International audienceWe show that the remainder in Hölder's inequality (Rogers, 1888; Hölder, 1889) may be computed exactly. It satisfies functional equations, and possesses monotonicity and scaling properties. We obtain as a consequence improvements of recent sharpenings (Aldaz, 2008) of the classical inequality

Kichenassamy, Satyanad

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Improving Hölder’s inequality
Satyanad Kichenassamy
To cite this version:
Satyanad Kichenassamy. Improving Hölder’s inequality. Houston Journal of Mathematics, 2010, 35
(1), pp.303-312. ￿hal-00826949￿
Houston Journal of Mathematics
c© 2010 University of Houston
Volume 36, No. 1, 2010
IMPROVING HO¨LDER’S INEQUALITY
SATYANAD KICHENASSAMY
Communicated by Ha¨ım Brezis
Abstract. We show that the remainder in Ho¨lder’s inequality may be com-
puted exactly. It satisfies functional equations, and possesses monotonicity
and scaling properties. Improved inequalities follow.
1. Introduction
For any two measurable functions f(x), g(x) on a measure space (Ω,A, µ), and
any p ∈ (1,+∞), Ho¨lder’s inequality [8, 5, 7, 4] states that
(1.1)
∫
Ω
fg dµ ≤ ‖f‖p‖g‖q,
where q = p/(p − 1), and ‖f‖p =
(∫
Ω
|f(x)|p dµ)1/p is the Lp norm. It plays in
Nonlinear Analysis the roˆle of the Cauchy-Schwarz inequality. It is a consequence
of the inequality
(1.2) aαbβ ≤ αa+ βb,
for a, b, α, β nonnegative, with 0 < α < 1 and β = 1−α. Indeed, it follows from
(1.2) with α = 1/p, a = F and b = G, where
F = |f(x)|p/‖f‖pp, G = |g(x)|q/‖g‖qq.
Inequality (1.2) may be deduced from the convexity inequality j(αt + βs) ≤
αj(t) + βj(s) by taking j to be the exponential, t = ln a and s = ln b. However,
this does not give a very precise expression for the difference between the two
2000 Mathematics Subject Classification. 26D15,46E30.
Key words and phrases. Ho¨lder’s inequality; Lp spaces.
303
304 S. KICHENASSAMY
sides of the inequality. On the other hand, for α = β = 1/2, (1.2) follows from
the identity
√
ab =
1
2
(a+ b)− 1
2
(
√
a−
√
b)2.
This suggests the search for identities that imply (1.2). Define the remainder
function R by
(1.3) R(α ; a, b) = αa+ βb− aαbβ ,
where a and b are positive, and β = 1− α. Thus, for t > 0,
tα = 1 + α(t− 1)−R(α ; t, 1).
In particular, R(0 ; a, b) = R(1 ; a, b) = 0 and
(1.4) R(
1
2
; a, b) =
1
2
(
√
a−
√
b)2.
The function R extends by continuity to a = 0 if α > 0, and to b = 0 if β > 0.
The convexity of the exponential and the concavity of tα yield, for 0 < α < 1,
1 + α ln t ≤ tα ≤ 1 + α(t− 1),
hence the inequality
(1.5) 0 ≤ R(α ; t, 1) ≤ α(t− 1− ln t).
We are interested in expressing R as a manifestly nonnegative expression that
vanishes for a = b.
In terms of the remainder function, Ho¨lder’s inequality follows from the identity
(1.6)
∫
Ω
|fg| dµ
‖f‖p‖g‖q = 1−
∫
Ω
R(
1
p
; F,G) dµ,
and the positivity of R. Any lower bound on R yields a sharpening of Ho¨lder’s
inequality. Such lower bounds will be obtained as a consequence of the following
properties of the remainder, proved in Section 2.
• It satisfies scaling, monotonicity and symmetry properties (Theorem 2.1).
These properties are valid also for α > 1, or negative. In particular, it
follows from the symmetry properties that R is nonpositive for α 6∈ [0, 1].
• It may be computed for exponents αα′ and α + α′ if it is known for
exponents α and α′ (Theorem 2.2).
• It may be computed in closed form for rational α (Theorem 2.3). In
addition, it admits an integral representation (Theorem 2.4).
HO¨LDER’S INEQUALITY 305
It is convenient to relax, in some of the general results, the restriction 0 < α < 1;
due to the symmetry relation (2.3) below, inequalities for 0 < α < 1 imply
counterparts for α > 1, corresponding to 0 < p < 1. Section 3 develops the
consequences of the properties of R:
• A two-sided estimate of R derived from the addition theorem (§ 3.1).
