inequality of same distance points

Question

I got this problem:

Let $x_i=\frac{i}{n}$ for $i=0,1,...,n$. Prove that for all $$ x\in[0,1]:|\prod_{k=0}^n (x-x_k)| \le \frac{n!h^{n+1}}{4} $$ where $h:=\frac{1}{n}$.

I tried to find maximum of each factor but it did not help. Any suggestions? thanks for helpers!

Answer 1

Let $[x_l, x_{l+1}]$ be the interval containing $x$. Then

1. $|(x-x_l)(x-x_{l+1})| = (x-x_l)(x_{l+1} - x)\le (\frac h2)^2 $ because the quadratic polynomial attains its maximum exactly between the two zeros.
2. For $k = l+1, \dots, n$, $|x - x_k| \le x_k - x_l = (k-l)h$. Therefore $$ \bigl|\prod_{k=l+2}^n (x-x_k) \bigr| \le 2\cdot 3 \cdots (n-l) \cdot h^{n-l-1} \quad . $$
3. Similarly $$ \bigl|\prod_{k=0}^{l-1} (x-x_k) \bigr| \le 2\cdot 3 \cdots (l+1) \cdot h^{l} \quad . $$

Putting all together: $$ \bigl|\prod_{k=0}^{n} (x-x_k) \bigr| \le \frac 14 \left( 2\cdot 3 \cdots (n-l) \right ) \left(2\cdot 3 \cdots (l+1) \right ) h^{n+1} \le \frac {n!}4 h^{n+1} \quad . $$