Inequalities for eigenfunctions of the p-Laplacian Текст научной статьи по специальности «Математика»

Issues of Analysis

Vol. 2(20), No. 1, 2013

B. A. BHAYQ, M. VUQRINEN

INEQUALITIES FOR EIGENFUNCTIONS OF THE

P -LAPLACIAN

Abstract. Motivated by the work of P. Lindqvist, we study eigenfunctions of the one-dimensional p-Laplace operator, the sinp functions, and prove several inequalities for these and p-analogues of other trigonometric functions and their inverse functions. Similar inequalities are given also for the p-analogues of the hyperbolic functions and their inverses.

Key words: eigenfunctions of p-Laplacian, sinp, generalized

trigonometric function.

2010 Mathematical Subject Classification: 33C99, 33B99.

§ 1. Introduction

In a highly cited paper P. Lindqvist [1] studied generalized trigonometric functions depending on a parameter p > 1 which for the case p = 2 reduce to the familiar functions. Numerous other authors, see e.g. [2], [3, 4], [5], [6], [7] and the bibliographies of these papers, have extended this work in various directions including the study of generalized hyperbolic functions and their inverses. Our goal here to study these p-trigonometric and p-hyperbolic functions and to prove several inequalities for them.

For the statement of some of our main results we introduce some notation and terminology for classical special functions, such as the classical gamma function r(x), the psi function -0(x) and the beta function B(x,y). For Rex > 0, Rey > 0, these functions are defined by

r(x)=r ^dt, «x)=n • b (x,y)=rx+yf

© Bhayo B. A., Yuorinen M., 2013

Given complex numbers a, b and c with c = 0, —1, — 2,..the Gaussian hypergeometric function is the analytic continuation to the slit place C \ [1 , to) of the series

F(a, b; c; z) = 2F1(a,b; c; z) = ^ (a,(n)(b>n) z__, ^ < L

(c, n) n!

n=0

Here (a, 0) = 1 for a = 0, and (a, n) is the shifted factorial function or the Appell symbol

(a, n) = a(a + 1)(a + 2) • • • (a + n — 1)

for n E N \ {0}, where N = {0,1, 2,..The hypergeometric function has numerous special functions as its special or limiting cases, see [8].

We start by discussing eigenfunctions of the so-called one-dimensional p-Laplacian Ap on (0,1), p E (1, to). The eigenvalue problem [5]

I

—Apu = — (\u \p-2u j = A\u\p-2u, u(0) = u(1) = 0, has eigenvalues

A_ = (p — 1)(n^p)p,

and eigenfunctions

sinp(nnp t), n E N,

where sinp is the inverse function of arcsinp, which is defined below and (cf. [7])

2 fLf_ , 2 „A 1 1\ 2n

np —

. (1 — s)-1/ps1/p-1ds = - b( 1 — 1, ^ = ,

p J 0 p \ p p J p sin(n/p)

Motivated by P. Lindqvist’s work, P. J. Bushell and D. E. Edmunds [9] found recently many new results for these generalized trigonometric functions. Some authors also considered various other p-analogues of trigonometric and hyperbolic functions and their inverses. In particular, they considered the following homeomorphisms

sinp : (0, ap) — I, cosp : (0, ap) — I, tanp : (0, bp) — I, sinhp : (0, cp) — I, tanhp : (0, to) — I,

where I = (0,1) and

- ’ 1 — ^2-^ =2~i/pF(1 1 i ■ 1 1

1\ 1/p^(. 1 „ 11

ap = ^,bp = — — tf —l = 2-1/pFl ; 1 + -

p 2 p 2p \ I 2p ) \2p) ) \p p p 2

Cp F I1 ’p;1 + p ;2

For x E I, their inverse functions are defined as

x /1 1 1 \

arcsinpx = f(1 — tp)-1/pdt = xF[ ; 1+—; xp I =

o \p p p )

= x(1 — xp)(p-1)/pF ^1,1; 1 + 1 xp^ ,

x / 1 1 \

arctanpx = f (1 + tp)-1dt = xF I 1, -; 1 ■—; —xp) ,

o V p p )

X / 1 1 1 \

arsinhp x = f(1 + tp)-1/pdt = xF I - , -; 1 ■—; —xp) ,

0 V p p p )

x { 1 1 \

artanhp x = f (1 — tp )-1 dt = xF I 1 , -; 1 ■—; xp ) ,

o V p p J

and by [9, Prop. 2.2] arccosp x = arcsinp((1 — xp)1/p). See also S. Takeuchi

[10] for two-parameter generalizations. For the particular case p = 2 one obtains the familiar elementary functions.

