Carleman's inequality

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

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:

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 a_n with a^1/p
_n, and letting p → ∞.