# A problem on polynomials

Let $$P(z)$$ be a polynomial of degree $$n$$ with $$|P(z)|\leq 1$$ on $$|z|=1$$ and $$P_m(z)$$ be a partial sum of $$P(z).$$ How large $$P_m(z)$$ can be on $$|z|=1?$$

• Trivial upper bound is $\|1+e^{i \theta}+...+e^{im \theta}\|_{L^{1}(\mathbb{T})}$. – Paata Ivanishvili May 25 at 4:43
• What here is fixed and what is allowed to vary? – user44191 May 25 at 4:46

The trivial upper bound $$\max_{|w|=1}|P_{m}(w)|\leq \|1+z+...+z^{m}\|_{L^{1}(\mathbb{T})} \asymp C \log(m)$$ that I wrote in the comment is actually sharp in the regime $$m=n/2$$. Here is the proof.
Notice that $$P_{m}(z)$$ is convolution of $$P(z)$$ with $$D_{m}(z) = 1+z+...+z^{m}$$ on the unit circle, therefore, by the triangle inequality $$\max_{|z|=1}|P_{m}(z)| \leq \frac{1}{2\pi}\int_{-\pi}^{\pi}|1+e^{i\theta}+...+e^{i m \theta}| d\theta \asymp C \log(m)$$
Next, let us show that the upper bound is sharp in the regime $$m=\frac{n}{2}$$ where $$n$$ is large.
Indeed, consider the polynomial $$P(z) = z^{n} \overline{\left( 1+\frac{z}{1}+...+\frac{z^{n}}{n}\right)} - z^{n}\left(1+\frac{z}{1}+...+\frac{z^{n}}{n} \right) = z^{n}+\frac{z^{n-1}}{1}+...+\frac{1}{n}-z^{n}-\frac{z^{n+1}}{1}-...-\frac{z^{2n}}{n}$$ it is of degree $$2n$$, and $$\max_{|z|=1}|P_{n-1}(z)|\geq |P_{n-1}(1)| \asymp \log(n)$$. On the other hand let us show that $$\max_{|z|=1}|P(z)|\asymp 1$$. Indeed, for $$z=e^{ix}$$ we have
$$|P(z)| = 2\left| \sum_{k=1}^{n} \frac{\sin(kx)}{k}\right|\leq 2 \int_{0}^{\pi} \frac{\sin(s)}{s}ds \asymp 1 \quad \text{for all} \quad n \geq 1, \; x \in [0, 2\pi)$$ The last inequality I guess is known "since Democritus".
• I haven't looked closely, but it feels like this argument might symmetrize to get an upper bound of $O(\log(\min(m,n-m)))$? – Steven Stadnicki May 26 at 15:17
• Yes, this can be done by writing $P_{m}(z) = P(z)-(P(z)-P_{m}(z))$ the first term is estimated by $1$, and the term $(P(z)-P_{m}(z))$ is upper bounded by $\|z^{m+1}+...+z^{n}\|_{L^{1}(\mathbb{T})}\asymp \log(n-m)$. – Paata Ivanishvili May 27 at 2:56