Choose normalized eigenvectors $u_n$ of the finite compressions and embed them as $v_n=P_n^*u_n\in H$. Then
$$
\langle Av_n,v_n\rangle=\lambda_n.
$$
The bounded sequence $(v_n)$ has weakly convergent subsequences. If $v_{n_j}\rightharpoonup v$, then for every $y\in H$, strong convergence $P_n^*P_ny\to y$ and the compressed eigenvalue equation give
$$
\langle(A-\lambda I)v,y\rangle
=\lim_j\langle(A-\lambda_{n_j}I)v_{n_j},P_{n_j}^*P_{n_j}y\rangle=0.
$$
Because $\lambda\notin\sigma(A)$, this forces $v=0$. Every weak cluster point is zero, so $v_n\rightharpoonup0$. Since $\langle Av_n,v_n\rangle=\lambda_n\to\lambda$, the weak-null characterization gives
$$
\boxed{\lambda\in W_e(A)}.
$$
Thus finite-section <spectral pollution> of a bounded operator can occur only in its <essential numerical range>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 358 / 2 / a / Solution

{wiki}

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.


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