Discrete spectrum (mathematics)

In mathematics, specifically in spectral theory, a discrete spectrum of a closed linear operator is defined as the set of isolated points of its spectrum such that the rank of the corresponding Riesz projector is finite.

Definition
A point $$\lambda\in\C$$ in the spectrum $$\sigma(A)$$ of a closed linear operator $$A:\,\mathfrak{B}\to\mathfrak{B}$$ in the Banach space $$\mathfrak{B}$$ with domain $$\mathfrak{D}(A)\subset\mathfrak{B}$$ is said to belong to discrete spectrum $$\sigma_{\mathrm{disc}}(A)$$ of $$A$$ if the following two conditions are satisfied:
 * 1) $$\lambda$$ is an isolated point in $$\sigma(A)$$;
 * 2) The rank of the corresponding Riesz projector $$P_\lambda=\frac{-1}{2\pi\mathrm{i}}\oint_\Gamma(A-z I_{\mathfrak{B}})^{-1}\,dz$$ is finite.

Here $$I_{\mathfrak{B}}$$ is the identity operator in the Banach space $$\mathfrak{B}$$ and $$\Gamma\subset\C$$ is a smooth simple closed counterclockwise-oriented curve bounding an open region $$\Omega\subset\C$$ such that $$\lambda$$ is the only point of the spectrum of $$A$$ in the closure of $$\Omega$$; that is, $$\sigma(A)\cap\overline{\Omega}=\{\lambda\}.$$

Relation to normal eigenvalues
The discrete spectrum $$\sigma_{\mathrm{disc}}(A)$$ coincides with the set of normal eigenvalues of $$A$$:
 * $$\sigma_{\mathrm{disc}}(A)=\{\mbox{normal eigenvalues of }A\}.$$

Relation to isolated eigenvalues of finite algebraic multiplicity
In general, the rank of the Riesz projector can be larger than the dimension of the root lineal $$\mathfrak{L}_\lambda$$ of the corresponding eigenvalue, and in particular it is possible to have $$\mathrm{dim}\,\mathfrak{L}_\lambda<\infty$$, $$\mathrm{rank}\,P_\lambda=\infty$$. So, there is the following inclusion:
 * $$\sigma_{\mathrm{disc}}(A)\subset\{\mbox{isolated points of the spectrum of }A\mbox{ with finite algebraic multiplicity}\}.$$

In particular, for a quasinilpotent operator
 * $$Q:\,l^2(\N)\to l^2(\N),\qquad Q:\,(a_1,a_2,a_3,\dots)\mapsto (0,a_1/2,a_2/2^2,a_3/2^3,\dots),$$

one has $$\mathfrak{L}_\lambda(Q)=\{0\}$$, $$\mathrm{rank}\,P_\lambda=\infty$$, $$\sigma(Q)=\{0\}$$, $$\sigma_{\mathrm{disc}}(Q)=\emptyset$$.

Relation to the point spectrum
The discrete spectrum $$\sigma_{\mathrm{disc}}(A)$$ of an operator $$A$$ is not to be confused with the point spectrum $$\sigma_{\mathrm{p}}(A)$$, which is defined as the set of eigenvalues of $$A$$. While each point of the discrete spectrum belongs to the point spectrum,
 * $$\sigma_{\mathrm{disc}}(A)\subset\sigma_{\mathrm{p}}(A),$$

the converse is not necessarily true: the point spectrum does not necessarily consist of isolated points of the spectrum, as one can see from the example of the left shift operator, $$ L:\,l^2(\N)\to l^2(\N), \quad L:\,(a_1,a_2,a_3,\dots)\mapsto (a_2,a_3,a_4,\dots). $$ For this operator, the point spectrum is the unit disc of the complex plane, the spectrum is the closure of the unit disc, while the discrete spectrum is empty:
 * $$\sigma_{\mathrm{p}}(L)=\mathbb{D}_1,

\qquad \sigma(L)=\overline{\mathbb{D}_1}; \qquad \sigma_{\mathrm{disc}}(L)=\emptyset. $$