How to prove this (difficult) fact about polynomials?

Question

This problem is kinda difficult, I haven't got a proof so far:

Let P be a polynomial of degree n, with real coefficients, such that P(x) is in [-1, 1] for every x in [-1, 1]. Show that P'(x) is in [-n^2 , n^2] for every x in [-1, 1].

I think the solution has something to do with Chebyshev's polynomials.

Answer 1

This is a classical inequality known as Markov's 1889 theorem. D. Mendeleyev, the inventor of the periodic table of elements, posed the question and gave solution for n=2. A.A. Markov, the inventor of Markov's chains, considered arbitrary n. His younger brother V.A. Markov extended this result for higher order derivatives (Markov's 1892 theorem, expressed in terms of Chebyshev polynomials). English translation of the paper by Markov 1889 is here:

http://www.math.technion.ac.il/hat/fpapers/markov4.pdf

The original Markov's proof has been simplified in the modern literature, but it is still lengthy. These are the main steps.

Step 1. Theorem from Fourier analysis (Fejer):
If the function g(z) is periodic in [-a, a] and continuous, then the arithmetic mean of the partial sums of its Fourier series σ_n (Cesaro sums) are all bounded by M=max( |g|). Cesaro sums are equal to Fejer's integrals, i.e. to convolutions of g with Fejer kernel (see Eq. (9) in the link below). Integral of Fejer kernel is equal to 2a (see Eq. (7)). Replace g by M, take it out of the integral (9) and get

Eq. (1)
|σ_n|<=M.

http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-100CSpring-2006/1C675012-F10F-4851-A331-7937096CB516/0/ricketson.pdf

Step 2. Bernstein inequality:
Consider a polynomial of n-th degree, f = sum_{from 0 to n} b_k z^k, such as |f|
Eq. (2)
|t_n'(θ)| <=nM.

Then you consider polynomials p_n(x) with x=cos(θ) to get

Eq. (3)
|p_n'(x)| <=nM/ √(1-x^2).

Step 3. Schur inequality for polynomials.
Consider polynomial of (n+1) degree t_n(θ)=r_{n-1} (cos θ) sin θ (r_{n-1} is a polynomial of degree (n-1)). If x=cos θ and max(|r_{n-1}(x)| √(1-x^2))=Q , then |t_n|<= Q. |t'| at θ=0 is equal to |r(1)|. Using Eq. (2) we see that |r_{n-1}(1)| <= n Q. Apply this relation to the function v_{n-1}(x) = r_{n-1}(y*x) with |y|<=1. Then |r_{n-1}(y)| <= n max( |v_{n-1}(x)| √(1-x^2)) <= n max( |v_{n-1}(x)| √(1-y^2 x^2)) = n Q. Thus

Eq. (4)
|r_{n-1}(y)| <= n max(|r_{n-1}(x)| √(1-x^2))


Set r_{n-1} = p_n'(x) in Eq. (4), which gives |p_n'(x)|/( n √(1-x^2)) <= |p_n'(x)|. Use Eq. (5) for M=1 and we are there:

|p_n'(x)| <= n^2.


Equalities are achieved if p_n are Chebyshev polynomilas. Basically above results mean that Chebyshev polynomials provide upper bounds for polynomial derivatives.

Answer 2

Greetings,

It might be an application of Cauchy's inequality,

If we let f(x) = the polynomial

| fn(a) | < = M*n!/r^n

where fn is the nth derivative, M is a constant such that | f(x) | < M on circle C of radius r center at x=a , i.e. M is an upper bound.

Regards

Answer 3

If you can factor the complex function into the product of two simpler functions, the zeroes of each of the simpler functions are also zeros of the complex function. For example, f(x) = 2x^3 + 2x^2 - 8x - 8 (Complex function) f(x) = (2x^2 - 8) (x + 1) (Product of simpler functions) So we have two solutions when (2x^2 - 8) is zero, and a third solution when (x + 1) is zero. The three solutions are x = 2, x = -2 and x = -1.