Prove that
$$
|\langle\psi|U^\dagger MU|\psi \rangle-\langle\psi|V^\dagger MV|\psi \rangle|\leq |||\Delta\rangle||+|||\Delta\rangle||
$$
where $|\psi\rangle,|\Delta\rangle=(U-V)|\psi\rangle$ are complex column vectors, $\langle\psi|=(|\psi\rangle)^\dagger,\langle\Delta|=(|\Delta\rangle)^\dagger$ and $M$ is positive definite and $U,V$ are unitary matrices and $|||\psi\rangle||=1$
My reference says it is proved using basic linear algebra and Cauchy-Schwarz inequality.
My Attempt
Cauchy-Schwarz inequality : $|\langle\psi|\phi\rangle|\leq |||\psi\rangle||.|||\phi\rangle||$
Matrix Norm, $||A||=\max_{|\psi\rangle\neq 0}\dfrac{||A|\psi\rangle||}{|||\psi\rangle||}=\sigma_{\max}$ where $\sigma_{\max}$ is the largest singular value of $A$.
$||U|\psi\rangle||=|||\psi\rangle||$ for any unitary matrix $U$
$M$ is positive definite, i.e., $M^\dagger=M$ and $\lambda_i\geq 0$
$$
|\langle\psi|U^\dagger MU|\psi \rangle-\langle\psi|V^\dagger MV|\psi \rangle|=|\langle\psi|U^\dagger MU|\psi \rangle-|\langle\psi|U^\dagger MV|\psi \rangle+|\langle\psi|U^\dagger MV|\psi \rangle-\langle\psi|V^\dagger MV|\psi \rangle|\\
=|\langle\psi|U^\dagger M(U-V)|\psi \rangle+\langle\psi|(U-V)^\dagger MV|\psi \rangle|=|\langle\psi|U^\dagger M|\Delta \rangle+\langle\Delta| MV|\psi \rangle|
\\\leq |\langle\psi|U^\dagger M|\Delta \rangle|+|\langle\Delta| MV|\psi \rangle|
$$
Now,
$$
|\langle\psi|U^\dagger M|\Delta \rangle|\leq ||\langle\psi|U^\dagger||.|| M|\Delta \rangle||=||\langle\psi|||.|| M|\Delta \rangle||=|||\psi\rangle||.|| M|\Delta \rangle||=|| M|\Delta \rangle||
$$
$$
|\langle\Delta| MV|\psi \rangle|\leq ||\langle\Delta| M||.||V|\psi \rangle||=||\langle\Delta| M||.|||\psi \rangle||=||\langle\Delta| M||=|| M|\Delta \rangle||
$$
$$
\implies |\langle\psi|U^\dagger MU|\psi \rangle-\langle\psi|V^\dagger MV|\psi \rangle|\leq ||M|\Delta\rangle||+||M|\Delta\rangle||
$$
How do I proceed further to prove the required statement ?
Original Reference
Best Answer
Because $M$ is an element of a POVM, both $M$ and $I-M$ are positive semidefinite, which means that the eigenvalues of $M$ are between $0$ and $1$, which means that the eigenvalues of $M^2$ are between $0$ and $1$, which means that $M^2$ and $I - M^2$ are positive semidefinite, which means that $$ \begin{align} \| M |\Delta \rangle \|^2 &= \langle \Delta|M^2 |\Delta \rangle \\ & \leq \langle \Delta |M^2 |\Delta \rangle + \langle \Delta |I - M^2 |\Delta \rangle \\ & = \langle \Delta |M^2 + (I - M^2)|\Delta\rangle = \|\Delta\|^2. \end{align} $$ With that, you can proceed from your work to the desired conclusion.