CUBO, A Mathematical Journal
Vol. 23, no. 03, pp. 457–468, December 2021

DOI: 10.4067/S0719-06462021000300457

Some integral inequalities related to Wirtinger’s
result for p-norms

S. S. Dragomir1,2

1 Mathematics, College of Engineering &

Science, Victoria University,

PO Box 14428 Melbourne City,

MC 8001, Australia.

sever.dragomir@vu.edu.au

2 DST-NRF Centre of Excellence in the

Mathematical and Statistical Sciences,

School of Computer Science & Applied

Mathematics, University of the

Witwatersrand, Private Bag 3,

Johannesburg 2050, South Africa.

ABSTRACT

In this paper we establish several natural consequences of
some Wirtinger type integral inequalities for p-norms. Ap-
plications related to the trapezoid unweighted inequalities,
of Grüss’ type inequalities and reverses of Jensen’s inequality
are also provided.

RESUMEN

En este artículo establecemos varias consecuencias naturales
de algunas desigualdades integrales de tipo Wirtinger para
p-normas. También se entregan aplicaciones relacionadas a
desigualdades trapezoidales sin peso, desigualdades de tipo
Grüss y reversos de la desigualdad de Jensen.

Keywords and Phrases: Wirtinger’s inequality, trapezoid inequality, Grüss’ inequality, Jensen’s inequality.

2020 AMS Mathematics Subject Classification: 26D15; 26D10.

Accepted: 05 October, 2021

Received: 02 May, 2021

©2021 S. S. Dragomir. This open access article is licensed under a Creative Commons

Attribution-NonCommercial 4.0 International License.

http://cubo.ufro.cl/
http://dx.doi.org/10.4067/S0719-06462021000300457
https://orcid.org/0000-0003-2902-6805


458 S. S. Dragomir CUBO
23, 3 (2021)

1 Introduction

The following Wirtinger type inequalities are well known∫ b
a

u2(t)dt ≤
(b − a)2

π2

∫ b
a

[u′ (t)]
2
dt (1.1)

provided u ∈ C1 ([a, b] , R) and u (a) = u (b) = 0 with equality holding if and only if u (t) =
K sin

[
π(t−a)
b−a

]
for some constant K, and, similarly, if u ∈ C1 ([a, b] , R) satisfies u (a) = 0, then

∫ b
a

u2(t)dt ≤
4 (b − a)2

π2

∫ b
a

[u′ (t)]
2
dt. (1.2)

The equality holds in (1.2) if and only if u (t) = K sin
[
π(t−a)
2(b−a)

]
for some constant K.

For p > 1 put πp := 2πp sin
(

π
p

)
. In [11], J. Jaroš obtained, as a particular case of a more general

inequality, the following generalization of (1.1)∫ b
a

|u (t)|p dt ≤
(b − a)p

(p − 1) πpp

∫ b
a

|u′ (t)|p dt (1.3)

provided u ∈ C1 ([a, b] , R) and u (a) = u (b) = 0, with equality if and only if u (t) = K sinp
[
πp(t−a)

b−a

]
for some K ∈ R, where sinp is the 2πp-periodic generalized sine function, see [18] or [5].

If u (a) = 0 and u ∈ C1 ([a, b] , R) , then∫ b
a

|u (t)|p dt ≤
[2 (b − a)]p

(p − 1) πpp

∫ b
a

|u′ (t)|p dt (1.4)

with equality iff u (t) = K sinp
[
πp(t−a)
2(b−a)

]
for some K ∈ R.

The inequalities (1.3) and (1.4) were obtained for a = 0, b = 1 and q = p > 1 in [17] by the use of

an elementary proof.

For some related Wirtinger type integral inequalities see [1, 2, 4, 8, 9, 11, 12] and [15]-[17].

These inequalities are used in various fields of Mathematical Analysis, Approximation Theory,

Integral Operator Theory and Analytic Inequalities Theory since they provide connections between

the Lebesgue norms of a function and the corresponding Lebesgue norms of the derivative under

some natural assumptions at the endpoints.

Motivated by the above results, in this paper we establish some natural consequences of the

Wirtinger type integral inequalities for p-norms (1.3) and (1.4). Applications related to the trape-

zoid unweighted inequalities, of Grüss’ type inequalities and reverses of Jensen’s inequality are also

provided.



