Remez inequality explained

In mathematics, the Remez inequality, discovered by the Soviet mathematician Evgeny Yakovlevich Remez, gives a bound on the sup norms of certain polynomials, the bound being attained by the Chebyshev polynomials.

The inequality

Let σ be an arbitrary fixed positive number. Define the class of polynomials πn(σ) to be those polynomials p of degree n for which

|p(x)|\le1

on some set of measure ≥ 2 contained in the closed interval [−1, 1+''σ'']. Then the Remez inequality states that

\sup
p\in\pin(\sigma)

\left\|p\right\|infty=\left\|Tn\right\|infty

where Tn(x) is the Chebyshev polynomial of degree n, and the supremum norm is taken over the interval [−1, 1+''σ''].

Observe that Tn is increasing on

[1,+infty]

, hence

\|Tn\|infty=Tn(1+\sigma).

The R.i., combined with an estimate on Chebyshev polynomials, implies the following corollary: If J ⊂ R is a finite interval, and E ⊂ J is an arbitrary measurable set, thenfor any polynomial p of degree n.

Extensions: Nazarov–Turán lemma

Inequalities similar to have been proved for different classes of functions, and are known as Remez-type inequalities. One important example is Nazarov's inequality for exponential sums :

Nazarov's inequality. Let

p(x)=

n
\sum
k=1

ak

λkx
e

be an exponential sum (with arbitrary λk ∈C), and let J ⊂ R be a finite interval, E ⊂ J—an arbitrary measurable set. Then

maxx|p(x)|\leq

maxk|\Reλk|\operatorname{mes
e

J}\left(

C\operatorname{mes
J}{\operatorname{mes}E}

\right)n-1\supx|p(x)|~,

where C > 0 is a numerical constant.

In the special case when λk are pure imaginary and integer, and the subset E is itself an interval, the inequality was proved by Pál Turán and is known as Turán's lemma.

This inequality also extends to

Lp(T), 0\leqp\leq2

in the following way
\|p\|
Lp(T)

\leqeA(n-1)(T\setminusE)}

\|p\|
Lp(E)

for some A > 0 independent of p, E, and n. When

\operatorname{mes}E<1-

logn
n

a similar inequality holds for p > 2. For p = ∞ there is an extension to multidimensional polynomials.

Proof: Applying Nazarov's lemma to

E=Eλ=\{x:|p(x)|\leqλ\}, λ>0

leads to

maxx|p(x)|\leq

maxk|\Reλk|\operatorname{mes
e

J}\left(

C\operatorname{mes
J}{\operatorname{mes}

Eλ}\right)n-1

\sup
x\inEλ

|p(x)|\leq

maxk|\Reλk|\operatorname{mes
e

J}\left(

C\operatorname{mes
J}{\operatorname{mes}

Eλ}\right)n-1λ

thus

\operatorname{mes}Eλ\leqC\operatorname{mes}J\left(

λ
maxk|\Reλk|\operatorname{mes
e
J
} \right)^

Now fix a set

E

and choose

λ

such that

\operatorname{mes}Eλ\leq\tfrac{1}{2}\operatorname{mes}E

, that is

λ=\left(

\operatorname{mes
E}{2C

\operatorname{mes}J}\right)n-1

-maxk|\Reλk|\operatorname{mes
e

J}maxx|p(x)|

Note that this implies:

\operatorname{mes}E\setminusEλ\ge\tfrac{1}{2}\operatorname{mes}E.

\forallx\inE\setminusEλ:|p(x)|>λ.

Now

\begin{align} \intx\in|p(x)|pdx&\geq

\int
x\inE\setminusEλ

|p(x)|pdx\\[6pt] &\geq

p1
2
λ

\operatorname{mes}E\\[6pt] &=

1
2

\operatorname{mes}E\left(

\operatorname{mes
E}{2C

\operatorname{mes}J}\right)p(n-1)

-pmaxk|\Reλk|\operatorname{mes
e

J}maxx|p(x)|p\\[6pt] &\geq

1
2
\operatorname{mes
E}{\operatorname{mes}J}\left(\operatorname{mes
E}{2C

\operatorname{mes}J}\right)p(n-1)

-pmaxk|\Reλk|\operatorname{mes
e

J}\intx|p(x)|pdx, \end{align}

which completes the proof.

Pólya inequality

One of the corollaries of the Remez inequality is the Pólya inequality, which was proved by George Pólya, and states that the Lebesgue measure of a sub-level set of a polynomial p of degree n is bounded in terms of the leading coefficient LC(p) as follows:

\operatorname{mes}\left\{x\in\R:\left|P(x)\right|\leqa\right\}\leq4\left(

a
2LC(p)

\right)1/n,a>0~.

References