The paper is organized into sections as follows. Section 1, the introduction, contains the statements of our main results. In Section 2 we give some inequalities for the p-analogues of trigonometric and hyperbolic functions. Section 3 contains the proofs of our main results and some identities. Finally in Section 4 we give some functional inequalities for elementary functions and Section 5 contains two small tables with a few values of the function sinp and related functions compiled with the Mathematica® software.

Some of the main results are the following theorems.

Theorem 1. For p > 1 and x E (0,1), we have

xp P . nr.

1) ( 1 ■-----—^---- ) x < arcsinp x < — x,

p(1 ■ p) p 2

f 1 xp \ n

2) I 1 +—(1+—) J (1 — xp)1/p < arccosp x < -2 (1 — xp)1/p,

1/p

, (p(1 + p)(1 + xp) + xp )x 1/pi (

3) —t;------rz;-/— < arctanp x < 2 /p bp I

p(1+ p)(1 + xp)1+1/p p p V

Theorem 2. For p > 1 and x E (0,1), we have

x

z 1 ■-----------------< arsinhp x < z 1 ■— log(1 + xp)

1 ■ p p p

x

z =

( xp \

V 1 + xp)

P 1/p

(1)

x ^ 1 — 1 + log(1 — xp^ < artanhp x < x ^ 1 — - log(1 — xp^ . (2)

The next result provides several families of inequalities for elementary functions.

Theorem 3. For x > 0 and z = nx/2, the function g(p) = f (zp)1/p is decreasing in p E (0, to), where

f (z) E jarcsin(2z/n), arcosh(z), artanh(2z/n)}.

Acknowledgments. The first author is indebted to the Graduate School of Mathematical Analysis and its Applications for support. The second author was, in part supported by the Academy of Finland, Project 2600066611. Both authors wish to acknowledge the expert help of Dr. H. Ruskeepaa in the use of the Mathematica® software [11] and Prof. P. Hasto for providing simplified versions of some of our proofs. Authors thank to anonymous referee for his valuable comments.

§ 2. Preliminaries and definitions

For convenience, we use the notation R+ = (0, to) .

Lemma 1. [13, Thm 2.1] Let f : R+ ^ R+ be a differentiable, log-convex function and let a > 1. Then g(x) = (f (x))a/f (ax) decreases on its domain. In particular, if 0 < x < y, then the following inequalities

(f (y))a < (f(x))a < (f(0))a-1

f(ay) < f(ax) <

hold true. If 0 < a < 1, then the function g is an increasing function on R+ and inequalities are reversed.

For easy reference we recall the following identity [8, 15.3.5]

F(a, b; c; z) = (1 — z)-bF(b, c — a; c; —z/(1 — z)). (3)

For the following lemma see [15, Thms 1.19(10), 1.52(1), Lems, 1.33,

1.35].

Lemma 2.

1) For a,b,c> 0, c < a + b, and \x\ < 1,

F(a, b; c; x) = (1 — x)c-a-bF(c — a,c — b; c; x).

2) For a,x E (0,1), and b,c E (0, to)

ab

F(—a, b; c; x) < 1------x .

c

3) For a,x E (0,1), and b,c E (0, to)

F(a, b; c; x) + F(—a, b; c; x) > 2 .

4) Let a,b,c E (0, to) and c> a + b. Then for x E [0,1],

f (“,b; c; x) < r(c)r(c)r;—b) ■

r(c — a)r(c — b)

5) For a, b > 0, the following function

F(a, b; a ■ b; x) — 1 f (x) = log(1/(1 — x))

is strictly increasing from (0,1) onto (ab/(a + b), 1/B(a, b)). Lemma 3. For p > 1 and x E (0,1), the functions

(arcsinp(xk ))1/k and (artanhp(xk ))1/k

are decreasing in k E (0, to), also

(arctanp(xk))1/k and (arsinhp(xk ))1/k

are increasing in k E (0, to).

In particular, for k > 1

karcsinp(xk) < arcsinp(x) < (arcsinp\fx)k ,

Let

We get

kartanhp(xk) < artanhp(x) < (artanhp tyx)k . (arsinhp \fx)k < arsinhp(x) < ^arsinhp(xk), (arctanp yx)k < arctanp(x) < ^arctanp(xk).

rx

f (k) := (E(xk))1/k,E(x) := I g(t) dt,E = E(xk).