CUBO
23, 3 (2021)

Some integral inequalities related to Wirtinger’s result for p-norms 459

2 Some applications for trapezoid inequality

We have:

Proposition 2.1. Let g ∈ C1([a, b], R). Then for p > 1 we have the trapezoid inequality

∣∣∣∣∣g (a) + g (b)2 − 1b − a
∫ b
a

g (t) dt

∣∣∣∣∣
≤

b − a
2 (p − 1)1/p πp

(
1

b − a

∫ b
a

|g′ (t) − g′ (a + b − t)|p dt

)1/p
. (2.1)

In particular, for p = 2, we have [7]

∣∣∣∣∣g (a) + g (b)2 − 1b − a
∫ b
a

g (t) dt

∣∣∣∣∣ ≤ b − a2π
(

1

b − a

∫ b
a

|g′ (t) − g′ (a + b − t)|2 dt

)1/2
. (2.2)

Proof. If g ∈ C1([a, b], R), then by taking

u (t) :=
g (t) + g (a + b − t)

2
−

g (a) + g (b)

2
, t ∈ [a, b]

we have u (a) = u (b) = 0 and by (1.3) we have

∫ b
a

∣∣∣∣g (t) + g (a + b − t)2 − g (a) + g (b)2
∣∣∣∣p dt ≤ (b − a)p(p − 1) 2pπpp

∫ b
a

|g′ (t) − g′ (a + b − t)|p dt, (2.3)

namely

(∫ b
a

∣∣∣∣g (t) + g (a + b − t)2 − g (a) + g (b)2
∣∣∣∣p dt

)1/p

≤
(b − a)

2 (p − 1)1/p πp

(∫ b
a

|g′ (t) − g′ (a + b − t)|p dt

)1/p
. (2.4)

By Hölder’s integral inequality we have for p, q > 1, 1
p
+ 1

q
= 1

∫ b
a

∣∣∣∣g (t) + g (a + b − t)2 − g (a) + g (b)2
∣∣∣∣dt

≤

(∫ b
a

dt

)1/q (∫ b
a

∣∣∣∣g (t) + g (a + b − t)2 − g (a) + g (b)2
∣∣∣∣p dt

)1/p

= (b − a)1/q
(∫ b

a

∣∣∣∣g (t) + g (a + b − t)2 − g (a) + g (b)2
∣∣∣∣p dt

)1/p

= (b − a)1−1/p
(∫ b

a

∣∣∣∣g (t) + g (a + b − t)2 − g (a) + g (b)2
∣∣∣∣p dt

)1/p
. (2.5)



460 S. S. Dragomir CUBO
23, 3 (2021)

By making use of the properties of modulus and integral, we also have∫ b
a

∣∣∣∣g (t) + g (a + b − t)2 − g (a) + g (b)2
∣∣∣∣dt ≥

∣∣∣∣∣
∫ b
a

[
g (t) + g (a + b − t)

2
−

g (a) + g (b)

2

]
dt

∣∣∣∣∣
=

∣∣∣∣∣
∫ b
a

g (t) dt −
g (a) + g (b)

2
(b − a)

∣∣∣∣∣ . (2.6)
By making use of (2.4)-(2.6) we get the desired result (2.1).

Further, we have:

Proposition 2.2. Let g ∈ C1([a, b], R). Then for p > 1 we have the trapezoid inequality∣∣∣∣∣g (a) + g (b)2 − 1b − a
∫ b
a

g (t) dt

∣∣∣∣∣
≤

b − a
(p − 1)1/p πp

(
1

b − a

∫ b
a

∣∣∣∣g′ (t) − g (b) − g (a)b − a
∣∣∣∣p dt

)1/p
. (2.7)

In particular, for p = 2, we have [7]∣∣∣∣∣g (a) + g (b)2 − 1b − a
∫ b
a

g (t) dt

∣∣∣∣∣ ≤ b − aπ
(

1

b − a

∫ b
a

∣∣∣∣g′ (t) − g (b) − g (a)b − a
∣∣∣∣2 dt

)1/2
. (2.8)

