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For n1n\ge1, 21nTn2^{1-n}T_n is the minimax monic polynomial of degree nn on [1,1][-1,1]

Statement

For n1n\ge1, the polynomial Pn=21nTnP_n=2^{1-n}T_n is monic of degree nn and for every monic real polynomial qq of degree nn, maxx[1,1]q(x)21n=maxx[1,1]Pn(x).\max_{x\in[-1,1]}|q(x)|\ge2^{1-n}=\max_{x\in[-1,1]}|P_n(x)|. Equality is attained by PnP_n. The conventions and prerequisite facts used below are recorded in Chebyshev polynomials of the first and second kinds by their three-term recurrences, Degrees and leading coefficients of the Chebyshev polynomials, Tn(cosθ)=cos(nθ)T_n(\cos\theta)=\cos(n\theta) and Un(cosθ)sinθ=sin((n+1)θ)U_n(\cos\theta)\sin\theta=\sin((n+1)\theta) for every nNn\in\mathbb N, Signs, monotonicity intervals, and ranges of sine and cosine, A nonzero real polynomial of degree nn has no more than nn distinct real roots, Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials, Extreme value theorem: a continuous real function on a nonempty compact subset of R\mathbb{R} attains a greatest and a least value, Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, and Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on [a,b][a,b] takes every value between f(a)f(a) and f(b)f(b).

Facts & Assumptions

Given: A natural n1n\ge1 and a monic polynomial qq of degree nn.

[L1]

Degrees and leading coefficients of the Chebyshev polynomials says that TnT_n has degree nn and leading coefficient 2n12^{n-1}.

[L2]

Tn(cosθ)=cos(nθ)T_n(\cos\theta)=\cos(n\theta) and Un(cosθ)sinθ=sin((n+1)θ)U_n(\cos\theta)\sin\theta=\sin((n+1)\theta) for every nNn\in\mathbb N gives Tn(cosθ)=cos(nθ)T_n(\cos\theta)=\cos(n\theta) and Tn(cos(jπ/n))=(1)jT_n(\cos(j\pi/n))=(-1)^j for 0jn0\le j\le n.

[L3]

Signs, monotonicity intervals, and ranges of sine and cosine says that cosine has range [1,1][-1,1] and is strictly decreasing on [0,π][0,\pi].

[L5]

A nonzero real polynomial of degree nn has no more than nn distinct real roots bounds the number of distinct real roots by the degree.

[L6]

Extreme value theorem: a continuous real function on a nonempty compact subset of R\mathbb{R} attains a greatest and a least value makes the maximum of q|q| on the nonempty compact interval [1,1][-1,1] exist once qq is continuous.

Proof

technique · contradiction
1.1

By [L1], Pn=21nTnP_n=2^{1-n}T_n is monic of degree nn. Put yj=cos((nj)π/n)y_j=\cos((n-j)\pi/n) for 0jn0\le j\le n. By [L3], y0<<yny_0<\cdots<y_n, and [L2] gives Pn(yj)=(1)nj21nP_n(y_j)=(-1)^{n-j}2^{1-n}.

L1L2L3
2.1

For each x[1,1]x\in[-1,1], [L3] supplies θ\theta with x=cosθx=\cos\theta; [L2] then gives Pn(x)=21ncos(nθ)21n|P_n(x)|=2^{1-n}|\cos(n\theta)|\le2^{1-n}. Equality holds at every yjy_j, so max[1,1]Pn=21n\max_{[-1,1]}|P_n|=2^{1-n}.

L2L3step 1.1
2.2

By [L7], qq and hence q|q| are continuous, so [L6] makes the displayed maximum well-defined. Suppose, for contradiction, that it is <21n<2^{1-n}. Then r:=qPnr:=q-P_n has degree at most n1n-1. At the successive points yjy_j, the values of rr have the opposite alternating signs to PnP_n, hence are nonzero and alternate.

L6L7assume-contrastep 1.1
3.1

By [L7], rr is continuous, so [L4] gives a root of rr in each disjoint interval (yj,yj+1)(y_j,y_{j+1}) (0j<n)(0\le j<n). Thus rr has at least nn distinct roots, contradicting [L5] because step 2.2 makes rr nonzero of degree at most n1n-1. The contradiction proves the lower bound, while step 2.1 proves equality for PnP_n.

L4L5L7step 2.1step 2.2discharge-contradiction

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