• A sequence of more and more precise estimates of R derived from the
multiplication theorem and the integral representation. The first of these
estimates recovers a result of [1] (§ 3.2).
• Another estimate derived from the multiplication formula, improving an
inequality discussed in [6] (§ 3.3).
• An improvement of Clarkson’s inequality suggested by the exact formulae
for R (§ 3.4).
2. Properties of R
In this section, a and b are positive. It is not assumed that α lies between 0
and 1.
2.1. Scaling, symmetry and monotonicity.
Theorem 2.1. The remainder function possesses the scaling property:
(2.1) R(α ; λa, λb) = λR(α ; a, b).
if λ > 0. It possesses two symmetry properties:
(2.2) R(α ; a, b) = R(β ; b, a),
and, for α 6= 0,
(2.3) R(
1
α
; aα, bα) = − 1
α
b−βR(α ; a, b).
Finally, it possesses the monotonicity property:
(2.4) α−10 R(α0 ; a, b) ≤ α−1R(α ; a, b)
for α ≤ α0, both nonzero.
Remark. Since R(α ; a, b) ≥ 0 for 0 < α < 1, (2.3) implies that R(α ; a, b) ≤ 0 for
α > 1; this in turn implies, by (2.2), that the same holds for α < 0.
Proof. Equations (2.1) and (2.2) are readily checked using the definition of R.
306 S. KICHENASSAMY
To prove (2.3), let t = a/b and write
R(
1
α
;
aα
bα
, 1) = R(
1
α
; tα, 1) =
tα
α
+ (1− 1
α
)− t
=
1
α
[tα − (1− α)− αt]
= − 1
α
R(α ; t, 1) = − 1
αb
R(α ; a, b)
by the scaling property. Multiplying through by bα, and using scaling once more,
the result follows.
To prove (2.4), consider ϕ(α, t) := R(α ; t, 1)/α = (t − 1) − α−1(tα − 1) for
α 6= 0, extended by continuity to α = 0, so that ϕ(0, t) = t − 1 − ln t. We have,
for α 6= 0,
∂ϕ
∂α
=
tα − 1
α2
− t
α
α
ln t = − t
α
α2
[
t−α − 1 + α ln t] .
The inequality eu ≥ 1 + u, with u = −α ln t, implies that ∂ϕ/∂α ≤ 0, as desired.

2.2. Functional equations.
Theorem 2.2. The remainder satisfies the functional equation
(2.5) R(αα′ ; a, b) = α′R(α ; a, b) + b1−αR(α′ ; aα, bα).
It also satisfies an addition theorem: writing Rα for R(α ; a, b), we have
(2.6) Rα+α′ =
[
1 + α′
a− b
b
]
Rα + (a/b)αRα′ − αα
′
b
(a− b)2.
Remark. Since the left-hand side of (2.5) is symmetric with respect to α and α′,
we have the identity
α′R(α ; a, b)− αR(α′ ; a, b) = b1−α′R(α ; aα′ , bα′)− b1−αR(α′ ; aα, bα).
The addition theorem may be given a form symmetric in α and α′, by replacing
(a/b)α = (aαbβ)/b by 1 + α(a− b)/b−Rα/b; the result is:
R(α+ α′ ; a, b) = Rα +Rα′ − b−1RαRα′ + a− b
b
[αRα′ + α′Rα]− αα
′
b
(a− b)2.
Proof. Writing tαα
′
= (tα)α
′
, we obtain
tαα
′
= 1 + α′(tα − 1)−R(α′ ; tα, 1)
= 1 + α′ [α(t− 1)−R(α ; t, 1)]−R(α′ ; tα, 1)
= 1 + αα′(t− 1)− α′R(α ; t, 1)−R(α′ ; tα, 1).
HO¨LDER’S INEQUALITY 307
Therefore,
(2.7) R(αα′ ; t, 1) = α′R(α ; t, 1) +R(α′ ; tα, 1).
Replacing t by a/b and using the scaling property, equation (2.5) follows.
For the addition theorem, one writes, letting Rα = R(α ; t, 1) and Rα′ =
R(α′ ; t, 1),
1− tα+α′ = 1 + tα(−tα′)
= 1 + tα(Rα′ − 1− α′(t− 1))
= 1 + tαRα′ − tα(1 + α′(t− 1))
= 1 + tαRα′ − (1 + α(t− 1)−Rα)(1 + α′(t− 1))
=
[
Rα(1 + α′(t− 1)) + tαRα′ − αα′(t− 1)2
]− (α+ α′)(t− 1).
