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Monotonicity of the Fekete diameters and the transfinite diameter
Statement
Let be nonempty and compact, with the Fekete diameters and transfinite diameter of Fekete points and the transfinite diameter of a compact set. Then
the sequence therefore converges, and
No choice principle is used.
Facts & Assumptions
Given: a nonempty compact and the quantities , , , and Fekete tuples of Fekete points and the transfinite diameter of a compact set.
For and one has and ; the maximum of over the nonempty compact is attained and ; and (Fekete points and the transfinite diameter of a compact set). In particular for every , because is increasing on and is the reciprocal of .
and for all , and exactly for (Conjugation is an involutive real-field automorphism, , and modulus is definite, multiplicative, and subadditive).
Every has a unique -th root , the map is strictly increasing on for , and for one has exactly when (Existence and uniqueness of -th roots: a unique with , Monotonicity of and of ).
A monotone sequence of reals converges if and only if it is bounded (A monotone sequence converges if and only if it is bounded).
If a sequence of reals converges to and for all , then ; if for all , then (Limits preserve non-strict inequalities).
If is nonempty and bounded below, then is a lower bound of and no larger lower bound exists; in particular a lower bound of equals when every lower bound is (Greatest lower bound (infimum), Every nonempty set bounded below has an infimum).
Proof
Fix and a tuple . For let be the -tuple obtained by deleting the -th entry of . Each lies in , so by [F1] is at most ; that is, .
Expanding the factors by [F1] and [F2] gives ; the unordered pair contributes the factor for exactly those , that is for of the indices , so the last product equals , the last equality again by [F1] and [F2].
Multiplying the inequalities of step 1.1 and substituting step 2.1 yields for every . Taking the maximum over and using that is increasing on together with [F3] gives , since is the exponent obtained from . The common exponent is a positive integer (as ), so [F3] applied to the two nonnegative numbers and gives .
The sequence is nonincreasing by step 3.1 and bounded below by because by [F1]; hence it converges, with limit , by [F4]. Since for all , [F5] applied to the tail from gives for every , so is a lower bound of ; and if is any lower bound of that set, then for every and [F5] gives . Thus is the greatest lower bound and by [F6] and [F1].
Combining steps 3.1 and 4.1, for every and , which is the statement.
Remarks
Where the normalization enters. The exponent in the definition of is exactly what makes the exponents on the two sides of step 3.1 agree: the pair-count of the -tuple side equals the pair-count of the -tuple side, both equal to .
Choice. The argument uses only real algebra and order-completeness facts; no choice principle is involved, and the extremal tuples are maxima of continuous functions on compact product spaces supplied by Fekete points and the transfinite diameter of a compact set.
Depends on
- Fekete points and the transfinite diameter of a compact set
- Conjugation is an involutive real-field automorphism, $z\overline z=|z|^2$, and modulus is definite, multiplicative, and subadditive
- Existence and uniqueness of $n$-th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$
- Monotonicity of $x \mapsto x^n$ and of $n \mapsto a^n$
- A monotone sequence converges if and only if it is bounded
- Limits preserve non-strict inequalities
- Greatest lower bound (infimum)
- Every nonempty set bounded below has an infimum
Used by
Dependency tree · two levels
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Sources
- E. B. Saff, Logarithmic Potential Theory with Applications to Approximation Theory, §1 (standard reference, not scraped)
- B. Khoruzhenko, LTCC Potential Theory notes, §5 (standard reference, not scraped)