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Almost-additive sequences have a normalized limit

Statement

Let (an)n≥1 be a real sequence for which there is a finite constant K with ∣an∣≤Kn for every n≥1. Suppose there is a finite constant C≥0 such that ∣am+n−am−an∣≤C for all positive integers m,n. Then an/n has a finite limit as n→∞.

Statement references

Formal almost-additive result: Subsection “Almost additive sequences,” lemma almost_additive_converges.

Textbook Fekete proof: §5.1, Theorem 5.1.

Independent real-analysis treatment: Appendix B.1.3, Lemma B.5.

Facts & Assumptions

Given: The linear bound and the almost-additivity inequality in the Statement.

Proof

1.1givenalgebra

Set bn=an+C and cn=an−C. The upper half of the assumed inequality gives am+n≤am+an+C, hence bm+n≤bm+bn. The lower half gives am+n≥am+an−C, hence cm+n≥cm+cn. Also ∣bn∣,∣cn∣≤Kn+C, so both sequences are bounded linearly.

2.1step 1.1algebra

For a linearly bounded subadditive sequence b, fix k≥1 and write n=qk+r with 0≤r<k. Iterating subadditivity, with b0=0 for the remainder case, gives bn≤qbk+br. The finitely many possible remainder values have a finite maximum Bk, so bn/n≤qbk/n+Bk/n and therefore lim sup⁡nbn/n≤bk/k. Since this holds for every k and lim inf⁡nbn/n≥inf⁡kbk/k, the limit bn/n exists and is finite. Applying the same argument to −cn proves that cn/n has a finite limit.

3.1step 1.1step 2.1algebra∎

Since bn/n−cn/n=2C/n→0 and cn/n≤an/n≤bn/n, the two limits from step 2.1 coincide and force an/n to that same finite limit.

Used by

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Nothing. This result depends on no other item in the library.

Sources