Since Rα+α′ = 1+(α+α′)(t−1)−tα+α′ , the result follows, using again scaling. 
2.3. Closed-form representations of R. For rational α, we now show that the
remainder may be expressed using polynomials with positive integral coefficients.
Theorem 2.3. Let k and n be positive integers, with 1 ≤ k ≤ n, and α = k
n+ 1
.
We then have
(2.8) R(α ; an+1, bn+1) =
(a− b)2
n+ 1
(αIn,k + βJn,k),
where
In,k =
n−1∑
j=k−1
(n− j)ajbn−1−j ,
Jn,1 = 0 and, for 2 ≤ k ≤ n,
Jn,k =
k−2∑
j=0
(j + 1)ajbn−1−j .
Thus, the remainder is the product of (a−b)2/(n+1) by a polynomial with positive
integral coefficients.
Proof. By scaling, it suffices to determine (n + 1)R(α ; tn+1, 1), with α = kn+1
(so that β = (n+ 1− k)/(n+ 1)). It may be written
ktn+1 + (n+ 1− k)− (n+ 1)tk = k(tn+1 − 1)− (n+ 1)(tk − 1).
Consider first the case k = 1. We have
tn+1 − (n+ 1)t+ n = (t− 1)(tn + tn−1 + · · ·+ t− n).
308 S. KICHENASSAMY
Furthermore,
tn + tn−1 + · · ·+ t− n = (t− 1)(tn−1 + 2tn−2 + · · ·+ (n− 1)t+ n).
Since
In,1 =
n−1∑
j=0
(n− j)tj = tn−1 + 2tn−1 + · · ·+ (n− 1)t+ n,
we have
(2.9) tn+1 + n− (n+ 1)t = (t− 1)2In,1.
Since Jn,1 = 0, this proves the result for k = 1.
Assume now 2 ≤ k ≤ n. Equation (2.9), multiplied by k, may be written
k(tn+1 − 1)− k(n+ 1)(t− 1) = (t− 1)2
n−1∑
j=0
k(n− j)tj .
Exchanging the roˆles of k and n+ 1,
(n+ 1)(tk − 1)− (n+ 1)k(t− 1) = (t− 1)2
k−2∑
j=0
(n+ 1)(k − 1− j)tj .
Subtracting, we obtain, since k ≤ n,
k(tn+1 − 1)− (n+ 1)(tk − 1) = (t− 1)2
×

k−2∑
j=0
[k(n− j)− (n+ 1)(k − 1− j)] tj +
n−1∑
j=k−1
k(n− j)tj
 .
Since
k(n− j)− (n+ 1)(k − 1− j) = (n+ 1− k)(j + 1) = (n+ 1)β(j + 1) ≥ 0,
we have proved (2.8). 
For general exponents, we have:
Theorem 2.4. For 0 < α < 1, the remainder admits the general expression
(2.10) R(α ; a, b) =
(
ln
a
b
)2 ∫ 1
0
min[α(1− τ), βτ ]aτ b1−τdτ.
As a consequence,
(2.11) R(α ; a, b) ≥ 1
2
αβmin(a, b)
(
ln
a
b
)2
.
HO¨LDER’S INEQUALITY 309
Proof. Let j(t) be twice continuously differentiable for t > 0, and x, y positive.
Consider
u(t) := tj(x) + (1− t)j(y)− j(tx+ (1− t)y).
We have u(0) = u(1) = 0, and
−d
2u
dt2
= h(t) := (x− y)2j′′(tx+ (1− t)y).
This implies
u(t) =
∫ 1
0
g(t, τ)h(τ)dτ,
where g(t, τ) = min[(1 − t)τ, (1 − τ)t] is the Green’s function for this problem;
indeed, one checks directly that −∂2g(t, τ)/∂t2 = δ(t − τ), and that g vanishes
for t = 0 or 1. It follows that
αj(x) + βj(y)− j(αx+ βy) = (x− y)2
∫ 1
0
g(α, τ)j′′(τx+ (1− τ)y)dτ.
We now let x = ln a, y = ln b, and j(t) = exp(t) to obtain (2.10). For 0 < τ < 1,
aτ b1−τ ≥ min(a, b). Since g is nonnegative, and ∫ 1
0
g(α, τ)dτ = 12αβ, (2.11)
follows. 
