R E S E A R C H
Open Access
A note on Hardy’s inequality
Yu-Ming Chu
1*, Qian Xu
2and Xiao-Ming Zhang
3*Correspondence:
chuyuming2005@126.com 1School of Mathematics and
Computation Science, Hunan City University, Yiyang, 413000, China Full list of author information is available at the end of the article
Abstract
In this paper, we prove that the inequality ∞
n=1(1n n
k=1ak)p≤(
p p–1)
p∞ n=1(1 –
d(p) (n–1/2)1–1/p)a
p
nholds forp≤–1 and
d(p) = (1 + (2–1/p– 1)p)/[8(1 + (2–1/p– 1)p) + 2] ifan> 0 (n= 1, 2,. . .), and ∞
n=1a
p n< +∞. MSC: 26D15
Keywords: Hardy’s inequality; monotonicity; convergence
1 Introduction
Letp> andan> (n= , , . . .) with
∞
n=anp< +∞, then Hardy’s well-known inequality
[] is given by
∞
n=
n
n
k=
ak
p
<
p p–
p∞
n=
apn. (.)
Recently, the refinement, improvement, generalization, extension, and application for Hardy’s inequality have attracted the attention of many researchers [–].
Yang and Zhu [] presented an improvement of Hardy’s inequality (.) forp= as follows:
∞
n=
n
n
k=
ak
<
∞
n=
– √n+
an.
For /≤p≤, Huang [] proved that
∞
n=
n
n
k=
ak
p
<
p p–
p∞
n=
–
(n–/p+ ,)
apn.
In [], Wen and Zhang proved that the inequality
∞
n=
n
n
k=
ak
p
<
p p–
p∞
n=
– Cp
n–/p
ap
n (.)
holds forp> ifan> (n= , , . . .), with
∞
n=a
p
n< +∞, whereCp= – ( – /p)p–for
p≥ andCp= – /pfor <p≤.
Xuet al.[] gave a further improvement of the inequality (.):
∞
n=
n
n
k=
ak
p
<
p p–
p∞
n=
– Zp (n– )–/p
apn,
whereZp=p– –(p–)
p
/pfor <p≤ andZ
p= – (p–p )p–
p–
p forp> .
For the special parameterp= /, Denget al.[] established
∞
n=
n
n
k=
ak
/ ≤/
∞
n=
–
(n/+η /)
a/n ,
whereη/= //[(/– (
∞
n=n/ (
n
m=m–/)/)) – ] = .· · ·.
In [], Long and Linh discussed Hardy’s inequality with the parameterp< , and proved that
∞
n=
n
n
k=
ak
p
<
p p–
p∞
n=
apn (.)
forp≤– and
∞
n=
n
n
k=
ak
p
<
–p
–p
∞
n=
apn
for – <p< ifan> (n= , , . . .) with
∞
n=a
p n< +∞.
It is the aim of this paper to present an improvement of inequality (.) for the parameter
p≤–. Our main result is Theorem ..
Theorem . Let p≤–,d(p) = ( + (–/p– )p)/[( + (–/p– )p) + ]and a
n> (n=
, , . . .)with∞n=apn< +∞,then
∞
n=
n
n
k=
ak
p
≤
p p–
p∞
n=
– d(p) (n– /)–/p
ap n.
2 Lemmas
In order to establish our main result we need several lemmas, which we present in this section.
Lemma .(see [, Corollary .]) Suppose that a,b∈Rwith a<b,f : [a,b]n→Rhas
continuous partial derivatives and Dm=
x= (x,x, . . . ,xn) a≤ min
≤k≤n{xk}<xm=max≤k≤n{xk} ≤b
, m= , , . . . ,n.
If ∂∂fx(x)
m > holds for all x= (x,x, . . . ,xn)∈Dmand m= , , . . . ,n,then
f(x,x, . . . ,xn)≥f(xmin,xmin, . . . ,xmin)
Lemma . Let n∈Rbe a positive natural number and r∈Rwith r≥.Then
n
k=
k–
/r
≥ r
r+
n+/r– + –/r. (.)
