# Close in operator norm imply close in weak multiplicative sense?

Fix $$\epsilon > 0$$, and suppose $$U$$ and $$S$$ are $$n$$ qubit unitaries such that $$\| U - S \| \leq \epsilon$$ (operator norm). Furthermore, let $$P_U(y \mid x) = |\langle y | U | x \rangle|^2$$ be the probability of obtaining the output string $$y \in \{0,1\}^n$$ given the input string $$x \in \{0,1\}^n$$ w.r.t. the unitary $$U$$. ($$P_S(y \mid x)$$ is the same thing, but w.r.t. the unitary $$S$$.)

I am trying to prove that $$U$$ and $$S$$ are close in the weak multiplicative sense, i.e., that $$|P_U(y \mid x) - P_S(y \mid x)| \leq \epsilon P_U(y \mid x).$$ While I think this has to be true simply because $$U$$ and $$S$$ are effectively the same operation, I am struggling to get the inequalities to work out. Any help would be appreciated.

• This is not possible. Consider $U=I$ and $|x\rangle$ and $|y\rangle$ orthonormal. Then the RHS is zero, but the LHS may be positive (depending on $S$). Jan 11 at 23:48
• True, thank you! Jan 11 at 23:52