Proof. If g ∈ C1([a, b], R), then by taking

u (t) := g (t) −
g (a) (b − t) + g (b) (t − a)

b − a
, t ∈ [a, b]

we have u (a) = u (b) = 0 and by (1.3) we have∫ b
a

∣∣∣∣g (t) − g (a) (b − t) + g (b) (t − a)b − a
∣∣∣∣p dt ≤ (b − a)p(p − 1) πpp

∫ b
a

∣∣∣∣g′ (t) − g (b) − g (a)b − a
∣∣∣∣p dt, (2.9)

which gives that(∫ b
a

∣∣∣∣g (t) − g (a) (b − t) + g (b) (t − a)b − a
∣∣∣∣p dt

)1/p

≤
b − a

(p − 1)1/p πp

(∫ b
a

∣∣∣∣g′ (t) − g (b) − g (a)b − a
∣∣∣∣p dt

)1/p
. (2.10)

By Hölder’s integral inequality we have for p, q > 1, 1
p
+ 1

q
= 1 that

∫ b
a

∣∣∣∣g (t) − g (a) (b − t) + g (b) (t − a)b − a
∣∣∣∣dt

≤

(∫ b
a

dt

)1/q (∫ b
a

∣∣∣∣g (t) − g (a) (b − t) + g (b) (t − a)b − a
∣∣∣∣p dt

)1/p

= (b − a)1/q
(∫ b

a

∣∣∣∣g (t) − g (a) (b − t) + g (b) (t − a)b − a
∣∣∣∣p dt

)1/p
. (2.11)



CUBO
23, 3 (2021)

Some integral inequalities related to Wirtinger’s result for p-norms 461

By making use of the properties of modulus and integral, we also have

∫ b
a

∣∣∣∣g (t) − g (a) (b − t) + g (b) (t − a)b − a
∣∣∣∣dt ≥

∣∣∣∣∣
∫ b
a

[
g (t) −

g (a) (b − t) + g (b) (t − a)
b − a

]
dt

∣∣∣∣∣
=

∣∣∣∣∣
∫ b
a

g (t) dt −
g (a) + g (b)

2
(b − a)

∣∣∣∣∣ . (2.12)
By making use of (2.10)-(2.12) we get the desired result (2.7).

We also have:

Proposition 2.3. Let g ∈ C([a, b], R). Then for p > 1 we have the inequality

∣∣∣∣∣b + a2
∫ b
a

g (s) ds −
∫ b
a

tg (t) dt

∣∣∣∣∣
≤

(b − a)2

(p − 1)1/p πp

(
1

b − a

∫ b
a

∣∣∣∣∣g (t) − 1b − a
∫ b
a

g (s) ds

∣∣∣∣∣
p

dt

)1/p
. (2.13)

In particular, for p = 2, we have [7]

∣∣∣∣∣b + a2
∫ b
a

g (s) ds −
∫ b
a

tg (t) dt

∣∣∣∣∣
≤

(b − a)2

π


 1
b − a

∫ b
a

g2 (t) dt −

(
1

b − a

∫ b
a

g (s) ds

)21/2 . (2.14)

Proof. Assume that g : [a, b] → C is continuous, then by taking

u (t) :=

∫ t
a

g (s) ds −
t − a
b − a

∫ b
a

g (s) ds, t ∈ [a, b]

we have u (a) = u (b) = 0, u ∈ C1([a, b], C) and by (1.3) we get

∫ b
a

∣∣∣∣∣
∫ t
a

g (s) ds −
t − a
b − a

∫ b
a

g (s) ds

∣∣∣∣∣
p

dt ≤
(b − a)p

(p − 1) πpp

∫ b
a

∣∣∣∣∣g (t) − 1b − a
∫ b
a

g (s) ds

∣∣∣∣∣
p

dt.

This is equivalent to

(∫ b
a

∣∣∣∣∣
∫ t
a

g (s) ds −
t − a
b − a

∫ b
a

g (s) ds

∣∣∣∣∣
p

dt

)1/p

≤
b − a

(p − 1)1/p πp

(∫ b
a

∣∣∣∣∣g (t) − 1b − a
∫ b
a

g (s) ds

∣∣∣∣∣
p

dt

)1/p
. (2.15)



