0

I have the following homework question. Given vectors $$ \vec{v}=(v_1,\dots, v_n) \in \mathbb{R}^n\ \text{ and }\ \vec{u}=(u_1, \dots, u_n) \in \mathbb{R}^n $$ such that $|v_i|, |u_i| \le 1$ for all $i =\mathbb{N}\cap [1,n]$, prove that the following inequality holds true:

$$\Bigg\vert\prod_{i=1}^{n}(1 - |v_i|) - \prod_{i=1}^{n}(1 - |u_i|) \Bigg\vert \le n\max_{1 \le i \le n }|v_i - u_i| $$

I tried proving this by induction. The base case $n=1$ went something like this: $$ | (1 - |v_1|) - (1 - |u_1|)| = | |u_1|-|v_1||\le | u_1-v_1| = 1 \cdot\max_{1 \le i \le 1 }|v_i - u_i| $$ by the reverse triangle inequality. I then attempted to do the inductive step, where I did the following \begin{align*} \Bigg\vert\prod_{i=1}^{n+1}(1 - |v_i|) - \prod_{i=1}^{n+1}(1 - |u_i|) \Bigg\vert &=\Bigg\vert(1 - |v_{n+1}|)\prod_{i=1}^{n}(1 - |v_i|) - (1 - |u_{n+1}|)\prod_{i=1}^{n}(1 - |u_i|) \Bigg\vert \end{align*} and ideally I would want to separate the latter expression into someting where $\Bigg\vert\prod_{i=1}^{n}(1 - |v_i|) - \prod_{i=1}^{n}(1 - |u_i|) \Bigg\vert$ is isolated so I could apply the induction hypothesis, but I couldn't think of a way to separate the factors to get this.

  • Could anyone point me in the right direction to complete the induction step? Thanks!
Shadow56675
  • 3

1 Answers1

1

I think this is the way. Let $V_n=\prod_{i=1}^{n}(1 - |v_i|)$ and $U_n =\prod_{i=1}^{n}(1 - |u_i|)$ then as you wrote $$\begin{align*} \Bigg\vert\prod_{i=1}^{n+1}(1 - |v_i|) - \prod_{i=1}^{n+1}(1 - |u_i|) \Bigg\vert &=\Bigg\vert(1 - |v_{n+1}|)V_n - (1 - |u_{n+1}|)U_n \Bigg\vert \\ &= \Bigg\vert(1 - |u_{n+1}| + |u_{n+1}| - |v_{n+1}|)V_n - (1 - |u_{n+1}|)U_n \Bigg\vert \\ &=\Bigg\vert(1 - |u_{n+1}|)V_n + (|u_{n+1}|-|v_{n+1}|)V_n - (1 - |u_{n+1}|)U_n\Bigg\vert \end{align*}$$ don't read more, if you don't want see whole solution $$\begin{align*}...&\leq \left| (1 - |u_{n+1}|)(V_n - U_n)\right| + |(|u_{n+1}|-|v_{n+1}|)V_n| \\ &\leq |V_n - U_n| + ||u_{n+1}|-|v_{n+1}|| \leq n\max_{1 \le i \le n }|v_i - u_i|+|u_{n+1}-v_{n+1}|\\ &\leq (n+1)\max_{1 \le i \le n+1 } \end{align*}$$

jorlinski
  • 394