Jensen’s type inequalities for the finite Hilbert transform of convex functions

transform of convex functions

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Dragomir, Sever S (2020) Jensen’s type inequalities for the finite Hilbert transform of convex functions. Bulletin of the Transilvania University of Brasov, Series III: Mathematics, Informatics, Physics, 13 (2). pp. 509-520. ISSN 2065- 2151

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https://doi.org/10.31926/but.mif.2020.13.62.2.10

JENSEN’S TYPE INEQUALITIES FOR THE FINITE HILBERT TRANSFORM OF CONVEX FUNCTIONS

Silvestru Sever DRAGOMIR^∗,1

Abstract

In this paper we obtain some new inequalities for the finite Hilbert trans- form of convex functions by the use of Jensen’s integral inequality. Applica- tions for exponential function are provided as well.

2000Mathematics Subject Classification: 26D15; 26D10.

Key words: Finite Hilbert Transform, Convex functions, Inequalities.

1 Introduction

Allover this paper, we consider thefinite Hilbert transformon the open interval (a, b) defined by

(T f) (a, b;t) := 1 πP V

Z b a

f(τ)

τ−tdτ := lim

ε→0+

Z t−ε a

+ Z b

t+ε

f(τ) π(τ −t)dτ for t ∈ (a, b) and for various classes of functions f for which the above Cauchy Principal Value integral exists, see [13, Section 3.2] or [17, Lemma II.1.1].

For several recent papers devoted to inequalities for the finite Hilbert transform (T f), see [2]-[10], [14]-[16] and [18]-[19].

Now, if we assume that the mapping f : (a, b) → R is convex on (a, b), then it is locally Lipschitzian on (a, b) and then the finite Hilbert transform off exists in every pointt∈(a, b).

The following result concerning upper and lower bounds for the finite Hilbert transform of a convex function holds.

1∗Corresponding author, Mathematics, College of Engineering & Science,Victoria University Melbourne, Australia, e-mail: sever.dragomir@vu.edu.au

Theorem 1 (Dragomir et al., 2001 [1]). Let f : (a, b)→ R be a convex function on(a, b) and t∈(a, b).Then we have

1 π

f(t) ln

b−t t−a

+f(t)−f(a) +ϕ(t) (b−t)

(1)

≤(T f) (a, b;t)

≤ 1 π

f(t) ln

b−t t−a

+f(b)−f(t) +ϕ(t) (t−a)

,

where ϕ(t)∈

f₋⁰ (t), f₊⁰ (t)

, t∈(a, b).

Corollary 1. Let f : (a, b)→R be a differentiable convex function on (a, b) and t∈(a, b). Then we have

1 π

f(t) ln

b−t t−a

+f(t)−f(a) +f⁰(t) (b−t)

(2)

≤(T f) (a, b;t)

≤ 1 π

f(t) ln

b−t t−a

+f(b)−f(t) +f⁰(t) (t−a)

.

We observe that if we take t= ^a+b₂ ,then we get from (2) that 1

π

f

a+b 2

−f(a) +1 2f⁰

a+b 2

(b−a)

(3)

≤(T f)

a, b;a+b 2

≤ 1 π

f(b)−f

a+b 2

+1

2f⁰

a+b 2

(b−a)

.

More recently, we have established the following result as well:

Theorem 2 (Dragomir, 2018 [12]). Let f : (a, b) → R be a convex function on (a, b) with finite lateral derivativesf₊⁰ (a) andf−(b). Then fort∈(a, b) we have

1

π(b−a)f₊⁰ (a)≤ 1

π(b−a)f(t)−f(a)

t−a (4)

≤(T f) (a, b;t)−f(t) π ln

b−t t−a

≤ 1

π(b−a)f(b)−f(t) b−t ≤ 1

π (b−a)f−(b). In particular,

1

π(b−a)f₊⁰ (a)≤ 2

π(b−a)f ^a+b₂

−f(a)

b−a (5)

≤(T f)

a, b;a+b 2

≤ 2

π(b−a)f(b)−f ^a+b₂

b−a ≤ 1

π (b−a)f−(b).

