Let $T$ be a compact, self-adjoint operator on a separable Hilbert space H. Suppose that $f\in H$, $||f|| =1$ and $||(T-3)f||\leq 1/2$. Let P be the orthogonal projection onto the direct sum of all the eigenspace of T with eigenvalue $\lambda \in [2,4]$. Show that
$||Pf||\geq \frac{\sqrt{3}}{2}$.
I think what I need to do is use spectral theory to find an orthonormal basis of H and then find the projection using that basis.
Write $f=Pf+(I-P)f$; the summands are mutually orthogonal. From the assumption by the spectral theorem you get $$1/2\ge\|(T-3I)f\|\ge \|(T-3I)(I-P)f\|\ge\|(I-P)f\|$$ (the last inequality is true because $(I-P)f$ belongs to the spectral subspace of the set $\{x: \; |x-3|>1\}$, which is the complement of the segment $[2, 4]$). Now the claim follows from the identity $\|f\|^2=\|Pf\|^2+\|(I-P)f\|^2$.