Eigenvalues of compact operator don't have nonzero accumulation points

Question

In the book

Elements of the theory of functions and functional analysis

of Kolmogorov and Fomin, there is a proof of the following theorem,

Every compact operator $A$ on a Banach space $E$ has for arbitrary $\rho>0$ only a finite number of linearly independent eigenvectors which correspond to the eigenvalues whose absolute value are greater than $\rho$. The proof that it gives is later in the errata part of the book as wrong, but it doesn't say why and I don't know what is wrong with the following similar argument,

Assume that the theorem is false, that is, there exist a $\rho$ and a sequence $\{x_n\}$ of linearly independent vectors with $A x_n = \lambda_n x_n$ and $|\lambda_n|>\rho$. We can restrict $A$ to the span of this sequence, it is easy to check that it has a bounded inverse (if $x=a_1x_1+ \cdots +a_k x_k$ then $A^{-1}x=a_1 x_1/\lambda_1 + \cdots + a_kx_k/\lambda_k$ and $||A^{-1}|| \leq 1/\rho$) therefore on this space we have a compact operator with bounded inverse and consequently the identity map in this space is compact , arriving to a contradiction since the space is infinite dimensional.

Answer 1

The span of $\lbrace x_n \rbrace$ isn't obviously a Banach space (it's not clear why it's closed in $E$), and, as one consequence, it's not clear to me why the restriction of $A$ to this space should be "compact" (meaning, takes bounded sets to precompact sets).

To give a specific counterexample to show this is a real problem, let $H$ be a separable Hilbert space with an orthonormal Hilbert basis $\lbrace e_i \rbrace$, and let $B : H \to H$ be the operator defined by $B(e_k) = \tfrac{1}{k} e_k$. Then $B$ is compact by standard theorems.

Let $V \subset H$ be the finite span of the $e_i$ (that is, the space of finite linear combinations of the $e_i$). Then $B$ certainly fixes $V$. But the operator $B : V \to V$ does not take bounded sets to precompact sets; for instance, if $\mathcal{B}_1$ is the unit ball in $V$ then $B(\mathcal{B}_1)$ contains the Cauchy sequence $\lbrace u_n \rbrace$ defined by $$ u_n := \sum_{k=1}^N \frac{1}{k 2^k} e_k = B\Big( \sum_{k=1}^N \frac{1}{2^k} e_k \Big), $$ but this sequence obviously has no limit in $V$.

To try to fix this, one could in your proof replace the span of $\lbrace x_n \rbrace$ with its closure; but then one needs a better argument to show that $A$ is invertible with bounded inverse (which is, at least, not obvious to me at the moment).

Answer 2

We need

A lemma by Riesz: If $X$ is Banach and $M\subset X$ is a closed subspace then for every $\epsilon>0$ there is $x\in X$ such that $||x||=1$ and $d(x,M)>1-\epsilon$.

Proof: Let $1>\epsilon>0$. Pick $y\in X\setminus M$. Then $d(y,M)=a>0$. There exist $m\in M$ such that $$||y-m||\leq a\left(1+\frac{\epsilon}{1-\epsilon}\right)$$ Put $x:=\frac{y-m}{||y-m||}$. We can check that for $z\in M$ we have $$\begin{align}||x-z||&=\left|\left|\frac{y-m}{||y-m||}-z\right|\right|\\&=||y-m||^{-1}\cdot\left|\left|y-\color{red}{m-||y-m||\cdot z}\right|\right|\\&=||y-m||^{-1}\cdot||y-\color{red}{h}||\\&\geq||y-m||^{-1}\cdot a\ \ \ \ \text{ because }\color{red}{h}\in M\\&\geq 1-\epsilon\end{align}$$

---

Proving that eigenvalues of compact operators can only accumulate at $0$.

Assume that $\lambda\neq0$ is not an eigenvalue. Then $T:=A-\lambda$ is one-to-one.

Final goal: To show that $T^{-1}:X\to X$ exists and it is bounded. In particular this shows that $\lambda$ can't be a limit of eigenvalues.

Assume the image of $T$ is closed.

Then by the open mapping theorem $T^{-1}:\text{Im}(T)\to X$ exists and its continuous.

