Carleman's inequality

Carleman's inequality is an inequality in mathematics, named after Torsten Carleman, who proved it in 1923T. Carleman, Sur les fonctions quasi-analytiques, Conférences faites au cinquième congres des mathématiciens Scandinaves, Helsinki (1923), 181-196. and used it to prove the Denjoy–Carleman theorem on quasi-analytic classes.{{cite journal|mr=2040885|last1=Duncan|first1=John|last2=McGregor|first2=Colin M.|title=

Carleman's inequality|journal=Amer. Math. Monthly |volume=110|year=2003|issue=5|pages= 424–431|doi=10.2307/3647829}}{{cite journal|mr=1820809|last1=Pečarić|first1=Josip|last2=Stolarsky|first2=Kenneth B.|title=Carleman's inequality: history and new generalizations|journal=Aequationes Mathematicae|volume=61| year=2001|issue=1–2|pages=49–62|doi=10.1007/s000100050160}}

Statement

Let a_1,a_2,a_3,\dots be a sequence of non-negative real numbers, then

: \sum_{n=1}^\infty \left(a_1 a_2 \cdots a_n\right)^{1/n} \le \mathrm{e} \sum_{n=1}^\infty a_n.

The constant \mathrm{e} (euler number) in the inequality is optimal, that is, the inequality does not always hold if \mathrm{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

: \int_0^\infty \exp\left\{ \frac{1}{x} \int_0^x \ln f(t) \,\mathrm{d}t \right\} \,\mathrm{d}x \leq \mathrm{e} \int_0^\infty f(x) \,\mathrm{d}x

for any f ≥ 0.

Carleson's inequality

A generalisation, due to Lennart Carleson, states the following:{{cite journal|first=L.|last= Carleson|title=A proof of an inequality of Carleman|journal=Proc. Amer. Math. Soc.|volume=5|year=1954|pages=932–933|url=https://www.ams.org/journals/proc/1954-005-06/S0002-9939-1954-0065601-3/S0002-9939-1954-0065601-3.pdf|doi=10.1090/s0002-9939-1954-0065601-3|doi-access=free}}

for any convex function g with g(0) = 0, and for any -1 < p < ∞,

: \int_0^\infty x^p \mathrm{e}^{-g(x)/x} \,\mathrm{d}x \leq \mathrm{e}^{p+1} \int_0^\infty x^p \mathrm{e}^{-g'(x)} \,\mathrm{d}x.

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

Proof

=Direct proof=

An elementary proof is sketched below. From the inequality of arithmetic and geometric means applied to the numbers 1\cdot a_1,2\cdot a_2,\dots,n \cdot a_n

:\mathrm{MG}(a_1,\dots,a_n)=\mathrm{MG}(1a_1,2a_2,\dots,na_n)(n!)^{-1/n}\le \mathrm{MA}(1a_1,2a_2,\dots,na_n)(n!)^{-1/n}

where MG stands for geometric mean, and MA — for arithmetic mean. The Stirling-type inequality n!\ge \sqrt{2\pi n}\, n^n \mathrm{e}^{-n} applied to n+1 implies

:(n!)^{-1/n} \le \frac{\mathrm{e}}{n+1} for all n\ge1.

Therefore,

:MG(a_1,\dots,a_n) \le \frac{\mathrm{e}}{n(n+1)}\, \sum_{1\le k \le n} k a_k \, ,

whence

:\sum_{n\ge1}MG(a_1,\dots,a_n) \le\, \mathrm{e}\, \sum_{k\ge1} \bigg( \sum_{n\ge k} \frac{1}{n(n+1)}\bigg) \, k a_k =\, \mathrm{e}\, \sum_{k\ge1}\, a_k \, ,

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 a_k= C/k for k=1,\dots,n. As a consequence, Carleman's inequality is never an equality for a convergent series, unless all a_n vanish, just because the harmonic series is divergent.

=By Hardy’s inequality=

One can also prove Carleman's inequality by starting with Hardy's inequality{{cite book |last1=Hardy |first1=G. H. |last2=Littlewood |first2=J.E. |last3=Pólya |first3=G. |title=Inequalities |date=1952 |location=Cambridge, UK |edition=Second}}{{rp|§334}}

:\sum_{n=1}^\infty \left (\frac{a_1+a_2+\cdots +a_n}{n}\right )^p\le \left (\frac{p}{p-1}\right )^p\sum_{n=1}^\infty a_n^p

for the non-negative numbers a_1, a_2,… and p > 1, replacing each a_n with a_n^{1/p}, and letting p \to \infty.

Versions for specific sequences

Christian Axler and Mehdi Hassani investigated Carleman's inequality for the specific cases of a_i= p_i where p_i is the ith prime number. They also investigated the case where a_i=\frac{1}{p_i}.{{cite journal |last1=Christian Axler, Medhi Hassani |title=Carleman's Inequality over prime numbers |journal=Integers |volume=21, Article A53 |url=http://math.colgate.edu/~integers/v53/v53.pdf |access-date=13 November 2022}} They found that if a_i=p_i one can replace e with \frac{1}{e} in Carleman's inequality, but that if a_i=\frac{1}{p_i} then e remained the best possible constant.

Notes

References

  • {{cite book

| last = Hardy

| first = G. H. |author2=Littlewood J.E. |author3=Pólya, G.

| title = Inequalities, 2nd ed

| publisher = Cambridge University Press

| year = 1952

| pages =

| isbn = 0-521-35880-9

}}

  • {{cite book

| editor-last = Rassias

| editor-first = Thermistocles M.

| title = Survey on classical inequalities

| publisher = Kluwer Academic

| year = 2000

| pages =

| isbn = 0-7923-6483-X

}}

  • {{cite book

| last = Hörmander

| first = Lars

| title = The analysis of linear partial differential operators I: distribution theory and Fourier analysis, 2nd ed

| publisher = Springer

| year = 1990

| pages =

| isbn = 3-540-52343-X

}}