Help proving bound on POVM measurement probabilities I am trying to follow Nielsen and Chuang's 1 proof that the difference in measurement probabilities is bounded by the difference between two unitary operators applied to a given state.
Can someone show me how to get from Equation 4.66 to 4.67 in the proof below (see page 195 in the 10th anniversary edition):

1 Nielsen and Chuang, "Quantum Computation and Quantum Information"
 A: I'll elaborate on the answer, since it took me some effort and I'm glad to share all the steps.
Applying the Cauchy-Schwarz inequality to $|\langle \psi| U^\dagger M |\Delta\rangle | $ it follows that:
$$|\langle \psi| U^\dagger M |\Delta\rangle | \leq ||\psi|| \: ||U^\dagger M \Delta || $$
Note that angle brackets and vertical bar of the bra-ket notation were omitted to make the notation more readable.
The next step is:
$$||\psi|| \: ||U^\dagger M \Delta ||\leq||\psi||\:||U^\dagger M||\: ||\Delta || $$
This inequality derives from the definition of operator norm. If $A$ is a linear operator, we define the norm of $A$ as:
$$\|A  \|_{op} = \sup \{\frac {\| Ax \|} {\| x \|} \ \forall \: x \neq 0 \}$$
Then it follows that for any $\Delta $ it holds $\| U^\dagger M\Delta \|\leq\| U^\dagger M\|_{op}\|\Delta \|$. We'll omit the $\:_{op }$ subscript hereinafter.
From the properties of the norm, it follows that $$||\psi|| \:\| U^\dagger M\|\:||\Delta ||\leq||\psi|| \:\| U^\dagger \|\|M\|\:||\Delta ||$$
Now remember that $||\psi||= 1$ and that $\| U^\dagger\|=1$ since $U$ is a unitary matrix.
To prove that $||M||\leq 1$ remember that $M$ is an element of a POVM. A POVM is a set of elements $\{M_i\}$ of $n X n$ matrices which are hermitian, positive definite and complete; note that typically they are non-projective and non-orthogonal.
From completeness it holds:
$$\sum_{i=0}^{n} M_i = I$$
Since $M_i$ is hermitian, then it is a normal operator and any normal operator is diagonal with respect to some orthonormal basis (spectral decomposition theorem); since $M_i$ is hermitian and positive definite all the eigenvalues $\lambda_j$ are real and positive and then $M_i=\sum_{j=0}^{m} \lambda_j |j \rangle\langle j|$.
Since $||\sum_{i=0}^{n} M_i ||=|| I||$ then $||M_i||\leq1$.
So we proved that $|\langle \psi| U^\dagger M |\Delta\rangle | \leq||\Delta || $.
Applying all the previous steps to $|\langle \psi| MV |\Delta\rangle|$ it follows that $|\langle \psi| MV |\Delta\rangle | \leq||\Delta || $.
Then we proved that:
$$|\langle \psi| U^\dagger M |\Delta\rangle |+|\langle \psi| MV |\Delta\rangle| \leq \||\Delta\rangle ||+\||\Delta\rangle \| $$
I hope this can help someone else.
A: $$|\langle \psi| AB \phi\rangle | \leq ||\psi|| \: ||AB \phi|| \leq ||\psi||\: ||AB|| \: ||\phi|| \leq ||\psi|| \: ||A||\:||B||\: ||\phi||$$
In our case $||\psi||=1$ and $||A||, ||B|| \leq 1$.
because one of $A$ and $B$ is unitary and the other is part of a POVM.
