Numerical range of an operator ($W(A)$)

The numerical range is $W(A)=\{⟨Ax,x⟩:x\in D(A),\Vert x\Vert =1\}$.

For a bounded self-adjoint operator, $z$ outside the closed numerical range yields $\Vert (T-z)u\Vert \ge \mathrm{dist}(z,\overline{W(T)})\Vert u\Vert$. The same inequality for $T-\overline{z}$ makes the range dense, and the first makes it closed. Thus $z$ is in the resolvent. The enclosure remains valid for general bounded Hilbert-space operators by the analogous adjoint numerical-range argument.

For bounded self-adjoint $T$, put $B=\beta I-T\ge 0$ with $\beta =\sup W(T)$. Unit vectors with $⟨B{u}_n,u_n⟩\to 0$ satisfy $\Vert B{u}_n{\Vert }^2\le \Vert B\Vert ⟨B{u}_n,u_n⟩$ by positive-form Cauchy-Schwarz inequality. Thus $\beta$ is an approximate eigenvalue. Applying the same argument to $T-\alpha I$ proves the result at $\alpha =\inf W(T)$.

The radius is a scalar derived from the numerical range of an operator, rather than the range set itself.

For a bounded linear operator on a complex Hilbert space, the numerical radius is the supremum of the absolute values in its numerical range of an operator. It is a norm equivalent to the operator norm, with $w(T)\le \Vert T\Vert \le 2w(T)$. For a self-adjoint operator the two norms agree.

The essential numerical range consists of limits $\lim ⟨A{v}_n,v_n⟩$ along unit vectors converging weakly to zero. It is stable under compact perturbations and may be empty for an unbounded operator.

Spectral pollution is convergence of finite-section eigenvalues to points outside the spectrum of the limiting operator. For bounded operators, such pollution lies in the essential numerical range.

For nonempty closed subsets $F_i,F$ of a metric space, Attouch--Wets topology convergence means uniform convergence of their distance functions on every bounded set:It captures approximation of closed sets across each bounded observation region, even when the sets themselves are unbounded.

If $F_n$ decrease to a nonempty intersection $F$ in a proper metric space, then $F_n\to F$ in this sense. For a fixed $x$, almost-minimizing points in $F_n$ stay in one closed bounded ball, using any fixed point of $F$ as a comparison. Compactness of that ball and nestedness show $d(x,F_n)\uparrow d(x,F)$. These distance functions are continuous, so Dini's theorem makes convergence uniform on each compact ball.

This fails in an infinite-dimensional Hilbert space. In ${\ell }^2$, put $F_n=\overline{\mathrm{span}}\{e_k:k\ge n\}$. Their intersection is $\{0\}$, but $d(e_n,F_n)=0$ and $d(e_n,\{0\})=1$. The error on the unit ball stays at least1. Pointwise distance convergence of nested closed sets does not supply uniform convergence on bounded sets without an appropriate compactness hypothesis.