Carleman's inequality

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Carleman's inequality is an inequality in mathematics, named after Torsten Carleman, who proved it in 1923[1] and used it to prove the Denjoy–Carleman theorem on quasi-analytic classes.[2][3]

Statement

Let a1,a2,a3, be a sequence of non-negative real numbers, then

n=1(a1a2an)1/nen=1an.

The constant e (euler number) in the inequality is optimal, that is, the inequality does not always hold if e is replaced by a smaller number. The inequality is strict (it holds with "<" instead of "≤") if some element in the sequence is non-zero.

Integral version

Carleman's inequality has an integral version, which states that

0exp{1x0xlnf(t)dt}dxe0f(x)dx

for any f ≥ 0.

Carleson's inequality

A generalisation, due to Lennart Carleson, states the following:[4] for any convex function g with g(0) = 0, and for any -1 < p < ∞,

0xpeg(x)/xdxep+10xpeg(x)dx.

Carleman's inequality follows from the case p = 0.

Proof

An elementary proof is sketched below. From the inequality of arithmetic and geometric means applied to the numbers 1a1,2a2,,nan

MG(a1,,an)=MG(1a1,2a2,,nan)(n!)1/nMA(1a1,2a2,,nan)(n!)1/n

where MG stands for geometric mean, and MA — for arithmetic mean. The Stirling-type inequality n!2πnnnen applied to n+1 implies

(n!)1/nen+1 for all n1.

Therefore,

MG(a1,,an)en(n+1)1knkak,

whence

n1MG(a1,,an)ek1(nk1n(n+1))kak=ek1ak,

proving the inequality. Moreover, the inequality of arithmetic and geometric means of n non-negative numbers is known to be an equality if and only if all the numbers coincide, that is, in the present case, if and only if ak=C/k for k=1,,n. As a consequence, Carleman's inequality is never an equality for a convergent series, unless all an vanish, just because the harmonic series is divergent. One can also prove Carleman's inequality by starting with Hardy's inequality

n=1(a1+a2++ann)p(pp1)pn=1anp

for the non-negative numbers a1,a2,... and p > 1, replacing each an with a1/p
n
, and letting p → ∞.

Versions for specific sequences

Christian Axler and Mehdi Hassani investigated Carleman's inequality for the specific cases of ai=pi where pi is the ith prime number. They also investigated the case where ai=1pi.[5] They found that if ai=pi one can replace e with 1e in Carleman's inequality, but that if ai=1pi then e remained the best possible constant.

Notes

  1. T. Carleman, Sur les fonctions quasi-analytiques, Conférences faites au cinquième congres des mathématiciens Scandinaves, Helsinki (1923), 181-196.
  2. Duncan, John; McGregor, Colin M. (2003). "Carleman's inequality". Amer. Math. Monthly. 110 (5): 424–431. doi:10.2307/3647829. MR 2040885.
  3. Pečarić, Josip; Stolarsky, Kenneth B. (2001). "Carleman's inequality: history and new generalizations". Aequationes Mathematicae. 61 (1–2): 49–62. doi:10.1007/s000100050160. MR 1820809.
  4. Carleson, L. (1954). "A proof of an inequality of Carleman" (PDF). Proc. Amer. Math. Soc. 5: 932–933. doi:10.1090/s0002-9939-1954-0065601-3.
  5. Christian Axler, Medhi Hassani. "Carleman's Inequality over prime numbers" (PDF). Integers. 21, Article A53. Retrieved 13 November 2022.

References

  • Hardy, G. H.; Littlewood J.E.; Pólya, G. (1952). Inequalities, 2nd ed. Cambridge University Press. ISBN 0-521-35880-9.
  • Rassias, Thermistocles M., ed. (2000). Survey on classical inequalities. Kluwer Academic. ISBN 0-7923-6483-X.
  • Hörmander, Lars (1990). The analysis of linear partial differential operators I: distribution theory and Fourier analysis, 2nd ed. Springer. ISBN 3-540-52343-X.

External links