Consistent heuristic

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 a₁, a₂, a₃, ... be a sequence of non-negative real numbers, then

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

The constant e in the inequality is optimal, that is, the inequality does not always hold if 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

$\int _{0}^{\infty }\exp \left\{{\frac {1}{x}}\int _{0}^{x}\ln f(t)dt\right\}dx\leq e\int _{0}^{\infty }f(x)dx$

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 < ∞,

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

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 $1\cdot a_{1},2\cdot a_{2},\dots ,n\cdot a_{n}$

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

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

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

Therefore

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

whence

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

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.

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

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

for the non-negative numbers a₁,a₂,... and p > 1, replacing each a_n with aTemplate:Su, and letting p → ∞.

References

• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
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• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534

External links

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