Carleman's inequality explained

Carleman's inequality is an inequality in mathematics, named after Torsten Carleman, who proved it in 1923[1] and used it to prove the Denjoy - Carleman theorem on quasi-analytic classes.[2] [3]

Statement

Let

a1,a2,a3,...

be a sequence of non-negative real numbers, then
infty
\sum
n=1

\left(a1a2

1/n
a
n\right)

\lee

infty
\sum
n=1

an.

The constant

e

(euler number) 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

infty
\int
0

\exp\left\{

1
x
x
\int
0

lnf(t)dt\right\}dx\leqe

infty
\int
0

f(x)dx

for any f ≥ 0.

Carleson's inequality

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

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

infty
\int
0

xpe-g(x)/xdx\leqep+1

infty
\int
0

xpe-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 ⋅ a1,2 ⋅ a2,...,nan

MG(a1,...,an)=MG(1a1,2a2,...,na

-1/n
n)(n!)

\leMA(1a1,2a2,...,na

-1/n
n)(n!)

where MG stands for geometric mean, and MA - for arithmetic mean. The Stirling-type inequality

n!\ge\sqrt{2\pin}nne-n

applied to

n+1

implies

(n!)-1/n\le

e
n+1
for all

n\ge1.

Therefore,

MG(a1,...,an)\le

e
n(n+1)

\sum1\lekak,

whence

\sumn\ge1MG(a1,...,an)\lee\sumk\ge1(\sumn\ge

1
n(n+1)

)kak=e\sumk\ge1ak,

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

ak=C/k

for

k=1,...,n

. As a consequence, Carleman's inequality is never an equality for a convergent series, unless all

an

vanish, just because the harmonic series is divergent.

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

infty
\sum
n=1

\left(

a1+a2+ … +an
n

\right)p\le\left(

p
p-1

\right

infty
)
n=1
p
a
n

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

Versions for specific sequences

Christian Axler and Mehdi Hassani investigated Carleman's inequality for the specific cases of

ai=pi

where

pi

is the

i

th prime number. They also investigated the case where
a
i=1
pi
.[5] They found that if

ai=pi

one can replace

e

with
1
e
in Carleman's inequality, but that if
a
i=1
pi
then

e

remained the best possible constant.

Notes

  1. T. Carleman, Sur les fonctions quasi-analytiques, Conférences faites au cinquième congres des mathématiciens Scandinaves, Helsinki (1923), 181-196.
  2. 2040885. Duncan. John. McGregor. Colin M.. Carleman's inequality. Amer. Math. Monthly . 110. 2003. 5. 424 - 431. 10.2307/3647829.
  3. 1820809. Pečarić. Josip. Stolarsky. Kenneth B.. Carleman's inequality: history and new generalizations. Aequationes Mathematicae. 61. 2001. 1 - 2. 49 - 62. 10.1007/s000100050160.
  4. L.. Carleson. A proof of an inequality of Carleman. Proc. Amer. Math. Soc.. 5. 1954. 932 - 933. 10.1090/s0002-9939-1954-0065601-3. free.
  5. Christian Axler, Medhi Hassani . Carleman's Inequality over prime numbers . Integers . 21, Article A53 . 13 November 2022.

References