Show $||U_i-V_i||\leq\epsilon\ \ \forall i=1,...,m\Longrightarrow||U_m...U_1-V_m...V_1||\leq m\epsilon$ (spectral norm)

Question

Let $\{U_i\},\{V_i\}$ be sets of $m$ unitary operators with $||U_i-V_i||\leq\epsilon\ \ \forall i=1,...,m$.

Then $||U_m...U_1-V_m...V_1||\leq m\epsilon$ with $||\cdot||$ being the spectral norm.

Can someone help me show this conclusion? I'm going through some lecture notes and they omit the proof because it is "easy."

Answer 1

We have $$\begin{split} ||U_m...U_1-V_m...V_1|| &= \left \| \sum_{i=1}^m U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1 \right\|\\ &\leq \sum_{i=1}^m \|U_m...U_iV_{i-1}...V_1-U_m...U_{i+1}V_i...V_1\| \\ &\leq \sum_{i=1}^m \| U_m...U_{i+1}(U_i-V_i)V_{i-1}...V_1\|\\ &\leq \sum_{i=1}^m \|U_m...U_{i+1}\|\|U_i-V_i\|\|V_{i-1}...V_1\|\\ \end{split}$$ where the last inequality comes from the fact that with the spectral norm, $\|AB\|\leq\|A\|\|B\|$. Also, any unitary transform $T$ has spectral norm $1$, so $\|U_m...U_{i+1}\|=\|V_{i-1}...V_1\|=1$.

Finally, one can conclude that $$||U_m...U_1-V_m...V_1|| \leq \sum_{i=1}^m \| U_i-V_i\|\leq m\varepsilon $$