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Let $p=a_{0}+a_{1}z+\ldots+a_{n}z^{n}$ be a polynomial. Consider the following truncated $\ell^{1}$-seminorm of the coefficients of $p$: $$\|p\|_{\ell^{1},\text{trun.}}:=\sum_{k=1}^{n}|a_{k}|=\|p-a_{0}\|_{\ell^{1}}.$$

Does there exist $C>0$ (independent of $n=\deg(p)$) such that $$\|p\|_{\ell^{1},\text{trun.}}\leq C\|p\|_{\infty}\qquad\text{where}\qquad \|p\|_{\infty}:=\sup_{|z|\leq1}|p(z)|?\tag{$*$}$$

Remark 1: $\sup_{|z|\leq1}|p(z)|=\sup_{|z|=1}|p(z)|$ by the maximum modulus principle.

Remark 2: If $p$ is a polynomial with $a_{j}\geq0$, then the inequality clearly holds with $C=1$. Indeed, $\|p\|_{\ell^{1},\text{trun.}}\leq p(1)\leq\|p\|_{\infty}$.

Remark 3: Due to this MSE post I believe that estimation ($*$) does $C$ does not exist independent of $n$ if $\|\cdot\|_{\ell^{1},\text{trun.}}$ is replaced by $\|\cdot\|_{\ell^{1}}$.

Remark 4: The same MSE post also provides an example that shows $C>1$ if it exists. Namely, if $p(z)=z^{2}+2iz+1$, then $\|p\|_{\ell^{1},\text{trun.}}=|1|+|2i|=3$ and $\|p\|_{\infty}=2\sqrt{2}$ (I haven't checked this last fact).

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    $\begingroup$ The polynomials $p(z)$ and $zp(z)$ have the same $\infty$-norm. Thus a counterexample from Remark 3 becomes a counterexample for the present question after multiplying by $z$. $\endgroup$ Jul 20, 2022 at 14:10
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    $\begingroup$ By Fedor's remark and by setting $z=e^{it}$, this is the same as asking if the Fourier transform is bounded $L^{\infty}\to\ell^1$. This is false, and you can probably find this classical result in many books. Or see Exercise 8.6 of my lecture notes here (the exercise comes with detailed hints): math.ou.edu/~cremling/teaching/lecturenotes/fa-new/ln8.pdf $\endgroup$ Jul 20, 2022 at 14:20

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