I've been driven up a wall by the following question: let p be a complex polynomial of degree d. Suppose that |p(z)|≤1 for all z such that |z|=1 and |z-1|≥δ (for some small δ>0). Then what's the best upper bound one can prove on |p(1)|? (I only care about the asymptotic dependence on d and δ, not the constants.)


For the analogous question where p is a degree-d *real* polynomial such that |p(x)|≤1 for all 0≤x≤1-δ, I know that the right upper bound on |p(1)| is |p(1)|≤exp(d√δ). The extremal example here is p(x)=Td((1+δ)x), where Td is the dth Chebyshev polynomial.


Indeed, by using the Chebyshev polynomial, it's not hard to construct a polynomial p in z *as well as its complex conjugate z\**, such that


(i) |p(z)|≤1 for all z such that |z|=1 and |z-1|≥δ, and


(ii) p(1) ~ exp(dδ).


One can also show that this is optimal, for polynomials in both z and its complex conjugate.


The question is whether one can get a better upper bound on |p(1)| by exploiting the fact that p is really a polynomial in z. The fastest-growing example I could find has the form p(z)=Cd,δ(1+z)d. Here, if we choose the constant Cd,δ so that |p(z)|≤1 whenever |z|=1 and |z-1|≥δ, we find that


p(1) ~ exp(dδ2)


For my application, the difference between exp(dδ) and exp(dδ2) is all the difference in the world!


I searched about 6 approximation theory books---and as often the case, they answer every conceivable question except the one I want. If anyone versed in approximation theory can give me a pointer, I'd be incredibly grateful.


Thanks so much!
--Scott Aaronson


PS. The question is answered below by David Speyer. For anyone who wants to see explicitly the polynomial implied by David's argument, here it is:


pd,δ(z) = zd Td((z+z-1)(1+δ)/2+δ),


where Td is the dth Chebyshev polynomial.



Analogue of the Chebyshev polynomials over C?
I may be missing something obvious here. Let $f(z, z^{\*})$ be the polynomial in $z$ and $z^{\*}$ of degree $d$ which achieves $\exp(d \delta)$. Let $g(z)$ be the Laurent polynomial obtained from $f$ by replacing $z^{\*}$ by $z^{-1}$. On the unit circle, we have $f=g$.


Now, let $h$ be the polynomial $z^d g$. This is an honest polynomial, because we multiplied by a high enough power of $z$ to clear out all the denominators and, for $z$ on the unit circle, we have $|h|=|f|$.


Doesn't this mean that $h$ is a polynomial of degree $2d$, obeying your conditions, with $|h(1)| \sim \exp(d \delta)$?


