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 $z=1?$

1$\begingroup$ Trivial upper bound is $\1+e^{i \theta}+...+e^{im \theta}\_{L^{1}(\mathbb{T})}$. $\endgroup$ – Paata Ivanishvili May 25 at 4:43

3$\begingroup$ What here is fixed and what is allowed to vary? $\endgroup$ – 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
$$
P(z) = z^{n} \overline{\left( 1+\frac{z}{1}+...+\frac{z^{n}}{n}\right)}  z^{n}\left(1+\frac{z}{1}+...+\frac{z^{n}}{n} \right) = z^{n}+\frac{z^{n1}}{1}+...+\frac{1}{n}z^{n}\frac{z^{n+1}}{1}...\frac{z^{2n}}{n}
$$
it is of degree $2n$, and $\max_{z=1}P_{n1}(z)\geq P_{n1}(1) \asymp \log(n)$. 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".

1$\begingroup$ @user159888 I do not see typos. One small thing is that instead of P_{n1} I should be writing P_{n}. But both are fine it just gives example of degree 2n with m=n1 $\endgroup$ – Paata Ivanishvili May 26 at 12:43

1$\begingroup$ I haven't looked closely, but it feels like this argument might symmetrize to get an upper bound of $O(\log(\min(m,nm)))$? $\endgroup$ – Steven Stadnicki May 26 at 15:17

1$\begingroup$ 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 $\z^{m+1}+...+z^{n}\_{L^{1}(\mathbb{T})}\asymp \log(nm)$. $\endgroup$ – Paata Ivanishvili May 27 at 2:56