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Polynomial coefficients are bounded by values at fixed distinct nodes

Statement

Let n≥0 and let t0,…,tn be distinct real numbers. For 0≤j≤n, define Lj(x):=∏k≠jx−tktj−tk. If p(x)=∑k=0nckxk has degree at most n, then p(x)=∑j=0np(tj)Lj(x). Consequently, writing Lj(x)=∑k=0nℓjkxk, every coefficient satisfies ∣ck∣≤(∑j=0n∣ℓjk∣)max⁡0≤j≤n∣p(tj)∣. The empty product convention gives L0=1 when n=0.

Facts & Assumptions

Given: An integer n≥0, distinct real nodes t0,…,tn, and a real polynomial p of degree at most n.

[F1]

Formal real polynomials, evaluation, degree, leading coefficient, and monic polynomials: a polynomial of degree at most n is a finite coefficient sum ∑k=0nckxk.

[F2]

A nonzero real polynomial of degree n has no more than n distinct real roots: a nonzero polynomial of degree at most n has at most n distinct real roots.

Proof

1.1F1F2givenconstructalgebra

Each denominator tj−tk is nonzero, so every Lj is a polynomial of degree at most n and Lj(ti) is 1 for i=j and 0 otherwise. The polynomial p−∑jp(tj)Lj has degree at most n and vanishes at the n+1 distinct nodes; [F2] makes it the zero polynomial. This also covers n=0, when L0=1.

2.1givenstep 1.1algebra∎

Comparing the coefficients of xk in the identity of step 1.1 gives ck=∑jp(tj)ℓjk. The triangle inequality yields the stated coefficient bound.

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