462 S. S. Dragomir CUBO
23, 3 (2021)

By Hölder’s integral inequality we also have for p, q > 1, 1
p
+ 1

q
= 1 that

(b − a)1/q
(∫ b

a

∣∣∣∣∣
∫ t
a

g (s) ds −
t − a
b − a

∫ b
a

g (s) ds

∣∣∣∣∣
p

dt

)1/p

≥
∫ b
a

∣∣∣∣∣
(∫ t

a

g (s) ds −
t − a
b − a

∫ b
a

g (s) ds

)∣∣∣∣∣dt
≥

∣∣∣∣∣
∫ b
a

(∫ t
a

g (s) ds −
t − a
b − a

∫ b
a

g (s) ds

)
dt

∣∣∣∣∣ . (2.16)
Observe that, integrating by parts, we have∫ b

a

(∫ t
a

g (s) ds −
t − a
b − a

∫ b
a

g (s) ds

)
dt =

∫ b
a

(∫ t
a

g (s) ds

)
dt −

b − a
2

∫ b
a

g (s) ds

= b

∫ b
a

g (s) ds −
∫ b
a

tg (t) dt −
b − a
2

∫ b
a

g (s) ds

=
b + a

2

∫ b
a

g (s) ds −
∫ b
a

tg (t) dt. (2.17)

By making use of (2.15)-(2.17) we get the desired result (2.13).

3 Inequalities for the Čebyšev functional

For two Lebesgue integrable functions f, g : [a, b] → R, consider the Čebyšev functional:

C (f, g) :=
1

b − a

∫ b
a

f(t)g(t)dt −
1

(b − a)2
∫ b
a

f(t)dt

∫ b
a

g(t)dt. (3.1)

In 1935, Grüss [10] showed that

|C (f, g)| ≤
1

4
(M − m) (N − n) , (3.2)

provided that there exist real numbers m, M, n, N such that

m ≤ f (t) ≤ M and n ≤ g (t) ≤ N for a. e. t ∈ [a, b] . (3.3)

The constant 1
4

is the best possible in (3.1) in the sense that it cannot be replaced by a smaller

quantity.

Another, however less known result, even though it was obtained by Čebyšev in 1882, [3], states

that

|C (f, g)| ≤
1

12
(b − a)2 ∥f′∥∞ ∥g

′∥∞ , (3.4)

provided that f′, g′ exist and are continuous on [a, b] and ∥f′∥∞ = supt∈[a,b] |f
′ (t)| . The constant

1
12

cannot be improved in the general case.



CUBO
23, 3 (2021)

Some integral inequalities related to Wirtinger’s result for p-norms 463

The Čebyšev inequality (3.4) also holds if f, g : [a, b] → R are assumed to be absolutely continuous
and f′, g′ ∈ L∞ [a, b] while ∥f′∥∞ = ess supt∈[a,b] |f

′ (t)| .

A mixture between Grüss’ result (3.2) and Čebyšev’s one (3.4) is the following inequality obtained

by Ostrowski in 1970, [14]:

|C (f, g)| ≤
1

8
(b − a) (M − m) ∥g′∥∞ , (3.5)

provided that f is Lebesgue integrable and satisfies (3.3) while g is absolutely continuous and

g′ ∈ L∞ [a, b] . The constant 18 is the best possible in (3.5).

The case of euclidean norms of the derivative was considered by A. Lupaş in [13] in which he

proved that

|C (f, g)| ≤
1

π2
(b − a) ∥f′∥2 ∥g

′∥2 , (3.6)

provided that f, g are absolutely continuous and f′, g′ ∈ L2 [a, b] . The constant 1π2 is the best
possible.

We have:

Theorem 3.1. If f : [a, b] → R is continuous, p, q > 1 with 1
p
+ 1

q
= 1 and g : [a, b] → C is

absolutely continuous with g′ ∈ Lq [a, b] , then

|C (f, g)| ≤
(b − a)1/p

(p − 1)1/p πp

(
1

b − a

∫ b
a

∣∣∣∣∣f (t) − 1b − a
∫ b
a

f (s) ds

∣∣∣∣∣
p

dt

)1/p

×

(∫ b
a

|g′ (t)|q dt

)1/q
. (3.7)

In particular, for p = q = 2, we get

|C (f, g)| ≤
(b − a)1/2

π


 1
b − a

∫ b
a

f2 (t) dt −

(
1

b − a

∫ b
a

f (s) ds

)21/2

×

(∫ b
a

|g′ (t)|2 dt

)1/2
. (3.8)

Proof. Integrating by parts, we have

1

b − a

∫ b
a

(∫ x
a

f (t) dt −
x − a
b − a

∫ b
a

f (s) ds

)
g′ (x) dx

=
1

b − a


(∫ x

a

f (t) dt −
x − a
b − a

∫ b
a

f (s) ds

)
g (x)

∣∣∣∣∣
b

a

−
∫ b
a

g (x)

