and $d(x_n,v)>1/2$ for any $v\in \mathcal{N}(T_{\lambda }^{n-1})$. For any $m<n$:

$$\Vert T{x}_n-T{x}_m\Vert =\Vert \lambda x_n+(T_{\lambda }{x}_n-T_{\lambda }{x}_m-\lambda x_m)\Vert >|\lambda |/2$$

as $T_{\lambda }{x}_n-T_{\lambda }{x}_m-\lambda x_m$ is clearly an element of $\mathcal{N}(T_{\lambda }^{n-1})$. Thus, $(T{x}_n)$ cannot have a convergent subsequence, contradicting compactness.

Theorem 6: Let $T:X\to X$ be a compact linear operator on a normed space $X$, and let $\lambda \ne 0$ be an arbitrary complex number. There exists a smallest integer $q$ (dependent on $\lambda$) such that from $n=q$ onwards the ranges $T_{\lambda }^n(X)$ are all equal. Furthermore, for all $n<q$, the inclusions $T_{\lambda }^n(X)\supset T_{\lambda }^{n+1}(X)$ are all proper.

Proof) It suffices to show a $q$ exists such that $T_{\lambda }^q(X)=T_{\lambda }^{q+1}(X)$. Assume $T_{\lambda }^0(X)\supset T_{\lambda }^1(X)\supset \dots$ is an infinite sequence of proper inclusions. We can similarly take a sequence $(x_n)$ in $X$ such that $x_n\in T_{\lambda }^n(X)$, $\Vert x_n\Vert =1$, and $\Vert x_n-v\Vert >1/2$ for all $v\in T_{\lambda }^{n+1}(X)$. For any $m>n$:

$$\Vert T{x}_n-T{x}_m\Vert =\Vert \lambda x_n+(T_{\lambda }{x}_n-T{x}_m)\Vert >|\lambda |/2$$

as $T_{\lambda }{x}_n-T{x}_m$ is clearly in $T_{\lambda }^{n+1}(X)$. $(x_n)$ is therefore a bounded sequence but $(T{x}_n)$ has no convergent subsequence, contradicting compactness.

Theorem 7: Let $T:X\to X$ be a compact linear operator on a normed space $X$, and let $\lambda \ne 0$ be an arbitrary complex number. Let $T_{\lambda }^r$ be the power of $T_{\lambda }$ where the null spaces first stabilize (Thm 5) and $T_{\lambda }^q$ be the first power of $T_{\lambda }$ where the ranges first stabilize (Thm 6). Then, $r=q$.

Proof) Null space and range stabilization fit into the same geometric picture of sequentially applying $T_{\lambda }$ to $X$, and so it intuitively makes sense that $r=q$.

We first show that $\mathcal{N}(T_{\lambda }^q)=\mathcal{N}(T_{\lambda }^{q+1})$. First, note that if $x\in T_{\lambda }^q(X)$ and $T_{\lambda }x=0$, then $x=0$. This is because if $x\ne 0$, then for any arbitrarily large $n$, there exists some $y\ne 0$ such that $T_{\lambda }^{n-1}y=x$ and thus $y\in \mathcal{N}(T_{\lambda }^n)-\mathcal{N}(T_{\lambda }^{n-1})$. The nullspaces thus cannot stabilize, an immediate contradiction. For any $x$ such that $T_{\lambda }^{q+1}x=0$, $T_{\lambda }(T_{\lambda }^{q}x)=0$ and thus $T_{\lambda }^{q}x=0$ by above. Therefore, $\mathcal{N}(T_{\lambda }^q)=\mathcal{N}(T_{\lambda }^{q+1})$. If $q=0$, this implies $r\le 0$, and as $r$ is, by definition, $r\ge 0$, $r=q=0$and we are done.

For the case where $q>0$, we show that $\mathcal{N}(T_{\lambda }^q)\ne \mathcal{N}(T_{\lambda }^{q-1})$. Take an element $y\in T_{\lambda }^{q-1}(X)-T_{\lambda }^q(X)$. Then, there exists some $x\in X$ such that $T_{\lambda }^{q-1}x=y$. As the range stabilizes at $q$, there must exist some $z$ such that $T_{\lambda }^{q+1}z=T_{\lambda }y=T_{\lambda }^{q}x$. Thus, $T_{\lambda }^q(T_{\lambda }z-x)=0$, but $T_{\lambda }^{q-1}(T_{\lambda }z-x)=T_{\lambda }^{q}z-T_{\lambda }^{q-1}x=T_{\lambda }^{q}z-y\ne 0$. As $T_{\lambda }z-x\in \mathcal{N}(T_{\lambda }^q)-\mathcal{N}(T_{\lambda }^{q-1})$, we are done.