Trisha Shetty (Editor)

Carleman's inequality

Updated on
Edit
Like
Comment
Share on FacebookTweet on TwitterShare on LinkedInShare on Reddit

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.

Contents

Statement

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

∑ n = 1 ∞ ( a 1 a 2 ⋯ a n ) 1 / n ≤ e ∑ n = 1 ∞ 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

∫ 0 ∞ exp ⁡ { 1 x ∫ 0 x ln ⁡ f ( t ) d t } d x ≤ e ∫ 0 ∞ f ( x ) 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 < ∞,

∫ 0 ∞ x p e − g ( x ) / x d x ≤ e p + 1 ∫ 0 ∞ x p e − g ′ ( x ) d x .

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 ⋅ a 1 , 2 ⋅ a 2 , … , n ⋅ a n

M G ( a 1 , … , a n ) = M G ( 1 a 1 , 2 a 2 , … , n a n ) ( n ! ) − 1 / n ≤ M A ( 1 a 1 , 2 a 2 , … , n a n ) ( n ! ) − 1 / n

where MG stands for geometric mean, and MA — for arithmetic mean. The Stirling-type inequality n ! ≥ 2 π n n n e − n applied to n + 1 implies

( n ! ) − 1 / n ≤ e n + 1 for all n ≥ 1.

Therefore,

M G ( a 1 , … , a n ) ≤ e n ( n + 1 ) ∑ 1 ≤ k ≤ n k a k ,

whence

∑ n ≥ 1 M G ( a 1 , … , a n ) ≤ e ∑ k ≥ 1 ( ∑ n ≥ k 1 n ( n + 1 ) ) k a k = e ∑ k ≥ 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 , … , 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

∑ n = 1 ∞ ( a 1 + a 2 + ⋯ + a n n ) p ≤ ( p p − 1 ) p ∑ n = 1 ∞ a n p

for the non-negative numbers a1,a2,... and p > 1, replacing each an with a1/p
n, and letting p → ∞.

References

Carleman's inequality Wikipedia
(Text) CC BY-SA


Similar Topics