Carleman's inequality

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

∑ n = 1 ∞ ( a 1 a 2 ⋯ a n ) 1 / n ≤ e ∑ n = 1 ∞ a n . {\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}.} {\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 e {\displaystyle \mathrm {e} } {\displaystyle \mathrm {e} } (euler number) in the inequality is optimal, that is, the inequality does not always hold if e {\displaystyle \mathrm {e} } {\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.

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 {\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} {\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.

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 < ∞,

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

Proof

Direct 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 {\displaystyle 1\cdot a_{1},2\cdot a_{2},\dots ,n\cdot a_{n}} {\displaystyle 1\cdot a_{1},2\cdot a_{2},\dots ,n\cdot 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 {\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}} {\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 n ! ≥ 2 π n n n e − n {\displaystyle n!\geq {\sqrt {2\pi n}}\,n^{n}\mathrm {e} ^{-n}} {\displaystyle n!\geq {\sqrt {2\pi n}}\,n^{n}\mathrm {e} ^{-n}} applied to n + 1 {\displaystyle n+1} {\displaystyle n+1} implies

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

Therefore,

M G ( a 1 , … , a n ) ≤ e n ( n + 1 ) ∑ 1 ≤ k ≤ n k a k , {\displaystyle MG(a_{1},\dots ,a_{n})\leq {\frac {\mathrm {e} }{n(n+1)}}\,\sum _{1\leq k\leq n}ka_{k}\,,} {\displaystyle MG(a_{1},\dots ,a_{n})\leq {\frac {\mathrm {e} }{n(n+1)}}\,\sum _{1\leq k\leq n}ka_{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 , {\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}\,,} {\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 n {\displaystyle n} {\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 a k = C / k {\displaystyle a_{k}=C/k} {\displaystyle a_{k}=C/k} for k = 1 , … , n {\displaystyle k=1,\dots ,n} {\displaystyle k=1,\dots ,n}. As a consequence, Carleman's inequality is never an equality for a convergent series, unless all a n {\displaystyle a_{n}} {\displaystyle 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 inequality[5]: §334 

∑ n = 1 ∞ ( a 1 + a 2 + ⋯ + a n n ) p ≤ ( p p − 1 ) p ∑ n = 1 ∞ a n p {\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}} {\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 a 1 {\displaystyle a_{1}} {\displaystyle a_{1}}, a 2 {\displaystyle a_{2}} {\displaystyle a_{2}},… and p > 1 {\displaystyle p>1} {\displaystyle p>1}, replacing each a n {\displaystyle a_{n}} {\displaystyle a_{n}} with a n 1 / p {\displaystyle a_{n}^{1/p}} {\displaystyle a_{n}^{1/p}}, and letting p → ∞ {\displaystyle p\to \infty } {\displaystyle p\to \infty }.

Versions for specific sequences

Christian Axler and Mehdi Hassani investigated Carleman's inequality for the specific cases of a i = p i {\displaystyle a_{i}=p_{i}} {\displaystyle a_{i}=p_{i}} where p i {\displaystyle p_{i}} {\displaystyle p_{i}} is the i {\displaystyle i} {\displaystyle i}th prime number. They also investigated the case where a i = 1 p i {\displaystyle a_{i}={\frac {1}{p_{i}}}} {\displaystyle a_{i}={\frac {1}{p_{i}}}}.[6] They found that if a i = p i {\displaystyle a_{i}=p_{i}} {\displaystyle a_{i}=p_{i}} one can replace e {\displaystyle e} {\displaystyle e} with 1 e {\displaystyle {\frac {1}{e}}} {\displaystyle {\frac {1}{e}}} in Carleman's inequality, but that if a i = 1 p i {\displaystyle a_{i}={\frac {1}{p_{i}}}} {\displaystyle a_{i}={\frac {1}{p_{i}}}} then e {\displaystyle e} {\displaystyle 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. Duncan, John; McGregor, Colin M. (2003). "Carleman's inequality". Amer. Math. Monthly. 110 (5): 424–431. doi:10.2307/3647829. MR 2040885.
  3. Pečarić, Josip; Stolarsky, Kenneth B. (2001). "Carleman's inequality: history and new generalizations". Aequationes Mathematicae. 61 (1–2): 49–62. doi:10.1007/s000100050160. MR 1820809.
  4. Carleson, L. (1954). "A proof of an inequality of Carleman" (PDF). Proc. Amer. Math. Soc. 5: 932–933. doi:10.1090/s0002-9939-1954-0065601-3.
  5. Hardy, G. H.; Littlewood, J.E.; Pólya, G. (1952). Inequalities (Second ed.). Cambridge, UK.{{cite book}}: CS1 maint: location missing publisher (link)
  6. Christian Axler, Medhi Hassani. "Carleman's Inequality over prime numbers" (PDF). Integers. 21, Article A53. Retrieved 13 November 2022.

References

  • Hardy, G. H.; Littlewood J.E.; Pólya, G. (1952). Inequalities, 2nd ed. Cambridge University Press. ISBN 0-521-35880-9. {{cite book}}: ISBN / Date incompatibility (help)
  • 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.