Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 2G / b

Take $x_j=-1+j/4$, $0\le j\le 8$. These nine points satisfy $\cos (4\pi x_j)=(-1{)}^j$. Thus the zero polynomial has nine alternating error extrema of magnitude one and is minimax among polynomials of degree at most seven by the Chebyshev alternation theorem. Hence every such polynomial has uniform error at least one.

One can also prove the bound directly. If the error were strictly below one everywhere, $g(x_j)$ would have sign $(-1{)}^j$. The intermediate value theorem would give a distinct zero in each of the eight intervening intervals, contradicting the degree bound. Since the continuous error attains its maximum on the compact interval, there is an $x$ with $|f(x)-g(x)|\ge 1$.