If $X_1:=\text{Im}(T)\subsetneq X$ then all spaces $X_{n+1}:=\text{Im}(T)X_{n}$ are different, with each containing the next one.

Now, pick $x_n\in X_n$ such that $||x_n||$, and $d(x_n, X_{n+1})>\frac{1}{2}$. Then, for $m>n$

$$\begin{align}\left|\left|Ax_n-Ax_m\right|\right|&=|\lambda|\cdot\left|\left|x_n+\color{red}{-x_m-\frac{Tx_n-Tx_m}{\lambda}}\right|\right|\\&=|\lambda|\cdot||x_n-\color{red}{x}||\ \ \ \ \text{ where }\color{red}{x}\in X_{n+1}\\&\geq\frac{|\lambda|}{2}\end{align}$$

But this means that the sequence $Ax_n$ can't have convergent subsequences contradicting that $A$ is compact.

Therefore $\text{Im}(T)=X$ and $T^{-1}:X\to X$.

We can finish now because

The image of $T$ is closed.

Assume that $y_n=Tx_n$ converges to $y$.

Case 1: If $x_n$ is bounded then, since $A$ is compact there is a subsequence $x_{n_k}$ such that $Ax_{n_k}$ converges. But $x_{n_k}=\lambda^{-1}(Ax_{n_k}-y_{n_k})$, so $x_{n_k}$ is itself convergent (to some $x$). Then $y=Tx\in\text{Im}(T)$.

Case 2: If $x_n$ is unbounded (we can assume $||x_n||\to\infty$). Then $z_n:=x_n/||x_n||$ has norm $1$ and $Tz_n=y_n/||x_n||\to0$ (because $y_n$ is convergent and therefore bounded). Because $z_n$ is bounded and $A$ is compact, there is a subsequence $z_{n_k}$ such that $Az_{n_k}$ converges. But then $z_{n_k}=\lambda^{-1}(Az_{n_k}-Tz_{n_k})$ is itself convergent (to some $z$). Taking limit in this last equation we see that $\lambda z=Az$. Therefore $z=0$, since we assumed that $\lambda$ is not an eigenvalue. But this can't be because $||z_n||=1$ doesn't allow $z_n\to0$.

Answer 3

@inquisitor: I looked at a translation of Kolmogorov-Fomin into Spanish, can't differ too much from the original. Here is how they prove it, with some smoothing.

You take $E$ normed space and $A$ a linear operator, $x_n\in E$ linearly independent so that $A x_n = \lambda_n \cdot x_n$, and $|\lambda_n| \ge \delta > 0$. We will show that $A$ is not a compact operator.

First, some standard facts:

1. If $F$ is a closed subspace of $E$ and $x \not \in F$ then there exist $\alpha \in \mathbb{K}$ and $y \in F$ so that $||\alpha x + y || = 1$ and $\text{d}( \alpha x + y, F) \ge \frac{1}{2}$
2. If $F$ is a subspace of $E$ and $A x = \lambda x$ then for all vectors $z$ in the subspace $\mathbb{K} \cdot x + F$ we have $A z = \lambda z + g$ for some $g \in F$.

Consider now the sequence of finite dimensional subspaces $F_n = (x_1, \ldots, x_n)$. Take an $n$ and apply $1.$ to $x_n$ and $F_{n-1}$. We get $y_n\colon = \alpha_n x_n + f_{n-1}$ so that $||y_n||=1$ and $\text{d}(y_n, F_{n-1}) \ge 1/2$.

Note that the span of $y_1, \ldots , y_n$ is still $F_n$, for all $n$ (by induction). Moreover, from $2.$ we have $A y_n = \lambda_n y_n + g_n $ for some $g_n \in F_{n-1}$.

We are done now. For $n>m$ we have

$(Ay_n - A y_m) = \lambda_n\cdot (y_n - h)$ where $h \in F_{n-1}$ and so

$||Ay_n - A y_m|| = |\lambda_n| \cdot ||y_n - h||\ge \frac{1}{2} |\lambda_n|\ge \frac{1}{2} \delta$

while $||y_n|| =1$ for all $n$.