Denote the operator norm by $\|\cdot \|_\infty$. That is $\|X\|_\infty = \max\{\|Xv\|:\|v\| = 1\}$ and $\|\cdot\|$ is the Euclidean 2-norm for vectors.
Let $U_1, U_2$ be unitary matrices (i.e. $U_i^\star U_i = I$ where $\star$ denotes the transpose conjugate) such that
$$\|U_1 – U_2\|_\infty \leq \varepsilon$$
For positive semidefinite $X$, is there a bound on
$$\|U_1 XU_1^\star – U_2X U_2^\star\|_\infty$$
in terms of $\varepsilon$? If yes, how can one show this?
Best Answer
Note that the bound will have to scale with $\|X\|$, because the map $X\mapsto U_1XU_1^*-U_2XU_2^*$ is linear.
Assuming you are using the (Euclidean) operator norm, we have \begin{align*} \|U_1XU_1^*-U_2XU_2^*\|&\leq\|U_1XU_1^*-U_1XU_2^*\|+\|U_1XU_2^*-U_2XU_2^*\|\\ &\leq\|U_1X\|\|U_1^*-U_2^*\|+\|U_1-U_2\|\|XU_2^*\|\\ &\leq 2\|X\|\varepsilon, \end{align*} Since $\|UX\|=\|X\|$ and $\|X^*\|=\|X\|$ for any unitary $U$ and any matrix $X$.