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 $$L(f)\leq \sqrt{1+n} ||f||_\infty.$$

  • 1
    $\begingroup$ What paper would that be? $\endgroup$
    – Will Jagy
    Nov 11, 2012 at 22:10
  • $\begingroup$ Whats mean "$L(f)$ the length of a polynomial?! $\endgroup$
    – user48941
    Nov 11, 2012 at 22:30
  • $\begingroup$ This is the paper (page 7, lemma 3.2). @Ali see page 4. $\endgroup$
    – George
    Nov 11, 2012 at 22:31

1 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.


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