Carleman's inequality

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: for any convex function g with g(0) = 0, and for any -1  \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 :\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}. 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==