Operator-norm error bounds a basis-outcome probability change

Given: The space, orthonormal basis, unit vector, two unitary maps, and operator-norm bound in the Statement.

The measurement law for a finite continuous-time walk: for a unit vector $v$, the probability of basis outcome $y$ is $\mathrm{\mid }⟨v,e_y⟩{\mathrm{\mid }}^2$.

Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis: for one unit vector $e_y$, $\mathrm{\mid }⟨v,e_y⟩{\mathrm{\mid }}^2\le \mathrm{\parallel }v{\mathrm{\parallel }}^2$.

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv|| defines $\mathrm{\parallel }T{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ as the maximum of $\mathrm{\parallel }Tv\mathrm{\parallel }$ over unit $v$ on a nonzero finite-dimensional domain; The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector proves that this maximum exists and is attained.

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces: a unitary is a linear isometry and preserves every vector norm.

Real and imaginary parts, complex conjugation, and modulus: if $z=a+ib$, then $\mathrm{\mid }z\mathrm{\mid }=\sqrt{a^2+b^2}$.

Cauchy-Schwarz $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2$ with its equality case, the triangle inequality for $\parallel \cdot {\parallel }_2$, the parallelogram law and polarisation: the Euclidean norm on ${\mathbb{R}}^2$ satisfies the triangle inequality. Under the coordinates $z\mapsto (\mathrm{Re}z,\mathrm{Im}z)$ this gives $\mathrm{\mid }{z}_1+z_2\mathrm{\mid }\le \mathrm{\mid }{z}_1\mathrm{\mid }+\mathrm{\mid }{z}_2\mathrm{\mid }$ for complex numbers.