Suppose $H$ is a Hilbert space and $T: H \to H$ is a self-adjoint compact operator. Define the continuous spectrum of $T$ to be $c(T) := \{\lambda \in \mathbb{C} \mid T-\lambda I$ is injective, non-surjective, and $\mathrm{im}(T)$ is dense in $H\}$.

Is $c(T)$ necessarily empty? If not, what is an example of when $c(T)$ is non-empty? What conditions can one impose to ensure that $c(T)$ is empty?

This question came up when I am trying to write down explicitly the spectrum of integral operators on $L^2[0,1]$, defined by $f(x) \mapsto \int K(x,y)f(y) dy\,$, where $K(x,y)$ is bounded and continous on $[0,1]^2$. Clearly, the point spectrum of this operator is taken care of by the spectral theorem. Residual spectrum is empty in this case.


1 Answer 1


Every non-zero spectral value of a self-adjoint compact operator (see for instance Rudin, Functional Analysis, p109) is an eigenvalue, therefore, you have : $$c(T) \subset \{0\}$$

It is possible that $c(T)=\emptyset$ ($T=0$ for instance), but it can also be not empty. Indeed, suppose $\dim(H) = \infty$ and $H$ separable. Pick your favourite orthonormal basis $(e_n)$ and define $T$ by $$T(e_n)=\dfrac{1}{n+1}e_n$$ Then $T$ is compact (as a limit of finite rank operators) and injective, its range contains the orthonormal basis (hence is dence), but not surjective, thus $c(T)=\{0\}$.

If you impose $\ker(T) \neq \{0\}$ and $\dim(H)=\infty$, then $0$ must belong to the spectrum of $T$ but, since $T$ is not injective, $c(T)=\emptyset$.

  • $\begingroup$ Thank you for the Rudin reference and example! $\endgroup$
    – beeflavor
    Mar 12, 2015 at 21:01
  • $\begingroup$ I should add that it's theorem 4.25 of Functional Analysis by Rudin for those who are interested in the fact $c(T)\subseteq \{0\}$. $\endgroup$
    – beeflavor
    Mar 13, 2015 at 2:41

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