APPLICATIONS OF TAYLOR SERIES FOR

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CARLEMAN'S INEQUALITY THROUGH HARDY. INEQUALITY. Mohammed Muniru Iddrisu and Christopher Adjei Okpoti. Abstract. In this paper, we prove the ...
Korean J. Math. 23 (2015), No. 4, pp. 655–664 http://dx.doi.org/10.11568/kjm.2015.23.4.655

APPLICATIONS OF TAYLOR SERIES FOR CARLEMAN’S INEQUALITY THROUGH HARDY INEQUALITY Mohammed Muniru Iddrisu and Christopher Adjei Okpoti Abstract. In this paper, we prove the discrete Hardy inequality through the continuous case for decreasing functions using elementary properties of calculus. Also, we prove the Carleman’s inequality through limiting the discrete Hardy inequality with applications of Taylor series.

1. Introduction G. H. Hardy in [4] established the following discrete inequality: !p  p X n ∞ ∞ X p 1X ak ≤ apn , (1) n k=1 p − 1 n=1 n=1  p p where p > 1, ak ≥ 0 and the constant p−1 is the best possible (see also [3], p. 239). The continuous analogue of (1) is given as p  p Z ∞ Z ∞ Z x 1 p (2) f (t)dt dx ≤ f p (x)dx, x 0 p−1 0 0 where p > 1, x > 0, is a nonnegative measurable function on (0, ∞)  f  p p and the constant p−1 is the best possible. The inequality (2) is Received August 24, 2015. Revised October 10, 2015. Accepted October 15, 2015. 2010 Mathematics Subject Classification: 26D15. Key words and phrases: Hardy inequality, Carleman’s inequality, Applications, Taylor series. c The Kangwon-Kyungki Mathematical Society, 2015.

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Iddrisu, M.M. and Okpoti, C.A.

sometimes written as  p Z ∞ Z ∞ p p F (x)dx ≤ (3) f p (x)dx, p−1 0 0 R x where 0 < F (x) = x1 0 f (t)dt < ∞ and f > 0. These interesting results (1) to (3) are very popular in the research environment and are usually called the classical Hardy inequalities (see also [1], [2], [3], [7], [9], [10], [11], [12] and the references therein.) Let us also consider the inequality √ √ (4) a1 + a1 a2 + · · · + n a1 a2 · · · an < e(a1 + a2 + · · · ), P where a1 , a2 · · · , an are positive numbers and ∞ j=1 aj is convergent. This inequality (4) is due to a Swedish mathematician called Torsten Carleman who discovered it in 1922 (see [3], p.249). In order to agree further with T. Carleman, other Mathematicians also proved the inequality (4) by different methods: Thus by differentiation and the variations of the Arithmetic (An )–Geometric (Gn ) mean inequality (i.e. Gn ≤ An ) methods (See [5], [6] [13], [14], [15] and the references therein). The aim of this paper is first to provide a simple proof of the discrete Hardy inequality through the continuous case and then recover the Carleman’s inequality (4) through limiting the discrete Hardy inequality with applications of Taylor series. This approach here for the prove of the Carleman’s inequality is mainly to demonstrate the use of Taylor series in the evaluation of large expressions. 2. Results and Discussions We begin as follows: Theorem 2.1. Let p > 1 and ak be a non-increasing sequence of positive real numbers, then  p X ∞ ∞ X p p Ak ≤ apk , p − 1 k=1 k=1  p P p where Ak = k1 kj=1 aj and the constant p−1 is the best possible.

Applications of Taylor Series for Carleman’s Ineq. through Hardy Ineq.

657

Proof. Consider the non-increasing sequence ak . Let f (x) = ak on [k − 1, k]. From the integral inequality (3) we have p Z k ∞ Z k ∞  X X p p (5) F (x)dx ≤ f p (x)dx p − 1 k−1 k=1 k−1 k=1 where

Z 1 x F (x) = f (t)dt. x 0 Since the function F (x) is non-increasing, it follows that F (k) ≤ F (x) ≤ F (k − 1).

Thus Z

k

k

Z

p

F (k)dx ≤ k−1

Z

p

k

F p (k − 1)dx.