J 0

f' = —E1/k log E-1 + 1E1/k-1 E'xk log x =

E1/k “k^

k2 k

E

E'

{-log— e — Vlogxk

If g > 1, then

E 1 fx

~k = t g(t) dt >1

kk

x x J 0

If g is increasing, then

&

E1

E'— xk = g(xk) — xk I g(t) dt > 0

x0

x

so that xkEE — 1 > 0. Thus f' < 0 under these assumptions.

For arcsinp and artanhp, g is (1 — tp)-1/p and (1 — tp)-1, so the conditions are clearly satisfied. Additionally, we see that for arsinhp and arctanp the conditions g < 1 and g is decreasing and this conclude that f' > 0. This completes the proof.

Theorem 4. For p > 1 and r,s E (0,1), the following inequalities hold

1) arcsinp(rs) < -y/arcsinp(r2) arcsinp(s2) < arcsinp(r) arcsinp(s),

r, s E (0, 1) ,

2) artanhp(rs) < \Jartanhp(r2) artanhp(s2) < artanhp(r) artanhp(s), r, s E (0, 1) ,

3) arsinhp(r2) arsinhp(s2) < arsinhp(r2) arsinhp(s2) < arsinhp(rs),

4) artanhp(r) artanhp(s) < \Jartanhp(r2) artanhp(s2) < artanhp(r s).

Let h(x) := log f (ex). Then h is convex (in the C2 case) when h'' > 0, i. e. if

f (f' + yf'') > (f ')2,

y

where y = ex and the function is evaluated at y. If f'' > 0, then

- > f'(0), y

so a sufficient condition for convexity is f'(0)(f' + yf'') > (f')2. If f'' < 0, the reverse holds, so a sufficient condition for concavity is f '(0)(f'+yf'') <

< (f')2. Suppose

f (x) := / g(t) dt.

0

Then f' = g and f'' = g'. Then one easily checks that h is convex in case g is (1 — tp)-1/p and (1 — tp)-1, and concave for g equal to (1 + tp)-1/p and (1 + tp)-1. Now proof follows easily from Lemma 3.

Lemma 4. For k, p > 1 and r > s, we have

arcsm^A* < arcsinp(.k) ^,s € (0,1),

arcsinp (r) J arcsinp(rk)

/ artanhp(s^ k < artanhp(sk) e (0 1)

\artanhp(r) J ~ artanhp(rk)’ ’ ’ ’

arsinhp(sfe) < / arsinhp(s^X k r s e (0 1) arsinhp(rk) _ \arsinhp (r) J ’ ’ ’

For x > 0, the following functions

u(x) = arcsinp(e-x), v(x) = artanhp(e-x),

w1(x) = 1/arsinhp(e-x)

are log-convex by the proof of Theorem 4. Let x < y, e-x = r > s = e-y, now inequalities follow from Lemma 1.

Lemma 5. [17, Thm 2, p.151] Let J C R be an open interval, and let f : J ^ R be strictly monotonic function. Let f-1 : f (J) ^ J be the inverse to f then

1) if f is convex and increasing, then f-1 is concave,

2) if f is convex and decreasing, then f-1 is convex,

3) if f is concave and increasing, then f-1 is convex,

4) if f is concave and decreasing, then f-1 is concave.

Lemma 6. For k,p > 1 and r > s, we have

sinp(rnk < s:%(rk), r,se (0,1),

(

p p(

sinp(s)/ sinp (sk)

^tanhp(r) \k ^ tanhp(rk)

/tanhp (r)\ tanhp (rk)

ytanh^s)/ _ tanhp (sk)

/sinhp (r) \ k sinhp (rk)

Vsinhp(s) 7 > sinhp(sk), r, s E (0,1).

Inequalities reverse for k E (0,1).

It is clear from the proof of Theorem 4 that the functions

f (x) = log(arcsinp(e-x)), g(x) = log(artanhp(e-x)),

h(x) = log(1/arsinhp (ex)) are convex and decreasing, then Lemma 5(2) implies that

f-1(y) = log(1/sinp(ey)), g-1(y) = log(1/tanhp(ey)),

h-1 (y) = log(sinhp (e-y)), are convex, now the result follows from Lemma 1.

Theorem 5. For p> 1, the following inequalities hold

1)\Jsinp(r2) sinp(s2) < sinp(rs), r, s E (0,np/2),

2)y^tanhp(r2) tanhp(s2) < tanhp(rs), r, s E (0, to),

3)sinhp(rs) < sinhp(r2) sinhp(s2), r, s E (0, to).