Proof We use mathematical induction to prove inequality (.). We clearly see that in-equality (.) becomes in-equality forn= . We assume that inequality (.) holds forn=i
(i∈N,i≥), namely
i
k=
k–
/r
≥ r
r+
i+/r– + –/r.
Then forn=i+ we have
i+
k=
k–
/r
=
i
k=
k–
/r
+
i+
/r
≥ r
r+
i+/r– + –/r+
i+
/r
= r
r+
(i+ )+/r– + –/r+
i+
/r
– i+
i
x/rdx. (.)
Note thatx/r(r≥) is concave on (, +∞), therefore Hermite-Hadamard’s inequality
implies that
i+
/r
≥
i+
i
x/rdx. (.)
From (.) and (.) we know that inequality (.) holds forn=i+ . Remark . The inequality
–/r≥ r
r+ (.)
holds for allr≥ with equality if and only ifr= .
Proof We clearly see that inequality (.) becomes equality forr= .
Ifr> , then it is well known that the function ( + /r)ris strictly increasing on (, +∞),
so we get
+
r
r
> . (.)
Therefore, inequality (.) follows from (.).
Lemma . The inequality
(r+ )––/r
r+ >
–/r– r
r+
r+
r (.)
Proof Letr≥, then we clearly see that
(log – )r+r+ log – ≥(log – ) > . (.)
Inequality (.) leads to
e(r–r+)/(r+)< . (.)
It follows from the well-known inequality ( +x)/x<e(x> ) that
e>
+r
–r+
r+ r
(r+r)/(r–r+)
. (.)
From (.) and (.) we have
–/r< r
+ r
r+ r+r+ . (.)
Therefore, inequality (.) follows easily from (.).
Lemma . Let r≥and
f(x) = x
r
(r+r x+/r+ –/r– r r+)r+
. (.)
Then f is convex on[/, +∞).
Proof From (.) we have
f(x) = –
r+
r+x
r+/r+ (–/r– r r+)rx
r–
( r
r+x+/r+ –/r–
r r+)r+
,
(.)
f(x) =
(r+)(r+)
(r+) x+/r– (–/r–r+r )
(r+)(r+)
r(r+) x+/r+r(r– )(–/r–
r r+)
(r+r x+/r+ –/r– r r+)r+
xr–.
It follows from Lemma . and (.) that
f(x)≥
(r+)(r+)
(r+) ––/r– (–/r–r+r )(r +)(r+)
r(r+)
(r+r x+/r+ –/r– r r+)r+
xr+/r–> (.)
for allx∈[/, +∞).
Therefore, Lemma . follows from inequality (.).
Lemma . Let r≥, ≤t≤and c= (r+ – /rr)/[(r+ – /rr) + ],then
(r+ )
–/r– r
r+
( –ct)t≥ –
+
+r
/rr–
t
–r
Proof Ifr= , then we clearly see that inequality (.) becomes equality. Next, we assume thatr> . Let
f(t) = (r+ )
–/r– r
r+
( –ct)t– +
+
+r
/rr–
t
–r
. (.)
Then simple computations lead to
f() = , (.)
f(t) =(r+ )(
–/r– r r+)
[ + (r/+rr– )t]r+
( – ct)
+
r+ /rr–
t
r+
–
. (.)
Note that
– ct≥ – c=
(r+ – /rr) + > , (.)
+
r+ /rr–
t
r+
≥ + (r+ )
r+ /rr–
t. (.)
It follows from Remark . and (.)-(.) that
f(t)≥(r+ )(
–/r– r r+)
[ + (r+
/rr– )t]r+
( – ct)
+ (r+ )
r+ /rr–
t
–
. (.)
Let
g(t) = ( – ct)
+ (r+ )
r+ /rr–
t
– . (.)