(
f (x) −

1

b − a

∫ b
a

f (s) ds

)
dx




= −
1

b − a

∫ b
a

f (x) g (x) dx +
1

b − a

∫ b
a

f (s) ds
1

b − a

∫ b
a

g (x) dx,



464 S. S. Dragomir CUBO
23, 3 (2021)

which gives that

C (f, g) =
1

b − a

∫ b
a

(
x − a
b − a

∫ b
a

f (s) ds −
∫ x
a

f (t) dt

)
g′ (x) dx. (3.9)

Using Hölder’s integral inequality for p, q > 1 with 1
p
+ 1

q
= 1 we have

|C (f, g)| =

∣∣∣∣∣ 1b − a
∫ b
a

(
x − a
b − a

∫ b
a

f (s) ds −
∫ x
a

f (t) dt

)
g′ (x) dx

∣∣∣∣∣
≤

1

b − a

∫ b
a

∣∣∣∣∣x − ab − a
∫ b
a

f (s) ds −
∫ x
a

f (t) dt

∣∣∣∣∣ |g′ (x)| dx
≤

1

b − a

(∫ b
a

∣∣∣∣∣x − ab − a
∫ b
a

f (s) ds −
∫ x
a

f (t) dt

∣∣∣∣∣
p

dx

)1/p (∫ b
a

|g′ (x)|q dx

)1/q
=: I (3.10)

Using (2.15) we have

I ≤
1

b − a

(∫ b
a

∣∣∣∣∣
∫ t
a

f (s) ds −
t − a
b − a

∫ b
a

f (s) ds

∣∣∣∣∣
p

dt

)1/p (∫ b
a

|g′ (x)|q dx

)1/q

≤
1

b − a
b − a

(p − 1)1/p πp

(∫ b
a

∣∣∣∣∣f (t) − 1b − a
∫ b
a

f (s) ds

∣∣∣∣∣
p

dt

)1/p (∫ b
a

|g′ (x)|q dx

)1/q

=
(b − a)1/p

(p − 1)1/p πp

(
1

b − a

∫ b
a

∣∣∣∣∣f (t) − 1b − a
∫ b
a

f (s) ds

∣∣∣∣∣
p

dt

)1/p (∫ b
a

|g′ (x)|q dx

)1/q
(3.11)

for p, q > 1 with 1
p
+ 1

q
= 1, which proves (3.7).

Now, if we put p = q = 2 and take into account that

1

b − a

∫ b
a

∣∣∣∣∣f (t) − 1b − a
∫ b
a

f (s) ds

∣∣∣∣∣
2

dt =
1

b − a

∫ b
a

f2 (t) dt −

(
1

b − a

∫ b
a

f (s) ds

)2
,

then by (3.7) we derive (3.8).

This results can be used to obtain various inequalities by taking particular examples of functions

f and g as follows.

We have the following trapezoid type inequality:

Proposition 3.2. Assume that g : [a, b] → C has an absolutely continuous derivative with g′′ ∈
Lq [a, b] , where p, q > 1 and 1p +

1
q
= 1. Then∣∣∣∣∣g (a) + g (b)2 − 1b − a

∫ b
a

g (t) dt

∣∣∣∣∣ ≤ (b − a)
1+1/p

2 (p − 1)1/p (p + 1)1/p πp

(∫ b
a

|g′′ (t)|q dt

)1/q
. (3.12)

Proof. We use the following identity that can be proved integrating by parts

g (a) + g (b)

2
−

1

b − a

∫ b
a

g (t) dt =
1

b − a

∫ b
a

(
t −

a + b

2

)
g′ (t) dt = C

(
ℓ −

a + b

2
, g′
)
,



CUBO
23, 3 (2021)

Some integral inequalities related to Wirtinger’s result for p-norms 465

where ℓ (t) = t, t ∈ [a, b] .

Using (3.7) we have

∣∣∣∣C
(
ℓ −

a + b

2
, g′
)∣∣∣∣

≤
(b − a)1/p

(p − 1)1/p πp

(
1

b − a

∫ b
a

∣∣∣∣∣t − a + b2 − 1b − a
∫ b
a

(
s −

a + b

2

)
ds

∣∣∣∣∣
p

dt

)1/p (∫ b
a

|g′′ (x)|q dx

)1/q

=
(b − a)1/p

(p − 1)1/p πp

(
1

b − a

∫ b
a

∣∣∣∣t − a + b2
∣∣∣∣p dt

)1/p (∫ b
a

|g′′ (x)|q dx

)1/q

=
(b − a)1+1/p

2 (p − 1)1/p (p + 1)1/p πp

(∫ b
a

|g′′ (x)|q dx

)1/q
,

which proves the desired inequality (3.12).