F (x)dx ≤ k−1

k−1

This simplifies to Z

p

k

F p (x)dx ≤ F p (k − 1)

F (k) ≤ k−1

Thus (6)

∞ X

p

F (k) ≤

k=1

∞ Z X

k p

F (x)dx ≤

k−1

k=1

∞ X

F p (k − 1).

k=1

Considering the first inequality of (6) and then applying (5), we obtain ∞ Z k ∞ X X p F p (x)dx F (k) ≤ k=1 k−1 ∞  X

k=1

p Z k p ≤ f p (x)dx p − 1 k−1 k=1  p X ∞ p ≤ ap . p − 1 k=1 k Putting F (k) = Ak , we get ∞ X k=1

Apk

 ≤

p p−1

p X ∞ k=1

apk .

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Iddrisu, M.M. and Okpoti, C.A.

Theorem 2.1. Let ak be a sequence of distinct positive real numbers not necessarily non-increasing, then p X  ∞ ∞ X p p apk (7) Ak ≤ p − 1 k=1 k=1 holds for p > 1. Proof. The inequality (7) is valid for non-increasing sequence {ak } P P p as in Theorem 2.1, where Ak = k1 kn=1 an . Assume that ∞ k=1 ak < ∞ which implies ak → 0. Now let us consider re-arrangement of the sequence {ak } by setting sup{ak } = max{ak } and considering the sub∗ ∗ ∞ sequence akj such that {akj }∞ j=1 = {aj }j=1 . Denote by ({ak } \ {aj }), the difference between the two sequences. Thus a∗1 = max({ak }) = ak1 a∗2 = max({ak } \ {a∗1 }) = ak2 a∗3 = max({ak } \ {a∗1 , a∗2 }) = ak3 .. . a∗n = max({ak } \ {a∗1 , a∗2 , . . . , a∗(n−1) } = akn .. . Thus a∗1 = ak1 ≥ ak2 ≥ ak3 ≥ · · · ≥ akn = a∗n ≥ . . . This shows that the sequence {a∗j }∞ j=1 is non-increasing. Hence n X k=1

and

while

ak ≤

n X

a∗j



j=1 ∞ X

(a∗j )p

∞ X

a∗j

=

j=1

=

∞ X

ak

k=1

∞ X

j=1

k=1

n

n

apk

1X 1X ∗ ak ≤ a n k=1 n j=1 j for n > 0. increasing.

Thus (7) is valid for sequence ak not necessarily non-

Applications of Taylor Series for Carleman’s Ineq. through Hardy Ineq.

659

Let us now establish some results for the limiting Hardy inequality. First, we present for the case of two positive numbers followed by a generalization. Theorem 2.2. Let {ak } be a sequence of distinct positive real numbers, then !p 2 1 √ 1X = a1 a2 . lim (ak ) p p→+∞ 2 k=1 Proof. Let 2

Sp =

1 1X (ak ) p 2 k=1

!p

1

1

(a1 ) p + (a2 ) p 2

=

!p .

Applying logarithm to base e to both sides and using x = eln x , we obtain 1

1

ln Sp = p ln = p ln

(a1 ) p + (a2 ) p 2 e

1 p

ln a1

+e 2

1 p

!

ln a2

! .

Now consider the Taylor expansion   1 x x2 x 2 e = 1+x+x + + + ... 2 3! 4! (8) = 1 + x + x2 f (x)   2 where f (x) = 12 + 3!x + x4! + . . . → 21 when x → 0. Applying the Taylor series (8) to ln Sp , we obtain " # 2 1 1 1 1 X 2 ln Sp = p ln 1 + 1 + (ln a1 + ln a2 ) + 2 (ln ak ) f ( ln ak ) 2 p p k=1 p   1 p = p ln 1 + ln(a1 a2 ) + 2p p   p 1 √ = p ln 1 + ln a1 a2 + p p = p ln(1 + up )

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Iddrisu, M.M. and Okpoti, C.A.

