Let $r(t)$ be a real trigonometric polynomial of degree $n>1$. Assume it has zero at $t=0$ of multiplicity $k>0$. What can be said about the lower bound of the constant $c(k,n)$ such that $$ \max_{t \in \mathbb{R}}|r'| \geq c(k,n) \max_{t \in \mathbb{R}}|r|. $$
For example,
is it known that if $|n-k|<A$ (here $A$ is some fixed finite number), then $c(k,n)>C\sqrt{n}$ where $C=C(A)$ depends only on $A$?
In the latter case I think I can show that $c(k,n)>C(A,\varepsilon)\, n^{\frac{1}{2}-\varepsilon}$. But I am almost convinced that the right lower bound should be $C(A)\sqrt{n}$. I also think that this should be somethig well--known, so any reference would be helpful.
Remarks:
- there is a subtle result of Borwein-Erdelyi which shows that $\max_{t \in \mathbb{R}}|r'| \leq c_{1}\sqrt{n(n-k+1)}\max_{t \in \mathbb{R}}|r|$. So according to this estimate one possible guess could be that $c(k,n)\geq c_{0}\sqrt{\frac{n}{n-k+1}}$.
- there is a Turan's inequality which says that if all zeros of the trigonometric polynomial $t$ are real then $\|r'\|_{\infty} \geq \frac{\sqrt{n}}{6}\|r\|_{\infty}$. So maybe if we allow some finite number of zeros to be not necessarily real then such an estimate still persists, but I am not sure about this, probably some extra restrictions on the zeros are still needed.
- A well-known reverse Bernstein inequality (which I think goes back to Zigmund) says that if a periodic function $f(t) \in L^{\infty}([0,2\pi])$ lives on high frequencies, say $\hat{f}(k)=0$ for all $|k|\geq n>0$ then $\|f'\|_{\infty} \geq C n\|f\|_{\infty}$. This is standard Foruier analytic argument, one can construct a multiplier $g$ such that $f=f'*g$ with $\|g\|_{1}\leq C/n$, and, for example, Young's convolution inequality finishes the story. I am not sure if such multiplier argument will be relevant in our case, because, for example, when you multiply trigonometric polynomial by $\sin^{k}(t)$ its Fourier coefficients become not easy to control.
