6
$\begingroup$

Let $f$ be a polynomial of degree $d$. Of course $|f(z)|\sim C|z|^d$ as $|z|\rightarrow\infty$ but also, since any polynomial is completely determined by its values at any $d+1$ points, we may ask how the modulus of all polynomials at some fixed point $z$ is related to their maximum moduli at $d+1$ fixed points $\{z_k\}_{k=0}^{d}$ as $d\rightarrow\infty$.

The case I have in mind is that in which $z=-1$ and $z_k=k$ for $k\in[0,d]$. Using Lagrange's formula or linear dependence directly, we have

$$f(-1)=\sum_{k=0}^d{d+1\choose k+1}(-1)^kf(k),$$

which gives us $$|f(-1)|\leq (2^{d+1}-1)\max_{k\in[0,d]}|f(k)|.$$ This is the kind of inequality I am looking for, that is, an inequality of the form

$$|f(-1)|\leq g(d)\max_{k\in[0,d]}|f(k)|.$$

In general, the function $2^{d+1}-1$ is best possible because one can obtain an equality by fixing $f(k)=f(0)(-1)^k$ for $k\in[0,d]$. In fact equality occurs if and only $f$ is this polynomial.
So, putting $f(z)=\sum_{n=0}^{d}c_nz^n$, the maximum of

$$m(f)=\min_{k\in[0,d]}\left|\frac{\sum_{n=0}^{d}c_n(-1)^n}{\sum_{n=0}^{d}c_nk^n}\right|$$

over all sequences $\{c_n\}\in\mathbb{C}^d$ occurs if and only if $\{c_n\}$ is the unique (up to a scalar multiple) sequence of coefficients corresponding to the polynomial taking the values $f(k)=(-1)^k$, $k\in [0,d]$.

I would like to know:

Which mild assumptions on $f$ lead to the (marginally better) $\log m(f)=o(d)$?

For the application I have in mind I must allow $c_n=(-1)^n\nu_n$ where $\nu_n$ is a sequence of non-negative integers with $\nu_0=1$.

$\endgroup$

0

You must log in to answer this question.