In this paper, by the use of Jensen’s celebrated inequality, we obtain some new inequalities for the finite Hilbert transform of convex functions. Applications for exponential function are provided as well.

2 Main Results

We have:

Theorem 3. Letf : (a, b)→Rbe a convex function on(a, b)andx: (a, b)→Ra locally absolutely continuos function on (a, b) with x⁰(s)∈(a, b) for almost every s∈(a, b). Then we have

f 1

b−a

π(T x) (a, b;t)−x(t) ln

b−t t−a

(6)

≤ 1 b−a

πT

Z

a

f ◦x⁰(s)ds

(a, b;t)− Z t

a

f◦x⁰(s)ds

ln b−t

t−a

,

for any t∈(a, b). In particular, f

π

b−a(T x)

a, b;a+b 2

≤ π b−aT

Z

a

f◦x⁰(s)ds a, b;a+b 2

. (7) Proof. Using Jensen’s integral inequality for the convex functionf we have

f

x(τ)−x(t) τ −t

=f Rτ

t x⁰(s)ds τ−t

≤ Rτ

t f◦x⁰(s)ds

τ−t (8)

for any τ, t∈(a, b) withτ 6=t.

Lett∈(a, b) and t−a > ε >0, then by integrating (8) over τ ∈[a, t−ε] we get

Z t−ε a

f

x(τ)−x(t) τ −t

dτ ≤

Z t−ε a

Rτ

t f ◦x⁰(s)ds τ−t

dτ. (9)

Using Jensen’s integral inequality for the convex function f we get f

1 t−ε−a

Z t−ε a

x(τ)−x(t) τ−t dτ

≤ 1 t−ε−a

Z t−ε a

f

x(τ)−x(t) τ −t

dτ. (10) By using (9) and (10) we get

(t−ε−a)f

1 t−ε−a

Z t−ε a

x(τ)−x(t) τ −t dτ

≤ Z t−ε

a

R^τ

t f◦x⁰(s)ds τ −t

dτ (11) fort∈(a, b) andt−a > ε >0.

Let t∈(a, b) and b−t > ε >0, then by integrating (8) over τ ∈[t+ε, b] we obtain

Z b t+ε

f

x(τ)−x(t) τ −t

dτ ≤

Z b t+ε

Rτ

t f◦x⁰(s)ds τ −t

dτ. (12)

Using Jensen’s integral inequality for the convex function f we get f

1 b−t−ε

Z b t+ε

x(τ)−x(t) τ −t

dτ ≤ 1

b−t−ε Z b

t+ε

f

x(τ)−x(t) τ−t

dτ. (13) On utilising (12) and (13) we deduce

(b−t−ε)f

1 b−t−ε

Z _b

t+ε

x(τ)−x(t) τ−t

dτ ≤

Z _b

t+ε

Rτ

t f◦x⁰(s)ds τ−t

dτ (14) fort∈(a, b) and b−t > ε >0.

If we add the inequalities (11) and (14) we get (t−ε−a)f

1 t−ε−a

Z t−ε a

x(τ)−x(t) τ−t dτ

(15) + (b−t−ε)f

1 b−t−ε

Z b t+ε

x(τ)−x(t) τ−t

dτ

≤ Z t−ε

a

Rτ

t f◦x⁰(s)ds τ −t

dτ+

Z _b

t+ε

Rτ

t f◦x⁰(s)ds τ −t

dτ

fort∈(a, b) and min{b−t, t−a}> ε >0.

