# Suppose $U_1,\dots,U_k$ and $V_1,\dots,V_k$ are $n\times n$ unitary matrices. Show that $\|U_1\cdots U_k-V_1\cdots V_k\|\leq\sum_{i=1}^k\|U_i-V_i\|$

Let $$V,W$$ be complex inner product spaces. Suppose $$T: V \to W$$ is a linear map, then we define $$\|T\|:=\sup\{\|Tv\|_{W}:\|v\|_{V}=1\}$$ where $$\|v\_{V}\|:=\sqrt{\langle v,v\rangle}$$ and $$\|Tv\|_{W}:=\sqrt{\langle Tv,Tv\rangle}$$.

Question: Suppose $$U_1,\ldots,U_k$$ and $$V_1,\ldots,V_k$$ are $$n {\times} n$$ unitary matrices. Show that $$\|U_1\cdots U_k-V_1\cdots V_k\| \leq \sum_{i=1}^{k}\|U_i-V_i\|$$

I have tried to use triangle inequality for norms and induction but failed. Can anyone give some hints? Thank you!

For $$k = 1$$, the inequality becomes an identity. So, start with the special case $$k = 2$$: $$\begin{array}{ll} & ||U_1 U_2 - V_1 V_2||\\ \\ = & ||U_1 U_2 - V_1 U_2 + V_1 U_2 - V_1 V_2||\\ \\ = & || ( U_1 - V_1) U_2 + V_1 (U_2 - V_2) || \\ \\ \leq & ||( U_1 - V_1) U_2 || + ||V_1 (U_2 - V_2) ||.\\ \end{array}$$ (The last inequality is the triangle inequality.) Now, since $$U_2$$ is unitary, we have $$||U_{2}|| = 1$$, so $$|| ( U_1 - V_1) U_2 || \leq || U_1 - V_1 ||.$$ A similar bound obtains for $$||V_1 (U_2 - V_2) ||$$.