 Solution

Expand $v={\sum }_j{c}_j{x}_j$, where $c_j=x_j^{†}v$. By orthonormality,Since the spectrum avoids $[a,b]$, each ${\lambda }_j$ is either below $a$ or above $b$, making every product strictly positive. This proves positivity of a quadratic polynomial across a spectral gap:In particular $\beta >0$ for every nonzero $v$. For $v=0$, $\beta =0$; the printed strict claim for an arbitrary vector must exclude this case.