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The pointwise interpolation extremum
Source. Bernstein 1931, equation (1), printed p. 1025 / PDF p. 1, and its explanation on printed p. 1026 / PDF p. 2, in the complete source. The argument below uses degree at most and nodes; Bernstein calls the degree .
Let , let be real, and put
These are polynomials, including at their nodes. Away from the nodes, . For every real ,
The maximum is the same for real or complex coefficients. In particular, , , and is continuous.
Proof. The polynomial has degree at most and vanishes at distinct points, so it is zero. Thus
For fixed real , assign when , and assign any value of modulus at most one to the remaining node data. The interpolating polynomial has real coefficients and value at . This proves attainment and (E). Interpolating the constant polynomial gives , hence ; evaluation at a node and continuity give the other assertions.
Equality and endpoints. At a non-node , every is nonzero. Equality holds precisely when all node values have modulus one and the numbers have one common complex argument. For real coefficients this allows a common sign. At , equality requires only . Also, precisely when all the real numbers are nonnegative. All statements apply at when the nodes lie in ; the rational expression is never evaluated through a zero denominator.
Dependencies. Polynomial interpolation and the triangle inequality, both proved or applied explicitly above.
Proof scope. Complete rewritten elementary argument; independently reviewed on 6 September 2026 (component C1 of the local-chain review). No formal verification is claimed.
Bears on. Problem 1153; Problem 1129, common global minimax quantity.