Polynomial of degree 'n' related problem

Question

Suppose $p\left( x \right) = {a_0} + {a_1}x + {a_2}{x^2} + \cdots + {a_n}{x^n}$. If $\left| {p\left( x \right)} \right| \le \left| {{e^{x - 1}} - 1} \right|$ for $x\ge 0$, prove that $\left| {{a_1} + 2{a_2} + \cdots + n{a_n}} \right| \le 1$.

My approach is as follow $p'(1)=\ {{a_1} + 2{a_2} + \cdots + n{a_n}} $.

We need to prove that for positive $x$, $|p'(1)|\le 1 $ but not able to proceed.

Answer 1

First off, directly from the inequality you get that $p(1)=0$. Then one has $$\frac{|p(1+h) - p(1)|}{h} \le \frac{|e^{h}-1|}{h}.$$ Now pass with $h$ to $0$ in the expression above.