I'm solving the following (numerical analysis) problem.
Let $f(x)$ be a polynomial of degree $n$ such that there exist exactly $n+1$ distinct points $x_0,x_1\cdots,x_n$ in the interval $[-1,1]$ satisfying $|f(x_i)|$ $=\max_{x\in [-1,1]}(|f(x)|)$ for $i=0,1,\cdots,n.$ Show that the sign of the sequence $f(x_i)$ for $i=0,1,\cdots,n$ is alternating, that is $f(x_i)f(x_{i+1})<0.$
I've tried several methods (Chebyshev polynomials,...) but in vain. How should I solve this problem?
HINT: A polynomial of degree $n$ can have at most $n-1$ local extrema on $\Bbb R$; on a closed interval it can have two more, one at each endpoint, but that’s still only a maximum of $n+1$ local extrema on the interval. Here $f$ actually has that many local extrema on $[-1,1]$, so every local extremum of $f$ on $\Bbb R$ has to occur within that interval and have the same absolute value, and $f(-1)$ and $f(1)$ also have to have that absolute value. Can $f$ have two local maxima on $\Bbb R$ with no local minimum between them (or the opposite)? What about on $[-1,1]$?