Proof using Cauchy–Schwarz Inequality.

Question

This is a problem I am trying to solve for a couple of months without any success. I found it in a paper and according to the authors can be proved using Cauchy–Schwarz inequality.

Let $f(x)$ be a polynomial in $\mathbb{C}[x]$. Denote by $L(f)$ the length of the polynomial and by $||.||_\infty$ the sup norm on $\{|z|\leq 1\}$. So for $f(x)=a_0+a_1x+\ldots +a_nx^n , L(f)=|a_0|+|a_1|+\ldots+|a_n|$ and $||f||_\infty=\sup{\{|f(z)|:|z|\leq 1\}}.$ The problem is:

If $f(x)\in \mathbb{C}[x]$ is a polynomial of degree $n$ then $$L(f)\leq \sqrt{1+n} ||f||_\infty.$$

Answer 1

Cauchy-Schwarz inequality yields $L(f)^2\leqslant(n+1)\cdot K(f)$ where $K(f)=\sum\limits_{k=0}^n|a_k|^2$. One completes the proof thanks to the formula $$ K(f)=\oint_{S^1}\left|f(z)\right|^2\frac{\mathrm dz}{2\mathrm i\pi z}. $$ (To see that the integral in the RHS is indeed $K(f)$, expand $|f(z)|^2$ as a linear combination of powers of $z$ and compute the integral of each term using the residue theorem. Only the constant terms contribute.) Since $|f(z)|^2\leqslant\|f\|_\infty^2$ for every $z$ in $S^1$, $K(f)\leqslant\|f\|_\infty^2$ and we are done.