By the convexity of f we also have (t−ε−a)f

1 t−ε−a

Z t−ε a

x(τ)−x(t) τ −t dτ

+ (b−t−ε)f

1 b−t−ε

Z b t+ε

x(τ)−x(t) τ−t

dτ

≥(b−a−2ε)f

1 b−a−2ε

Z t−ε a

x(τ)−x(t) τ−t dτ+

Z b t+ε

x(τ)−x(t) τ −t

(16) fort∈(a, b) and min{b−t, t−a}> ε >0.

Therefore, by (15) and (16) we get (b−a−2ε)f

1 b−a−2ε

Z t−ε a

x(τ)−x(t) τ −t dτ +

Z b t+ε

x(τ)−x(t) τ −t

(17)

≤ Z t−ε

a

Rτ

t f◦x⁰(s)ds τ −t

dτ+

Z b t+ε

Rτ

t f◦x⁰(s)ds τ −t

dτ

= Z t−ε

a

Rτ

a f◦x⁰(s)ds−Rt

af ◦x⁰(s)ds τ−t

! dτ

+ Z b

t+ε

Rτ

a f◦x⁰(s)ds−Rτ

a f◦x⁰(s)ds τ −t

dτ

fort∈(a, b) and min{b−t, t−a}> ε >0.

Taking the limit over ε→0+ in (17) we get f

1 b−aP V

Z b a

x(τ)−x(t) τ−t dτ

(18)

≤ 1 b−aP V

Z b a

Rτ

a f ◦x⁰(s)ds−Rt

af ◦x⁰(s)ds τ−t

! dτ,

fort∈(a, b),which is an inequality of interest in itself.

As for the mapping ,1(t) = 1,t∈(a, b), we have (T1) (a, b;t) = 1

πP V Z b

a

1 τ −tdτ

= 1 π lim

ε→0+

Z t−ε a

1 τ −tdτ+

Z b t+ε

1 τ −tdτ

= 1 π lim

ε→0+

h

ln|τ −t||^t−ε_a + ln (τ−t)|^b_t+εi

= 1 π lim

ε→0+[lnε−ln (t−a) + ln (b−t)−lnε]

= 1 πln

b−t t−a

, t∈(a, b). Then, obviously, for g: (a, b)→R we have

(T g) (a, b;t) = 1 πP V

Z b a

g(τ)−g(t) +g(t)

τ −t dτ

= 1 πP V

Z b a

g(τ)−g(t)

τ−t dτ+g(t) π P V

Z b a

1 τ−tdτ from where we get the equality

(T g) (a, b;t)−g(t) π ln

b−t t−a

= 1 πP V

Z b a

g(τ)−g(t)

τ−t dτ (19)

for all t∈(a, b).

Using the equality (19) we have P V

Z b a

x(τ)−x(t) τ −t dτ

=π(T x) (a, b;t)−x(t) ln b−t

t−a

and

P V Z b

a

Rτ

a f◦x⁰(s)ds−Rt

af◦x⁰(s)ds τ −t

! dτ

=π

T Z

a

f ◦x⁰(s)ds

(a, b;t)− Z t

a

f◦x⁰(s)dsln b−t

t−a

and by (18) we get the desired result (6).

We also have the reverse inequality:

Theorem 4. Let f : [a, b]→ R be a convex function on [a, b] and x : [a, b]→R an absolutely continuos function on [a, b] with x⁰(s)∈ [a, b] for almost every s∈ (a, b). Then we have

1 b−a

πT

Z

a

f◦x⁰(s)ds

(a, b;t)− Z t

a

f◦x⁰(s)ds

ln

b−t t−a

(20)

≤ f(a) b−a

b− 1 b−a

π(T x) (a, b;t)−x(t) ln b−t

t−a

+ f(b) b−a

1 b−a

π(T x) (a, b;t)−x(t) ln b−t

t−a

−a

for any t∈(a, b). In particular,

π b−aT

Z

a

f◦x⁰(s)ds a, b;a+b 2

(21)

≤ f(a) b−a

b− π

b−a(T x)

a, b;a+b 2

+ f(b) b−a

π

b−a(T x)

a, b;a+b 2

−a

.

