Carleman's inequality

Let ${\displaystyle a_{1},a_{2},a_{3},\dots }$ be a sequence of non-negative real numbers, then

${\displaystyle \sum _{n=1}^{\infty }\left(a_{1}a_{2}\cdots a_{n}\right)^{1/n}\leq \mathrm {e} \sum _{n=1}^{\infty }a_{n}.}$

The constant ${\displaystyle \mathrm {e} }$ (euler number) in the inequality is optimal, that is, the inequality does not always hold if ${\displaystyle \mathrm {e} }$ is replaced by a smaller number. The inequality is strict (it holds with "<" instead of "≤") if some element in the sequence is non-zero.

Carleman's inequality has an integral version, which states that

${\displaystyle \int _{0}^{\infty }\exp \left\{{\frac {1}{x}}\int _{0}^{x}\ln f(t)\,\mathrm {d} t\right\}\,\mathrm {d} x\leq \mathrm {e} \int _{0}^{\infty }f(x)\,\mathrm {d} x}$

for any f ≥ 0.

A generalisation, due to Lennart Carleson, states the following:[4]

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

${\displaystyle \int _{0}^{\infty }x^{p}\mathrm {e} ^{-g(x)/x}\,\mathrm {d} x\leq \mathrm {e} ^{p+1}\int _{0}^{\infty }x^{p}\mathrm {e} ^{-g'(x)}\,\mathrm {d} x.}$

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

An elementary proof is sketched below. From the inequality of arithmetic and geometric means applied to the numbers ${\displaystyle 1\cdot a_{1},2\cdot a_{2},\dots ,n\cdot a_{n}}$

${\displaystyle \mathrm {MG} (a_{1},\dots ,a_{n})=\mathrm {MG} (1a_{1},2a_{2},\dots ,na_{n})(n!)^{-1/n}\leq \mathrm {MA} (1a_{1},2a_{2},\dots ,na_{n})(n!)^{-1/n}}$

where MG stands for geometric mean, and MA — for arithmetic mean. The Stirling-type inequality ${\displaystyle n!\geq {\sqrt {2\pi n}}\,n^{n}\mathrm {e} ^{-n}}$ applied to ${\displaystyle n+1}$ implies

${\displaystyle (n!)^{-1/n}\leq {\frac {\mathrm {e} }{n+1}}}$ for all ${\displaystyle n\geq 1.}$

Therefore,

${\displaystyle MG(a_{1},\dots ,a_{n})\leq {\frac {\mathrm {e} }{n(n+1)}}\,\sum _{1\leq k\leq n}ka_{k}\,,}$

whence

${\displaystyle \sum _{n\geq 1}MG(a_{1},\dots ,a_{n})\leq \,\mathrm {e} \,\sum _{k\geq 1}{\bigg (}\sum _{n\geq k}{\frac {1}{n(n+1)}}{\bigg )}\,ka_{k}=\,\mathrm {e} \,\sum _{k\geq 1}\,a_{k}\,,}$

proving the inequality. Moreover, the inequality of arithmetic and geometric means of ${\displaystyle n}$ 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 ${\displaystyle a_{k}=C/k}$ for ${\displaystyle k=1,\dots ,n}$. As a consequence, Carleman's inequality is never an equality for a convergent series, unless all ${\displaystyle a_{n}}$ vanish, just because the harmonic series is divergent.

One can also prove Carleman's inequality by starting with Hardy's inequality

${\displaystyle \sum _{n=1}^{\infty }\left({\frac {a_{1}+a_{2}+\cdots +a_{n}}{n}}\right)^{p}\leq \left({\frac {p}{p-1}}\right)^{p}\sum _{n=1}^{\infty }a_{n}^{p}}$

for the non-negative numbers a₁,a₂,... and p > 1, replacing each a_n with a^1/p
_n, and letting p → ∞.

Christian Axler and Mehdi Hassani investigated Carleman's inequality for the specific cases of ${\displaystyle a_{i}=p_{i}}$ where ${\displaystyle p_{i}}$ is the ${\displaystyle i}$th prime number. They also investigated the case where ${\displaystyle a_{i}={\frac {1}{p_{i}}}}$.[5] They found that if ${\displaystyle a_{i}=p_{i}}$ one can replace ${\displaystyle e}$ with ${\displaystyle {\frac {1}{e}}}$ in Carleman's inequality, but that if ${\displaystyle a_{i}={\frac {1}{p_{i}}}}$ then ${\displaystyle e}$ remained the best possible constant.

• Hardy, G. H.; Littlewood J.E.; Pólya, G. (1952). Inequalities, 2nd ed. Cambridge University Press. ISBN 0-521-35880-9.
• Rassias, Thermistocles M., ed. (2000). Survey on classical inequalities. Kluwer Academic. ISBN 0-7923-6483-X.
• Hörmander, Lars (1990). The analysis of linear partial differential operators I: distribution theory and Fourier analysis, 2nd ed. Springer. ISBN 3-540-52343-X.

• "Carleman inequality", Encyclopedia of Mathematics, EMS Press, 2001 [1994]