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 $$a_1,a_2,a_3,\dots$$ 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 "&lt;" instead of "&le;") 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:

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


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

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

Proof
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 &mdash; 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.

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 a$1/p n$, and letting p &rarr; &infin;.

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 $$i$$th prime number. They also investigated the case where $$a_i=\frac{1}{p_i}$$. 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.