Editing Openai/696b1675-eda0-800c-80f9-612b39635436 (section)

==== 1) Is ====
Υ
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≪
log
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Υ(X)≪logn for all 
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X?

Yes — and the right “reason” is that the max–min problem you wrote down is already a classical extremal problem in interpolation theory (Bernstein–Erdős). One formulation (in modern notation) is:
* Let M(X):=max⁡iλi(X)=max⁡x∈[−1,1]ΛX(x)M(X):=\max_i \lambda_i(X)=\max_{x\in[-1,1]}\Lambda_X(x)M(X):=maxiλi(X)=maxx∈[−1,1]ΛX(x) be the usual Lebesgue constant.
* Let m(X):=min⁡iλi(X)=Υ(X)m(X):=\min_i \lambda_i(X)=\Upsilon(X)m(X):=miniλi(X)=Υ(X) be the smallest local maximum.

Then there exists a distinguished point set X\''X^\''X\* (“Lebesgue–optimal nodes”) for which the local maxima are all equal (equioscillation), and for every node set XXX one has the sandwich inequality

m(X) ≤ m(X\'') = M(X\'') ≤ M(X),m(X)\ \le\ m(X^\'')\ =\ M(X^\'')\ \le\ M(X),m(X) ≤ m(X\'') = M(X\'') ≤ M(X),
i.e. the same set X\''X^\''X\* maximizes the smallest local maximum and minimizes the largest one. DNB Portal<ref>{{cite web|title=DNB Portal|url=https://d-nb.info/1090940718/34|publisher=DNB Portal|access-date=2026-01-18}}</ref>
Equivalently,

Υ(X)≤Υ(X\'')=M(X\'').\Upsilon(X)\le \Upsilon(X^\'') = M(X^\'').Υ(X)≤Υ(X\'')=M(X\'').
Now M(X\'')M(X^\'')M(X\*) is the optimal Lebesgue constant, and it is known to have logarithmic growth:

M(X\'')  =  2π(log⁡(n+1)+γ+log⁡4π)+o(1)(n→∞),M(X^\'') \;=\; \frac{2}{\pi}\Big(\log(n+1)+\gamma+\log\frac{4}{\pi}\Big)+o(1)
\qquad (n\to\infty),M(X\*)=π2(log(n+1)+γ+logπ4)+o(1)(n→∞),
in the standard indexing conventions (shifting n↦n−1n\mapsto n-1n↦n−1 just changes log⁡(n+1)\log(n+1)log(n+1) to log⁡n\log nlogn at this level). DNB Portal<ref>{{cite web|title=DNB Portal|url=https://d-nb.info/1090940718/34|publisher=DNB Portal|access-date=2026-01-18}}</ref>
Therefore,

Υ(X) ≤ Clog⁡n\Upsilon(X)\ \le\ C\log nΥ(X) ≤ Clogn
for an absolute constant CCC, uniformly over all XXX.

Moreover, this is sharp in order: at the extremal set X\''X^\''X\'', Υ(X\'')∼2πlog⁡n\Upsilon(X^\'')\sim \tfrac{2}{\pi}\log nΥ(X\'')∼π2logn.