Then fromg() = andg() = (r+ – /rr)/[/rr((r+ – /rr) + )]≥ together with
the fact thatg(t) is a concave parabola we know that
g(t)≥ (.)
fort∈[, ].
Therefore, Lemma . follows easily from (.) and (.) together with (.)-(.).
Lemma . Let r≥,c= (r+ – /rr)/[(r+ – /rr) + ],N is a positive natural number,
ak> (k= , , . . . ,N)and BN=min≤k≤N{(k– /)/rak},then
r+
r
r N
n=
– c
(n– /)+/r
arn–
N
n=
n
n k=/ak
r
≥BrN
r+
r
r N
n=
– c
(n– /)+/r
n– /
–
N
n=
n
n
k=(k– /)/r
r
Proof Letak=bk/(k– /)/r(k= , , . . . ,N), thenBN=min≤k≤N{bk}and inequality (.)
becomes
r+
r
r N
n=
– c
(n– /)+/r
br n
n– /–
N
n=
n
n k=
(k–/)/r
bk
r
≥BrN
r+
r
r N
n=
– c
(n– /)+/r
n– /
–
N
n=
n
n
k=(k– /)/r
r
. (.)
LetDm={b= (b,b, . . . ,bN)|bm=max≤k≤N{bk}>min≤k≤N{bk}}(m= , , . . . ,N), and
f(b,b, . . . ,bN) =
r+
r
r N
n=
– c
(n– /)+/r
brn n– /
–
N
n=
n
n k=
(k–/)/r
bk
r
. (.)
Then for anyb∈Dm(m= , , . . . ,N) we have
∂f(b) ∂bm
=
– c
(m– /)+/r
(r+ )rbr–
m
(m– /)rr–
–r(m– /)
/r
b
m N
n=m
nr
(nk=(k–/)b /r
k )
r+
>
– c
(m– /)+/r
(r+ )rbr–
m
(m– /)rr–
–r(m– /)/rbrm–
+∞
n=m
nr
(nk=(k– /)/r)r+. (.)
From Lemma . and (.) one has
r(m– /)/rbr–
m
∂f(b) ∂bm
>
r+
r
r
– c
(m– /)+/r
(m– /)+/r
–
+∞
n=m
nr
(r+r n+/r+ –/r– r r+)r+
. (.)
It clearly follows from Lemma . and the Hermite-Hadamard inequality that
m+/
m–/
xr
( r
r+x+/r+ –/r–
r r+)r+
≥ mr
( r
r+m+/r+ –/r–
r r+)r+
and
+∞
m–/
xr
(r+r x+/r+ –/r– r r+)r+
≥ +∞
n=m
nr
(r+r n+/r+ –/r– r r+)r+
Note that +∞
m–/
xr
(r+r x+/r+ –/r– r r+)r+
=
+r r
r
/r
r+ – /rr
–
+
r+ /rr–
(m– /)––/r –r
, (.)
< (m– /)––/r≤+/r≤. (.) From Lemma . and (.) one has
r+
r
r
– c
(m– /)+/r
(m– /)+/r
≥
+r
r
r
/r
r+ – /rr
–
+
r+ /rr–
(m– /)––/r –r
. (.)
Inequalities (.), (.), and (.) together with (.) lead to the conclusion that
∂f(b) ∂bm
> (.)
for anyb= (b,b, . . . ,bN)∈Dmandm= , , . . . ,N.
It follows from Lemma . and (.) that
f(b,b, . . . ,bN)≥f(BN,BN, . . . ,BN). (.)
Therefore, inequality (.) follows from (.) and (.).
Lemma . Let r≥,c= (r+ – /rr)/[(r+ – /rr) + ],then
r+
r
– +/rc> . (.)
Proof We clearly see that inequality (.) holds forr= . Next, we assume thatr> , let
t= +/r, then <t< and Lemma . leads to
r+
r
r
– +/rc– ≥ ( r+
r ) r
(r+ – /rr)
–
+(r+ –
/rr)
r
–r
–
≥ (r+r ) r
(r+ – /rr)
(r+ – /rr)
+ (r+ – /rr)–
= (
r+
r ) r
+ (r+ – /rr)– . (.)
