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.


2 Answers 2


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 $\mu(m^{-1}(\lambda)) > 0$.

  • 7
    $\begingroup$ 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
    Commented Jan 31, 2016 at 16:21
  • $\begingroup$ Can we find uncountably many self-adjoint operators on H with spectrum [0, 1] and empty point-spectrum that are pairwise unitarily inequivalent? $\endgroup$
    – NotaChoice
    Commented Feb 23, 2022 at 23:16

Denote by $I$ the identity operator on a complex Hilbert space $H$. Take any nonzero real number $r$, and $T=rI$ is a selfadjoint operator, and 0 is not in the point spectrum.

Take any selfadjoint operator $T$ on $H$, and any real $r>\|T\|$. Then $r-T$ is an invertible selfadjoint operator. In particular, 0 is not in the point spectrum.

Take any bounded linear operator $T$ on $H$, and any real $r>\|T^*T\|=\|T\|^2$. Then $r-T^*T$ is an invertible selfadjoint operator and 0 is not in the point spectrum.

  • 1
    $\begingroup$ All three statements are wrong. For example, the operator $T$ in the first example has point spectrum $\{r\}$. The operators in the second and third example may or may not have empty point spectrum, depending on $T$. You seem to be mixing up the point spectrum with the kernel of an operator. $\endgroup$
    – MaoWao
    Commented Apr 12, 2022 at 10:41
  • 1
    $\begingroup$ You are right. I miss read the question, and in my mind I had 0 is not in the point spectrum. Thanks for the correction. $\endgroup$
    – Abdel
    Commented Apr 13, 2022 at 7:06

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