Logarithmic bounds from a consecutive pair of nodes

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Source. Bernstein 1931, equations (29), (31), and (31 bis), printed pp. 1038--1039 / PDF pp. 14--15, in the complete source.

Use the distinct real nodes and nodal polynomial $A$ of the interpolation identity. For consecutive nodes $a=a_k<b=a_{k+1}$, let $\delta=b-a$, $m=(a+b)/2$, and, for $x\notin[a,b]$, define

$$I_k(x)=|A(m)| \left(\frac1{|x-a|\,|A'(a)|} +\frac1{|x-b|\,|A'(b)|}\right). \tag{29}$$

Then

$$I_k(x)>\frac12\log\frac{b-x}{a-x}\quad(x<a), \qquad I_k(x)>\frac12\log\frac{x-a}{x-b}\quad(x>b). \tag{31, 31 bis}$$

All logarithms are natural, and every displayed ratio is positive.

Proof. Suppose first that $x<a$, and set $r=\sqrt{|A'(b)|/|A'(a)|}>0$. The midpoint inequality and the arithmetic-geometric mean inequality give

$$I_k(x)\ge\frac{\delta}{4} \left(\frac r{a-x}+\frac{r^{-1}}{b-x}\right) \ge\frac{\delta}{2\sqrt{(a-x)(b-x)}}.$$

Writing $z=\delta/(a-x)>0$, the last expression is $z/(2\sqrt{1+z})$. For $t=\tfrac12\log(1+z)>0$,

$$\frac z{\sqrt{1+z}}=2\sinh t>2t=\log(1+z).$$

Here $\sinh t>t$ follows by integrating $\cosh s>1$ over $0<s<t$. This proves the first bound. Reversing the real line proves the second, or the same computation applies with $z=\delta/(x-b)$.

Equality and excluded points. The midpoint step can be an equality only for a quadratic nodal polynomial. The arithmetic-geometric mean step is an equality, on the left, exactly when $|A'(b)|/|A'(a)|=(a-x)/(b-x)$, and analogously on the right. The final hyperbolic inequality is strict for every positive gap and finite exterior $x$, so equality never occurs in (31) or (31 bis). At $x=a$ or $x=b$, $I_k$ has a zero denominator and these formulas are not used.

Dependencies. Equation (27), arithmetic-geometric mean, and the elementary hyperbolic identity proved above.

Proof scope. Complete rewritten proof; independently reviewed on 6 September 2026 (component C3 of the local-chain review).

Bears on. Problem 1153, local lower bound.