# Is there a way to find a good lower bound on $\Vert p_n \Vert_\infty$ without finding the extrema?

Let $$p_n(x):=x^n+c_{n-1}x^{n-1}+ \cdots + c_0$$ be defined over some interval $[a,b]$. Is there a way to find a good lower bound on $\max_{x\in [a,b]} | p_n (x) |$ without actually finding the extrema?

In particular, I'd like to show that $$\max_{x\in [-1,1]} | p_4 (x) | \geq \frac{1}{8}$$

Any hints are hugely appreciated.

• $$\lVert p_n\rVert_\infty \geqslant \frac{1}{2} \lVert p_n\rVert_2 \geqslant \frac{1}{2}\operatorname{dist}(x^n, P_{n-1})$$ where $P_k$ is the space of polynomials of degree $\leqslant k$. I don't know if that suffices to give you $\lVert p_4\rVert_\infty \geqslant \frac{1}{8}$, but one can try. – Daniel Fischer May 21 '15 at 21:29
• @DanielFischer: that looks very interesting. What exactly do you mean by $\operatorname{dist}(x^n, P_{n-1})$? – Leo May 21 '15 at 21:32
• The distance of a point to a subset (here subspace). In Hilbert spaces like $L^2([-1,1])$, there is a simple formula to get the distance to a closed subspace, it's the norm of the projection to the orthogonal complement. Thus let $F_0,F_1,\dotsc$ be an orthonormal sequence of polynomials with $\deg F_k = k$ for all $k$, then $\operatorname{dist}(x^n,P_{n-1}) = \langle x^n, F_n\rangle$. I can't remember whether these polynomials $F_k$ are Chebyshev, Legendre, or Some-other-person polynomials, but the family has a name. – Daniel Fischer May 21 '15 at 21:37
• Legendre, but needs some scaling to be orthonormal. – Daniel Fischer May 21 '15 at 21:43