I want to find a self-adjoint bounded operator on a Hilbert space with empty point spectrum i.e. $$ T = T^* ~\text{but}~ \sigma_p(T)= \emptyset $$

Some definitions and results of the lecture: (On a Hilbert space $X$ and let $T \in \mathscr{L}(X)$ i.e. a bounded linear operator on $X$)

  • $T=T^* \Leftrightarrow \sigma(T) \subset \mathbb{R} $
  • $TT^* = T^* T \Rightarrow \sigma_r(T)=\emptyset$ i.e. the residual spectrum is empty
  • $\sigma(T)=\sigma_r(T) \cup \sigma_p \cup \sigma_c(T)$ disjoint unions
  • If the space is finite then $\sigma_p(T)=\sigma(T)$
  • $\sigma(T)$ is non-empty

So, I have a self-adjoint operator i.e the residual spectrum is empty and I also want the point spectrum to be empty i.e. I want to achieve $\sigma(T)=\sigma_c(T)$ i.e $$ \{\lambda \in \mathbb{C} ~|~ (\lambda I - T) ~\text{not invertible} \}=$$ $$\{\lambda \in \mathbb{C} ~|~ \ker(\lambda I - T)=\emptyset ~\text{ and }~ \text{ran}(\lambda I - T) \neq \overline{\text{ran}(\lambda I - T)}=X \}$$ Additionally the space has to be infinite since otherwise $\sigma(T)=\sigma_p(T)$.

Does someone have such an example for me? And please explain why this example works in this way. This spectral theory is new for me.


Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e.

$$(Tf)(x) = x\cdot f(x).$$

Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint.

Clearly $T$ has no eigenvalues, since

$$(T - \lambda I)f = 0 \iff (x-\lambda)\cdot f(x) = 0 \quad \text{a.e.},$$

and since $x-\lambda$ is nonzero for all but at most one $x\in [0,1]$, it follows that $f(x) = 0$ for almost all $x$.

We have $\sigma(T) = [0,1]$, as is easily seen - for $\lambda \in \mathbb{C}\setminus [0,1]$, the function $x \mapsto \frac{1}{x-\lambda}$ is bounded on $[0,1]$.

The reason that $T$ has no eigenvalues is that for all $\lambda\in \mathbb{C}$ the set $\operatorname{id}_{[0,1]}^{-1}(\lambda)$ has measure $0$. If $m \colon [0,1] \to \mathbb{R}$ is a bounded measurable function such that $\mu(m^{-1}(\lambda)) > 0$ for some $\lambda$, then $\lambda$ is an eigenvalue of $T_m \colon f \mapsto m\cdot f$, any function $f\in H$ that vanishes outside $m^{-1}(\lambda)$ is an eigenfunction of $T_m$ then. Conversely, if $T_m$ has an eigenvalue $\lambda$ and $f \neq 0$ is an eigenfunction to that eigenvalue, then, since $(m(x) - \lambda)f(x) = 0$ almost everywhere, it follows that $f$ vanishes almost everywhere outside $m^{-1}(\lambda)$, and since $f\neq 0$, it further follows that $\mu(m^{-1}(\lambda)) > 0$.

  • 6
    $\begingroup$ You should have. ;) Every self-adjoint operator is unitarily equivalent to a multiplication operator on some $L^2$-space, so multiplication operators are a reasonable place to look for examples. $\endgroup$ – MaoWao Jan 31 '16 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.