Does every bounded operator on a complex Hilbert space have an eigenvalue?

Question

Is the following statment true?

Let $\mathscr{H}$ be a complex Hilbert space and let $\varphi: \mathscr{H} \to \mathscr{H}$ be a bounded operator. Does $\varphi$ have an eigenvalue in general? If yes, how to prove this? If not, what is a counterexample and what property does one need in order to ensure the existence of an eigenvalue.

In particular, I am interested in the case where $\mathscr{H}$ has a unitary representation $\pi: G \to U(\mathscr{H})$ such that $\varphi \circ \pi(g) = \pi(g) \circ \varphi$ for all $g \in G$.

Answer 1

Let $\mathscr{H}=L^2[0,1]$, and let $L : L^2[0,1]\rightarrow L^2[0,1]$ be defined by $$ (Lf)(x) = xf(x),\;\; f \in L^2[0,1]. $$ The operator $L$ has no eigenvalues. The spectrum of $L$ is $\sigma(L)=[0,1]$, but no point in the spectrum is an eigenvalue. Each point in the spectrum is an approximate eigenvalue, meaning that there is a sequence of unit vectors $\{ f_n \}$ such that $\|(L-\lambda I)f_n\|\rightarrow 0$. You can view $\mathscr{H}$ as $L^2$ on the unit circle, which fits into your context as well.

Answer 2

No, not all of them. In $\ell^2$ you can have $S(x_1,x_2,...)=(0,x_1,x_2,...)$

If $S(x)=\lambda x$. Then $x_1=x_2=...=0$.

Answer 3

To look at things in a little more detail:

The answer is in general "no", boundedness of $\varphi$ is not sufficient for the existence of eigenvalues, as the example of the shift operator introduced by saltandpepper in his answer shows. Indeed, taking $\mathscr H = \ell^2$, and

$\vec x = (x_1, x_2, x_3, \ldots ) \in \ell^2, \tag 0$

we have

$S\vec x = S(x_1, x_2, x_3, \ldots) = (0, x_1, x_2, x_3, \ldots ); \tag 1$

if

$S \vec x = \lambda \vec x, \; \lambda \in \Bbb C, \tag 2$

we see that

$S(x_1, x_2, x_3, \ldots) = (0, x_1, x_2, x_3, \ldots ) = \lambda (x_1, x_2, x_3, \ldots) = (\lambda x_1, \lambda x_2, \lambda x_3, \ldots), \tag 3$

which implies

$\lambda x_1 = 0, \tag 4$

$\lambda x_n = x_{n - 1}, \; n \ge 2; \tag 5$

if $\lambda \ne 0$, (4) and (5) force

$x_1 = 0, \tag 6$

and then

$x_2 = 0, \; x_3 = 0, \ldots, x_n = 0, \ldots, \; \forall n \in \Bbb N; \tag 7$

thus we must have

$\vec x = 0; \tag 8$

but eigenvectors are non-zero by definition, thus we rule out (8) and we see that $S$ has no eigenvalue $\lambda \ne 0$. If $\lambda = 0$, we immediately obtain $\vec x = 0$ from (3), so we may rule out this case as well. $S$ has no eigenvalues.

We can, however, by placing suitable conditions on $\varphi$, ensure that it does in fact have eigenvalues and eigenvectors. For example, if $\phi$ is bounded, self-adjoint, and compact, it has bona fide eigenvalues and eigenvectors; a proof of this fact may be found here, as well as in many books and places on the web.