Proof. By the convexity of f we have f x⁰(s)

= f

(b−x⁰(s))a+ (x⁰(s)−a)b b−a

(22)

≤ b−x⁰(s)

b−a f(a) +x⁰(s)−a b−a f(b) for almost everys∈[a, b].

Let t, τ ∈[a, b] witht6=τ.Taking the integral mean _τ−t¹ Rτ

t ,which is a linear and isotonic functional, in the inequality (22), we get

1 τ−t

Z τ t

f x⁰(s)

ds (23)

≤ 1 τ−t

Z τ t

b−x⁰(s)

b−a f(a) + x⁰(s)−a b−a f(b)

ds

= b−_τ−t¹ Rτ

t x⁰(s)ds

b−a f(a) +

1 τ−t

Rτ

t x⁰(s)ds−a b−a f(b)

= b−^x(τ)−x(t)_τ−t

b−a f(a) +

x(τ)−x(t)

τ−t −a

b−a f(b)

= f(a) b−a

b−x(τ)−x(t) τ −t

+ f(b) b−a

x(τ)−x(t) τ −t −a

for any t, τ ∈[a, b] with t6=τ.

Lett∈(a, b) andt−a > ε >0, then by integrating (23) overτ ∈[a, t−ε] we get

Z t−ε a

R^τ

t (f◦x⁰) (s)ds τ−t

dτ (24)

≤ Z t−ε

a

f(a) b−a

b− x(τ)−x(t) τ −t

+ f(b) b−a

x(τ)−x(t) τ−t −a

dτ

= f(a) b−a

b(t−ε−a)− Z t−ε

a

x(τ)−x(t) τ −t dτ

+ f(b) b−a

Z t−ε a

x(τ)−x(t)

τ −t dτ−a(t−ε−a)

.

Let t ∈ (a, b) and b−t > ε > 0, then by integrating (23) over τ ∈ [t+ε, b] we obtain

Z b t+ε

Rτ

t (f◦x⁰) (s)ds τ −t

dτ (25)

≤ Z b

t+ε

f(a) b−a

b−x(τ)−x(t) τ−t

+ f(b) b−a

x(τ)−x(t) τ −t −a

dτ

= f(a) b−a

b(b−t−ε)− Z b

t+ε

x(τ)−x(t) τ−t dτ

+ f(b) b−a

Z b t+ε

x(τ)−x(t)

τ −t dτ −a(b−t−ε)

.

Fort∈(a, b) and min{b−t, t−a}> ε >0,we have by adding the inequalities (24) and (25) that

Z t−ε a

Rτ

t (f ◦x⁰) (s)ds τ −t

dτ+

Z b t+ε

Rτ

t (f◦x⁰) (s)ds τ −t

dτ (26)

≤ f(a) b−a

b(t−ε−a)− Z t−ε

a

x(τ)−x(t) τ−t dτ

+b(b−t−ε)− Z b

t+ε

x(τ)−x(t) τ −t dτ

+ f(b) b−a

Z t−ε a

x(τ)−x(t)

τ −t dτ−a(t−ε−a) +

Z b t+ε

x(τ)−x(t)

τ −t dτ−a(b−t−ε)

.

If we take ε→0+ in (26), then we obtain P V

Z b a

R^τ

t (f ◦x⁰) (s)ds τ−t

dτ

≤ f(a) b−a

b(t−a) +b(b−t)−P V Z b

a

x(τ)−x(t) τ −t dτ

+ f(b) b−a

P V

Z b a

x(τ)−x(t)

τ −t dτ−a(t−a)−a(b−t)

= f(a) b−a

b(b−a)−P V Z _b

a

x(τ)−x(t) τ −t dτ

+ f(b) b−a

P V

Z b a

x(τ)−x(t)

τ −t dτ−a(b−a)

