# 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
    $\sum _{n=1}^{\infty }\left(a_{1}a_{2}\cdots a_{n}\right)^{1/n}\leq \mathrm {e} \sum _{n=1}^{\infty }a_{n}.$
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 < 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 $1\cdot a_{1},2\cdot a_{2},\dots ,n\cdot a_{n}$
    $\mathrm {MG} (a_{1},\dots ,a_{n})=\mathrm {MG} (1a_{1},2a_{2},\dots ,na_{n})(n!)^{-1/n}\leq \mathrm {MA} (1a_{1},2a_{2},\dots ,na_{n})(n!)^{-1/n}$
where MG stands for geometric mean, and MA — for arithmetic mean. The Stirling-type inequality $n!\geq {\sqrt {2\pi n}}\,n^{n}\mathrm {e} ^{-n}$ applied to $n+1$ implies
    $(n!)^{-1/n}\leq {\frac {\mathrm {e} }{n+1}}$ for all $n\geq 1.$
Therefore,
    $MG(a_{1},\dots ,a_{n})\leq {\frac {\mathrm {e} }{n(n+1)}}\,\sum _{1\leq k\leq n}ka_{k}\,,$
whence
    $\sum _{n\geq 1}MG(a_{1},\dots ,a_{n})\leq \,\mathrm {e} \,\sum _{k\geq 1}{\bigg\sum _{n\geq k}{\frac {1}{n(n+1)}}{\bigg )}\,ka_{k}=\,\mathrm {e} \,\sum _{k\geq 1}\,a_{k}\,,$
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$^{:§334}$
    $\sum _{n=1}^{\infty }\left({\frac {a_{1}+a_{2}+\cdots +a_{n}}{n}}\right)^{p}\leq \left({\frac {p}{p-1}}\right)^{p}\sum _{n=1}^{\infty }a_{n}^{p}$
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 $i$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.