Let Φ : [m, M] ⊂ R → R be a differentiable convex function on (m, M) and f : [a, b] → [m, M] be
absolutely continuous so that Φ ◦ f, f, Φ′ ◦ f, (Φ′ ◦ f) f ∈ L [a, b] . If f′ ∈ L∞ [a, b] , then we have
the Jensen’s reverse inequality [6]

0 ≤
1

b − a

∫ b
a

(Φ ◦ f) (t) dt − Φ

(
1

b − a

∫ b
a

f (t) dt

)

≤
1

b − a

∫ b
a

(Φ′ ◦ f) (t) f (t) dt −
1

b − a

∫ b
a

Φ′ ◦ f (t) dt
1

b − a

∫ b
a

f (t) dt = C (Φ′ ◦ f, f) . (3.13)

We have the following reverse of Jensen’s inequality:

Proposition 3.3. Let Φ : [m, M] ⊂ R → R be a differentiable convex function on (m, M) and
f : [a, b] → [m, M] be absolutely continuous so that Φ ◦ f, f, Φ′ ◦ f, (Φ′ ◦ f) f ∈ L [a, b] .

(i) If f′ ∈ Lq [a, b] , Φ′ ◦ f ∈ Lp [a, b] with p, q > 1 and 1p +
1
q
= 1, then

0 ≤
1

b − a

∫ b
a

(Φ ◦ f) (t) dt − Φ

(
1

b − a

∫ b
a

f (t) dt

)

≤
(b − a)1/p

(p − 1)1/p πp

(
1

b − a

∫ b
a

∣∣∣∣∣(Φ′ ◦ f) (t) − 1b − a
∫ b
a

(Φ′ ◦ f) (s) ds

∣∣∣∣∣
p

dt

)1/p

×

(∫ b
a

|f′ (t)|q dt

)1/q
. (3.14)



466 S. S. Dragomir CUBO
23, 3 (2021)

(ii) If Φ is twice differentiable and (Φ′′ ◦ f) f′ ∈ Lq [a, b] with p, q > 1 and 1p +
1
q
= 1, then

0 ≤
1

b − a

∫ b
a

(Φ ◦ f) (t) dt − Φ

(
1

b − a

∫ b
a

f (t) dt

)

≤
(b − a)1/p

(p − 1)1/p πp

(
1

b − a

∫ b
a

∣∣∣∣∣f (t) − 1b − a
∫ b
a

f (s) ds

∣∣∣∣∣
p

dt

)1/p

×

(∫ b
a

|(Φ′′ ◦ f) (t) f′ (t)|q dt

)1/q
. (3.15)

The proof follows by Theorem 3.1 for C (Φ′ ◦ f, f) and the inequality (3.13).

We have the following mid-point type inequalities:

Corollary 3.4. Let Φ : [a, b] ⊂ R → R be a differentiable convex function on (a, b) .

(i) If Φ′ ∈ Lp [a, b] with p > 1, then

0 ≤
1

b − a

∫ b
a

Φ (t) dt − Φ
(
a + b

2

)

≤
b − a

(p − 1)1/p πp

(
1

b − a

∫ b
a

∣∣∣∣Φ′ (t) − Φ (b) − Φ (a)b − a
∣∣∣∣p dt

)1/p
. (3.16)

(ii) If Φ is twice differentiable and Φ′′ ∈ Lq [a, b] with p, q > 1 and 1p +
1
q
= 1, then

0 ≤
1

b − a

∫ b
a

Φ (t) dt−Φ
(
a + b

2

)
≤

(b − a)1+1/p

2 (p − 1)1/p (p + 1)1/p πp

(∫ b
a

|Φ′′ (t)|q dt

)1/q
. (3.17)

Acknowledgement

The author would like to thank the anonymous referees for valuable suggestions that have been

implemented in the final version of the manuscript.



CUBO
23, 3 (2021)

Some integral inequalities related to Wirtinger’s result for p-norms 467

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	Introduction
	Some applications for trapezoid inequality
	Inequalities for the Čebyšev functional