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

$\sum _{n=1}^{\mathrm{\infty }}{(a_1{a}_2\cdots a_n)}^{1/n}\le \mathrm{e}\sum _{n=1}^{\mathrm{\infty }}{a}_n.$

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\exp \{\frac{1}{x}\underset{0}{\overset{x}{\int }}\ln f(t)\mathrm{d}t\}\mathrm{d}x\le \mathrm{e}\underset{0}{\overset{\mathrm{\infty }}{\int }}f(x)\mathrm{d}x$

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 < ∞,

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^p{\mathrm{e}}^{-g(x)/x}\mathrm{d}x\le {\mathrm{e}}^{p+1}\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^p{\mathrm{e}}^{-g^{\prime }(x)}\mathrm{d}x.$

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 $1\cdot a_1,2\cdot a_2,\dots ,n\cdot a_n$

$\mathrm{M}\mathrm{G}(a_1,\dots ,a_n)=\mathrm{M}\mathrm{G}(1a_1,2a_2,\dots ,n{a}_n)(n!{)}^{-1/n}\le \mathrm{M}\mathrm{A}(1a_1,2a_2,\dots ,n{a}_n)(n!{)}^{-1/n}$

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 $n\ge 1.$

Therefore,

$MG(a_1,\dots ,a_n)\le \frac{\mathrm{e}}{n(n+1)}\sum _{1\le k\le n}k{a}_k,$

whence

$\sum _{n\ge 1}MG(a_1,\dots ,a_n)\le \mathrm{e}\sum _{k\ge 1}\left(\sum _{n\ge k}\frac{1}{n(n+1)}\right)k{a}_k=\mathrm{e}\sum _{k\ge 1}{a}_k,$

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 inequalityTemplate:Rp

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{a_1+a_2+\cdots +a_n}{n}\right)}^p\le {\left(\frac{p}{p-1}\right)}^p\sum _{n=1}^{\mathrm{\infty }}{a}_n^p$

for the non-negative numbers $a_1$, $a_2$,… and $p>1$, replacing each $a_n$ with $a_n^{1/p}$, and letting $p\to \mathrm{\infty }$.

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.

References

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External links

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