Hello to everyone.This is problem 25 of Donald J Newmans book A Problem Seminar.I uploaded the problem and a part of the solution.I want to formally show that continuity of f implies the existence of $0=x_0<x_1<....<x_{n-1}<x_n=1$.
The function $g_0:[0,1]\mapsto R$ with $g_0(x)=(x-x_0)(f(x)-f(x_0))$$\forall x\in [0,1]$ is continuous,strictly increasing and onto $[0,1]$ thus the fact that $\frac{1}{n^2}\in [0,1]$ and the intermediate value theorem imply the existence of a $x_1\in (0,1):g_0(x_1)=\frac{1}{n^2}$
Now let $x_1<...<x_k$ have been chosen with $k\lt n-1$and similarly define $g_k:[x_k,1]\mapsto R$ with $g_k(x)=(x-x_k)(f(x)-f(x_k))$$\forall x\in [x_k,1]$.The function $g_k$ is continuous,strictly increasing with $g_k(x_k)=0$,so if i show that $g_k(1)\geq\frac{1}{n^2}$ then continuity of $g_k$ and the intermediate value theorem once again imply the existence of a $x_{k+1}\gt x_k$ such that $g_k(x_{k+1})=\frac{1}{n^2}$
Any idea?