Note that
r+ – /rr< –log (.)
for allr≥. In fact, letx≥ and
Then
f(x) = +
log
x –
/x, (.)
f(x) = –(log)
x
/x< . (.)
It follows from (.) and (.) that
f(x) > lim
x→+∞
+
log
x –
/x
= . (.)
Equation (.) and inequality (.) lead to the conclusion that
f(x) < lim
x→+∞
x– /xx+ = –log. (.)
From (.) and (.) together with the fact that [(r+ )/r]r≥ we have
r+
r
r
– +/rc– >
+ ( –log)– =
log –
– log > . (.)
Therefore, inequality (.) follows from (.).
Lemma . Let r≥,c= (r+ – /rr)/[(r+ – /rr) + ],N is a positive natural number,
ak> (k= , , . . . ,N)and BN=min≤k≤N{(k– /)/rak},then
r+
r
r N
n=
– c
(n– /)+/r
arn–
N
n=
n
n k=/ak
r
≥BrN
r+
r
r
– +/rc–
. (.)
Proof Letm∈ {, , . . . ,N},f() = and
f(m) =
r+
r
r m
n=
– c
(n– /)+/r
n– /–
m
n=
n
n
k=(k– /)/r
r
. (.)
Then
f() =
+r r
r
– +/rc–
, (.)
f(m) –f(m– ) = (
+r r )r
m– /
– c
(m– /)+/r
–
m
m
k=(k– /)/r
r
. (.)
It follows from Lemma . and (.) together with Remark . that
f(m) –f(m– )
≥ ( +r
r ) r
m– /
– c
(m– /)+/r
–
m
r
r+(m+/r– ) + –/r
≥ (+rr) r
m– /
– c
(m– /)+/r
–
m
r r+m+/r
r
=(
+r r )
r[((r+ – /rr) + )(m– /)+/r–m(r+ – /rr)]
m(m– /)+/r[(r+ – /rr) + ] . (.)
Let
g(t) =r+ – /rr+ (t– /)+/r–r+ – /rrt. (.)
Then
g() =r+ – /rr+ ––/r–r+ – /rr
>–/r– r+ – /rr≥,
(.)
g(t) =
+
r
r+ – /rr+ (t– /)/r–r+ – /rr
>–/r– r+ – /rr≥ (.)
fort≥.
From (.)-(.) we get
f() <f() <· · ·<f(N– ) <f(N). (.)
Therefore, Lemma . follows easily from Lemma ., (.), (.), and (.).
3 Proof of Theorem 1.1
Letr= –p,c=c(r) =d(–r) andbn= /an(n= , , . . .), thenr≥,c= (r+ – /rr)/[(r+
– /rr) + ],b
n> and∞n=brn< +∞.
It follows from Lemmas . and . that one has
N
n=
n
n k=/bk
r
≤
r+
r
r N
n=
– c
(n– /)+/r
brn. (.)
Lettingn→+∞, (.) leads to
∞
n=
n
n k=/bk
r
≤
r+
r
r∞
n=
– c
(n– /)+/r
brn. (.)
Therefore, Theorem . follows immediately from (.) andr= –ptogether withbn=
/an.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Author details
1School of Mathematics and Computation Science, Hunan City University, Yiyang, 413000, China.2Jiaxing Radio &
Television University, Jiaxing, 314000, China.3Haining College, Zhejiang Radio & Television University, Haining, 314000, China.
Acknowledgements
The authors would like to express their deep gratitude to the referees for giving many valuable suggestions. This research was supported by the Natural Science Foundation of China under Grants 11171307 and 61374086, and the Natural Science Foundation of Zhejiang Province under Grant Y13A01000.
Received: 1 December 2013 Accepted: 14 June 2014 Published:23 Jul 2014
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