3. Applications
3.1. A consequence of the monotonicity property. Monotonicity, together
with the exact formulae of Theorem 2.3, yields precise estimates of the remainder.
Indeed, if t > 0, and k and n are integers such that 1 ≤ k ≤ n and k/(n + 1) <
α < (k + 1)/(n+ 1), (2.4) yields
(n+ 1)α
k + 1
R(
k + 1
n+ 1
; t, 1) ≤ R(α ; t, 1) ≤ (n+ 1)α
k
R(
k
n+ 1
; t, 1).
One may also obtain an exact formula: write α = k/(n+ 1) + ε, Rα = R(α ; t, 1),
and Rk,n = R( kn+1 ; t, 1); the addition theorem (2.6) then yields
Rα = [1 + ε(t− 1)]Rk,n + tk/(n+1)Rε − kε
n+ 1
(t− 1)2.
Combining with estimate (1.5), we obtain
0 ≤ Rα − [1 + ε(t− 1)]Rk,n + kε
n+ 1
(t− 1)2 ≤ εtk/(n+1)(t− 1− ln t).
310 S. KICHENASSAMY
3.2. First consequence of the multiplication theorem. The functional equa-
tion (2.5) may be used in two different ways, depending on whether Rα or Rα′ is
known. This section discusses the first.
Since Theorem 2.3 gives the remainder for α = 1/m, with integral m, equation
(2.5), where the pair (α, α′) is replaced by ( 1m ,mα), yields
(3.1) R(α ; t, 1) = mαR(
1
m
; t, 1) +R(mα ; t1/m, 1).
For example, for m = 2, using (1.4), R(α ; t, 1) = α(
√
t− 1)2 +R(2α ; √t, 1), and
(2.11) gives
(3.2) R(α ; t, 1) ≥ α(√t− 1)2 + 1
4
α(1− 2α) min(√t, 1) (ln t)2
for 0 < α ≤ 1/2. This improves the inequality [1]:
R(α ; t, 1) ≥ α(√t− 1)2 for 0 < α ≤ 1
2
.
If 1/2 ≤ α < 1, one may apply this result to R(β ; t, 1) = R(α ; 1, t). More and
more precise expressions may be obtained by iterating the process: thus, applying
the same argument to R(2α ;
√
t, 1), we obtain
R(α ; t, 1) = α(
√
t− 1)2 + 2α(t1/4 − 1)2 +R(4α ; t1/4, 1).
Therefore, for 0 < α ≤ 14 ,
R(α ; t, 1) ≥ α(√t− 1)2 + 2α(t1/4 − 1)2 + 1
8
α(1− 4α) min(t1/4, 1)(ln t)2.
3.3. Second consequence of the multiplication theorem. The second way
to take advantage of (2.5) runs as follows. Since R( 12 ; a, b) is known (by (1.4)),
we apply (2.5), where the pair (α, α′) is replaced by (mα, 1m ); this yields, instead
of (3.1),
(3.3) R(α ; t, 1) =
1
m
R(mα ; t, 1) +R(
1
m
; tmα, 1).
Consider now two conjugate exponents p and q, and α = 1/q. Taking m = 2 and
α = 1/q, we obtain in particular
R(
1
q
; t, 1) =
1
2
R(
2
q
; t, 1) +R(
1
2
; t2/q, 1),
hence
R(
1
q
; a, b) =
1
2
R(
2
q
; a, b) + b1−2/qR(
1
2
; a2/q, b2/q).
HO¨LDER’S INEQUALITY 311
Therefore, for f and g positive, using (2.2) and the relation p/q = p− 1,
R(
1
p
; fp, gq) = R(
1
q
; gq, fp)
=
1
2
R(
2
q
; gq, fp) + fp−(2p/q)R(
1
2
; g2, f2p/q)
=
1
2
R(
2
q
; gq, fp) +
1
2
f2−p
(
g − fp−1)2 .
For 1 < q ≤ 2, the first term is nonpositive (see the remark following Th. 2.1),
and we recover the inequality [6]
fp
p
+
gq
q
− fg ≤ 1
2
f2−p(g − fp−1)2.
If q > 2, the inequality is reversed, since R changes sign. More precisely, applying
the symmetry property (2.3) with α = 2/q, we have
R(
q
2
; g2, f2p/q) = −q
2
f (2p/q)−pR(
2
q
; gq, fp),
hence, after substitution and simplification using the scaling property,
R(
1
p
; fp, gq) =
1
2
f2−p
(
g − fp−1)2 − 1
q
R(
q
2
; f2−pg2, fp).