Let f (z) = log(arcsinp(e 2)), z > 0. Then

f' (z) = —(1 — e-pz)-1/p/F(1/p, 1/p; 1 + 1/p; e-pz) < 0,

f is decreasing and by the proof of Theorem 4 f is convex. By Lemma 5(2), f-1 (y) = log(1/ sinp(ey)) is convex. This implies that

log Up (ex1/* e.y/2)) < 2 (loKsLn^^) + loKsinpW^ ,

letting r = ex/2 and s = ey/2, we get the first inequality.

For (2), let g(z) = log(artanhp(e-z)), z > 0 and

g (z) = —1/((1 — e-pz)F(1,1/p; 1 + 1/p; e-pz)) < 0,

hence g is decreasing and by Theorem 4 g is convex. Then g-1 (y) = = log(1/artanhp(ey)) is convex by Lemma 5(2), and (2) follows. Finally, let h1 (z) = log(1/arsinhp(ez)) and

epz

h1(z) = —1/F y1,1/p;1 + 1/p; 1 + epz) < 0.

Then h-1(y) = log(sinhp(e-y)) is decreasing and convex by Lemma 5(2). This implies that

log(1/sinhp(e-x/2e-y/2)) < (log(1/sinhp(e-x)) + log(1/sinhp(e-y)))/2, and (3) holds for r,s E (0, to). Again h2(z) = log(1/arsinhp(e-z)) and

h2(z) = (F(1,1/p;1 + 1/p;1/(1 + epz)))-1 > 0, similarly proof follows from Lemma 5(2), this completes the proof of (3). Lemma 7. For p > 1, the following relations hold

1) v/sinp(r) sinp(s) < sinp((r + s)/2), r, s E (0,np/2),

2) y/sinhp(r) sinhp(s) < sinhp((r + s)/2), r, s E (0, to) .

The proof follows easily from Lemma 5 and 2\JYs < r + s since the functions are increasing.

Lemma 8. For p> 1, the following inequalities hold

1) sinp(r + s)/2 < (sinp(r) + sinp(s))/2 , r, s E (0,np/2),

2) tanhp(r + s)/2 < (tanhp(r) + tanhp(s))2 , r, s E (0,bp/2),

3) tanp(r + s)/2 > (tanp(r) + tanp(s))2 , r, s E (0,bp/2),

4) sinhp(r + s)/2 > (sinhp(r) + sinhp(s))2 , r, s E (0, to) .

Let f (x) = arcsinp(x), x E (0,ap). We get

f' (x) = (1 — xp)-1/p ,

which is increasing, hence f is convex. Clearly, f is increasing. Therefore

f1 = f-1(y) = sinp (y)

is concave by Lemma 5(1). This implies that f1 is decreasing. Clearly f1(0) = 0, and by [15, Thm 1.25], f1(y)/y is decreasing. Now it follows from [15, Lem 1.24] that

f1(r + s) < f1 (r) + f1(s), and (1) follows. The proofs of the remaining claims follow similarly.

§ 3. Proof of main results

In this section we give the proof of our main results and few more relations.

Proof of Theorem 1. By Lemma 2(3), (2) we get

xp \ f 11 1

2 — ( 1 (1+ ) ) < F [~, ~ x p (1 + p) p p p

and the first inequality of part one holds. For the second one we get

1 1 1

arcsinp x = x F I -, -; 1 +—; xp } < p p p x r(1 + 1/p)r(1 + 1/p — 1/p — 1/p)

< r(1 + 1/p — 1/p)r(1 + 1/p — 1/p)

= x r f 1 + 1 \ r (1 — 1 \ = x 1B (1 — 1,1 \ = x —

\ pj \ pj p V p pJ 2

by Lemma 2(4). By [9, Prop. (2.11)], arccosp x = arcsinp ((1 — xp)1/p), and (2) follows from (1). For (3), if we replace b = 1, c — a = 1/p, c = = 1 + 1/p, xp = z/(1 — z) in (3) then we get

arctanpx = xF (1, -; 1 +—; —xp) =

( p\ ( p \

x V (1, 1; 1 + x'

1 + xp p 1 + xp

x \ ( 1 \1/p-1 ^ (11 1 xp

F - -; 1 ^ -;

