Prove $|\prod_{i=1}^n a_i - a_n^n|\leq 2n\delta$ if $0 \leq a_i \leq 1$ and $|a_i - a_{i+1}| \leq \delta$

Question

How to prove the following?

If $0 \leq a_i \leq 1$ and $|a_i - a_{i+1}| \leq \delta$ for all $i<n$, then $$\Big|\prod_{i=1}^n a_i - a_n^n \Big| \leq 2n\delta.$$

I have run numerical experiments and it seems the above is true.

Answer 1

The proposition is false. Take $a_1 = \frac{1}{2},\ a_2 = \frac{1}{2},\ n=2, \delta = 0,$ so that the condition $|a_i - a_{i+1}| \leq \delta$ for all $i<2$ is satisfied.

Then, $$ \Big|\prod_{i=1}^n a_i - a_n^n \Big| = \big\vert a_1 - a_2^2 \big\vert\ \cdot\ \big\vert a_2 - a_2^2 \big\vert = \Bigg\vert \frac{1}{2} - \left(\frac{1}{2}\right)^2 \Bigg \vert\ \cdot\ \Bigg\vert \frac{1}{2} - \left(\frac{1}{2}\right)^2 \Bigg\vert = \frac{1}{4}\cdot \frac{1}{4} \not\leq 2\cdot 2\cdot 0 = 0.$$