P2 √ 1 1 2 where up = p1 ln a1 a2 + pp and p = 2p k=1 (ln ak ) f ( p ln ak ) for which p → 0 as p → ∞ since p1 → 0 as p → ∞. Also consider the Taylor expansion u2 u3 u4 + − + ... ln(1 + u) = u − 3 4  2  u u2 u3 = u 1− + − + ... 2 3 4 (9) = ug(u)   2 3 where g(u) = 1 − u2 + u3 − u4 + . . . → 1 as u → 0. Applying (9), we find that ln Sp = p ln(1 + up ) = pup g(up ) p = (ln (a1 a2 ) + p ).g(up ). As p → ∞, p → 0 and up → 0 implies g(up ) → 1. Hence √ lim Sp = a1 a2 . p→+∞

Now we consider the general case: Theorem 2.3. Let {ak } be a sequence of distinct positive real numbers, then !p n 1 1 1X lim (ak ) p = (a1 a2 . . . an ) n . p→+∞ n k=1 Equivalently, n

lim

p→+∞

1 1X (ak ) p n k=1

!p

n

= exp

1X log ak n k=1

! .

Proof. Let n

Vp =

1 1X (ak ) p n k=1

!p

1

=

1

1

(a1 ) p + (a2 ) p + · · · + (an ) p n

!p .

Applications of Taylor Series for Carleman’s Ineq. through Hardy Ineq.

661

Applying t = exp ln t, we get

ln Vp = p ln

exp( p1 ln a1 ) + exp( p1 ln a2 ) + · · · + exp( p1 ln an ) n

Applying ex = 1 + x + x2 f (x) where f (x) →

ln Vp

!

1 2

when x → 0, we obtain

" # n 1 1 1 X 1 = p ln n + ln(a1 a2 . . . an ) + 2 (ln ak )2 f ( ln ak ) n p p k=1 p   1 hp = p ln 1 + ln(a1 a2 . . . an ) + np p = p ln(1 + mp )

1 1 ln(a1 a2 . . . an ) + hpp and hp = np where mp = np 0 as p → ∞. Applying the Taylor expansion

Pn

1 2 k=1 (ln ak ) f ( p

  m m2 m3 ln(1 + m) = m 1 − + − + . . . = mg(m) 2 3 4  where g(m) = 1 −

m 2

+

m2 3



m3 4

+ ...



→ 1 as m → 0. Thus

ln Vp = pmp g(mp )   1 hp ln(a1 a2 . . . an ) + = p g(mp ) np p i h 1 = ln(a1 a2 . . . an ) n + hp g(mp ) As p → ∞, hp → 0, mp → 0 and g(mp ) → 1. Hence 1

lim Vp = (a1 a2 . . . an ) n

p→+∞

as required.

ln ak ) →

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Iddrisu, M.M. and Okpoti, C.A.

Remark 1. By Theorem 2.3, we have !p n X 1 1 1 = (a1 a2 . . . an ) n lim (ak ) p p→+∞ n k=1 = exp

n Y 1 log( ak ) n k=1

1 (log a1 + log a2 + · · · + log an ) n ! n 1X = exp log ak . n k=1

= exp

Theorem 2.4. Let {ak } be a sequence of distinct positive real numbers, then ! ∞ n ∞ X X 1X (10) exp log ak ≤ e an . n n=1 n=1 k=1 1

Proof. Replace an with (an ) p in inequality (1). Then !p  p X ∞ n ∞ X 1 1X p (ak ) p an ≤ n p − 1 n=1 n=1 k=1 Take limits on both sides. Thus (11)

∞ X

n

exp

n=1

1X log ak n k=1

! ≤e

∞ X n=1

since  lim

p→∞

p p−1

p =e

and by application of Remark 1. Equivalently (12)

∞ ∞ X X 1 (a1 a2 . . . an ) n ≤ e an n=1

n=1

which is the well known Carleman’s inequality.

an

Applications of Taylor Series for Carleman’s Ineq. through Hardy Ineq.

663

3. Conclusion We proved the discrete Hardy inequality through the integral Hardy inequality for decreasing functions. We also established the well known Carleman’s inequality through limiting Discrete Hardy inequality with applications of Taylor series.

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[15] J. Peˇcari´c and K.B. Stolarsky, Carleman’s inequality: History and new generalizations, Aequat. Math. 61 (2001), 49–62.

Mohammed Muniru Iddrisu Department of Mathematics University for Development Studies, P. O. Box 24, Navrongo campus, Navrongo, Ghana E-mail : [email protected] or [email protected] Christopher Adjei Okpoti Department of Mathematics University of Education Winneba, Ghana E-mail : [email protected]