,

which produces the following inequality of interest P V

Z b a

Rτ

t (f◦x⁰) (s)ds τ −t

dτ ≤f(a)

b− 1 b−aP V

Z b a

x(τ)−x(t) τ −t dτ

(27) +f(b)

1 b−aP V

Z b a

x(τ)−x(t) τ−t dτ−a

,

for any t∈(a, b).Now, since P V

Z b a

x(τ)−x(t) τ −t dτ

=π(T x) (a, b;t)−x(t) ln b−t

t−a

and

P V Z b

a

Rτ

a f◦x⁰(s)ds−Rt

af◦x⁰(s)ds τ −t

! dτ

=π

T Z

a

f ◦x⁰(s)ds

(a, b;t)− Z t

a

f ◦x⁰(s)dsln b−t

t−a

,

hence by (27) we get π

T

Z

a

f◦x⁰(s)ds

(a, b;t)− Z t

a

f◦x⁰(s)dsln b−t

t−a

≤f(a)

b− 1 b−a

π(T x) (a, b;t)−x(t) ln b−t

t−a

+f(b) 1

b−a

π(T x) (a, b;t)−x(t) ln

b−t t−a

−a

,

which proves (20).

3 Examples

If we consider the function expt=e^t,t∈(a, b) a real interval, then (Texp) (a, b;t) = exp (t)

π [E_i(b−t)−E_i(a−t)], (28) where Ei is defined by

E_i(x) :=P V Z x

−∞

exp (s)

s ds, x∈R.

Indeed, we have

E_i(b−t)−E_i(a−t) =P V Z b−t

a−t

exp (s)

s ds=P V Z b

a

exp (τ −t) τ −t ds

= exp (−t)π(Texp) (a, b;t) and the equality (28) is proved.

Let f : (a, b) → R be a convex function on (a, b). Then we have by (6) and (20) for x(t) = expt, that

f

exp (t) b−a

E_i(b−t)−E_i(a−t)−ln b−t

t−a

(29)

≤ 1 b−a

πT

Z

a

f ◦exp (s)ds

(a, b;t)− Z t

a

f◦exp (s)ds

ln b−t

t−a

≤ f(a) b−a

b−exp (t) b−a

Ei(b−t)−Ei(a−t)−ln b−t

t−a

+ f(b) b−a

exp (t) b−a

E_i(b−t)−E_i(a−t)−ln b−t

t−a

−a

fort∈(a, b).

In particular, we have f exp ^a+b₂

b−a

E_i

b−a 2

−E_i

−b−a 2

!

(30)

≤ 1 b−a

πT

Z

a

f ◦exp (s)ds a, b;a+b 2

≤ f(a) b−a

"

b−exp ^a+b₂ b−a

Ei

b−a 2

−Ei

−b−a 2

#

+ f(b) b−a

"

exp ^a+b₂ b−a

Ei

b−a 2

−Ei

−b−a 2

−a

# .

If we take f :R→ R,f(x) = exp (αx), α6= 0,then for an absolutely contin- uous functionx: [a, b]⊂R→R we have from (6) and (20) that

exp α

b−a

π(T x) (a, b;t)−x(t) ln b−t

t−a

≤ 1 b−a

πT

Z

a

exp αx⁰(s) ds

(a, b;t)− Z t

a

exp αx⁰(s) ds

ln

b−t t−a

≤ expa b−a

b− 1

b−a

π(T x) (a, b;t)−x(t) ln b−t

t−a

+expb b−a

1 b−a

π(T x) (a, b;t)−x(t) ln b−t

t−a

−a

(31) for any t∈(a, b).

In particular, exp

α b−a

π(T x)

a, b;a+b 2

≤ 1 b−a

πT

Z

a

exp αx⁰(s)

ds a, b;a+b 2

≤ expa b−a

b− π

b−a(T x)

a, b;a+b 2

+expb b−a

π

b−a(T x)

a, b;a+b 2

−a

.

(32)

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