Carleman's inequality

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

Statement

Let a1, a2, a3, ... 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 e \sum_{n=1}^\infty a_n.

The constant e 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

 \int_0^\infty \exp\left\{ \frac{1}{x} \int_0^x \ln f(t) dt \right\} dx \leq e \int_0^\infty f(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 < ,

 \int_0^\infty x^p e^{-g(x)/x} dx \leq e^{p+1} \int_0^\infty x^p e^{-g'(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 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 e^{-n} applied to n+1 implies

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

Therefore

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

whence

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

One can also prove Carleman's inequality by starting with Hardy's inequality

\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 a1,a2,... and p > 1, replacing each an with a1/p
n
, and letting p  .

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): 424431. doi:10.2307/3647829. MR 2040885.
  3. Pečarić, Josip; Stolarsky, Kenneth B. (2001). "Carleman's inequality: history and new generalizations". Aequationes Math. 61 (12): 4962. doi:10.1007/s000100050160. MR 1820809.
  4. Carleson, L. (1954). "A proof of an inequality of Carleman" (PDF). Proc. Amer. Math. Soc. 5: 932933. doi:10.1090/s0002-9939-1954-0065601-3.

References

External links

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