As before, (2.11) leads to sharpened inequalities. For example, for q ≥ 2,
fp
p
+
gq
q
− fg = 1
2
R(
2
q
; gq, fp) +
1
2
f2−p
(
g − fp−1)2
≥ 1
2
f2−p
(
g − fp−1)2 + 1
q
(1− 2
q
) min(fp, gq)(ln
fp
gq
)2.
3.4. Consequences of the exact expression for R. We now turn to examples
illustrating the identities in Theorem 2.3. They suggest improvements of other
classical inequalities. We begin with the simplest examples. The formulae below
have been obtained by multiplying (2.8) by (n+1), and letting a = fn+1, b = gn+1.
For k = 1 = n, we recover 2fg = f2 + g2 − (f − g)2.
For k = 1 et n = 2, we obtain
(3.4) 3fg2 = f3 + 2g3 − (f − g)2(f + 2g).
For k = 1 or 2, and n = 4,
5fg4 = f5 + 4g5 − (f − g)2(f3 + 2f2g + 3fg2 + 4g3),
5f2g3 = 2f5 + 3g5 − (f − g)2(2f3 + 4f2g + 6fg2 + 3g3).
312 S. KICHENASSAMY
Therefore, R( 13 ; t, 1) =
1
3 (t
1/3− 1)2(t1/3 + 2). Equation (3.4) implies the identity
(f + g)3 + 3(f − g)3 = 4(f3 + g3)− 6g(f − g)2;
it may be compared with Clarkson’s inequality [3, 2]
|f + g|3 + |f − g|3 ≤ 4(|f |3 + |g|3).
More generally, we have the following.
Theorem 3.1. If p > 2 and f > g > 0,
(f + g)p + (2p−1 − 1)(f − g)p ≤ 2p−1(fp + gp).
Proof. Let t = f/g. Consider ϕ(t) =
[
2p−1(tp + 1)− (t+ 1)p] (t − 1)−p, for
t > 1. We obtain
1
p
(t− 1)p+1 dϕ
dt
= (t− 1){(2t)p−1 − (t+ 1)p−1}− {2p−1(tp + 1)− (t+ 1)p}
= (2t)p−1[t− 1− t]− 2p−1 + (t+ 1)p−1[−t+ 1 + t+ 1]
= 2
[
(t+ 1)p−1 − 2p−2(tp−1 + 1)] .
Since p − 1 > 1, (t + 1)p−1 ≤ 2p−2(tp−1 + 1), by Ho¨lder’s inequality. Therefore,
ϕ(t) is decreasing over (1,+∞), and its infimum is its limit as t → ∞, namely
2p−1 − 1. The theorem follows. 
References
[1] J. M. Aldaz. A stability version of Ho¨lder’s inequality. J. Math. Anal. Appl., 343 (2008)
842–852.
[2] H. Brezis. Analyse Fonctionnelle : The´orie et Applications, Masson, Paris 1983.
[3] J. A. Clarkson. Uniformly convex spaces. Trans. Amer. Math. Soc., 40 (3) (1936) 396–414.
[4] G. H. Hardy, J. E. Littlewood, G. Polya. Inequalities, 2nd. ed., Cambridge, 1952.
[5] O. Ho¨lder. Ueber einen Mittelwerthssatz. Nachr. von der Ko¨nigl. Gesellschaft der Wiss.
und der Georg-August-Universita¨t zu Go¨ttingen, (1889) 38–47.
[6] J. Pecˇaric´, V. Sˇimic´. A note on the Ho¨lder inequality. J. Inequalities in Pure and Appl.
Math., 7 (5) (2006) Article 176, 3 pp.
[7] F. Riesz. Untersuchungen u¨ber Systeme integrierbare Funktionen. Math. Ann., 69 (1910)
449–497.
[8] L. J. Rogers. An extension of a certain theorem in inequalities. Messenger of Math., 17
(1888) 145–150.
Received July 23, 2009
Laboratoire de Mathe´matiques, CNRS FRE 3111, Universite´ de Reims Champagne-
Ardenne, Moulin de la Housse, B.P. 1039, F-51687 Reims Cedex 2, France
E-mail address: satyanad.kichenassamy@univ-reims.fr


Improving Hölder's inequality

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