< 21/p b

1 + xp 1 + xp p p p 1 + xp

xp \ 1/p ( 1 1 1 xp \

1+ p 1 F \~i~; 1 + ; 1+ p 1 <

1 + xp p p p 1 + xp

p \ 1/p

x

p 1 + xp

third identily and inequality follow from Lemma 2(1), (4). For the lower bound we get

arctanpx > (^(2 — F (1,1;1 + 1; >

>

1 + xp p p p 1 + xp

(p(1 + p)(1 + xp) + xp )x p(1 + p)(1 + xp)1+1/p

from Lemma 2(3),(2). □

Proof of Theorem 2. For (1), we replace b = 1/p, c — a = 1/p, c = = 1 + 1/p and xp = z/(1 — z) in (3) and see that

arsinhp x = x F ( - , — ; 1 +— ; —xp ) =

p p p

xp \ 1/p ( 1 1 xp

Now we get

, F 1, -; 1 +—;

1 + xp p p 1 + xp

log (1 + xp) f xp \ 1/p <

1 + p 1 + xp

<xF{p,1p ;1 + p; —x'p) < I1" plo^1 — 1+xp)) (r+xp)

from Lemma 2(5) and observing that B(1,1/p) = p, this implies (1).

For (2) we get from Lemma 2(5)

1 log (+1 <F (1,1; 1 + 1; x^ < 1 log () +1,

1 + p 1 — xp p p p 1 — xp

which is equivalent to

x ^ 1 — 1 + log(1 — xp )^ < x F ^1 ,-;l + - ; xp^ <

< x ^ 1 — 1 log(1 — xp)^ ,

and the result follows. □

Remark. For the particular case p = 2. Zhu [18] has proved for x > 0

6 v/2( V1 + x2 — 1)1/2

4+ v/2^1 + x2 + 1)1/2

< arsinh(x).

When p = 2, our bound in Theorem 2(1) differs from this bound roughly

0.01 when x E (0, 1).

Lemma 9. For p > 1 and x E (0,1), the following inequalities hold:

1) arctanp(x) < arsinhp(x) < arcsinp(x) < artanhp(x),

2) tanhp (z) < sinp (z) < sinhp (z) < tanp(z),

the first and the second inequalities hold for z E (0,np/2), and the third one holds for z E (0, bp).

From the definition of the p-analogues functions we get (1), and (2) follows from (1).

Lemma 10. For p > 1, we have

6p2 12p2 2n

3p2 — 2 _ p _ 6p2 — n2 , p p sin(n/p)

By [19, Thm 3.1] we get

n ( n2 \ (n \ n ( 2

p v1 — 6p^J - siHp) - p I1 — sp2

and the result follows easily.

Lemma 11. For a E (0,1) and k,r,s E (1, to) , the following inequalities hold

1) nrs — \Jnr2 ns2 — \Jnr ns ,

2) nra si—a — anr + (1 — a)ns ,

ra s

k

ns \ _ n sk

3) ^ ^, r — s.

\ ^r J nrk

Let f (x) = log(nex), x > 0. We get

f (x) = e-2xn2(csc (e~xn))2 — e-xn cot (e-xn) ,

which is positive, because the function g(y) = y2(csc(y))2 — y cot(y) is positive. This implies that f is convex. Hence

log(ne(x+y)/2 ) — 2 (log(nex) + log(ney)) ,

setting r = ex/2 and r = ey/2, we get the first inequality of (1), and the

second one follows from the fact that np is decreasing in p E (1, to). Now

it is clear that nex is convex, and we get

nea x+(1 — a)y — anex + (1 — a)ney ,

and (2) follows easily. Let 0 — x — y, then we get

(ney )k < (nex )k

nek y nek x

from Lemma 1, and (3) follows if we set r = ex and r = ey.

Lemma 12. For p > 1 and x E (0,1), we have

x

p 1 + xp ,

arcsinp ( 1 = arctanp(x),

x

arcsinp(x) = arctan.

P 1 p 1 — xp J ’

V1 — xp '

arccosp(x) = arctanp

x

1

p 1+ xp

arccosp I 1 = arctanp(x).

We get

arctanp(x) = xF (-,-; 1 +—; —xp) =

\P P P )

x ^ 1 xp

F ( 1; 1 + ;

1 + xp\ p 1 + xp

x ( 1 \1/p_1 ^ (1 1 , 1 xp

F ; 1 + -;

1 + xp V 1 + xp J \P P P 1 + xp

x \ 1/p ( 11 1 ( x

F - -; 1 + -;

1 + xp J \p p p V(1 + xp)1/p

x

arcsinp

p 1 + xp

by (3) and Lemma 2(1). Write y = x/ p 1 — xp, and second follows from first one. For the third identity, we get

arctanp ^ ^ — x ^ = x P1 — xpF (1, 1; 1 + 1; (1 — xp)^ =

x ) \p p p

= arcsinp((1 — xp)1/p) = arccosp(x)

by (3), Lemma 2(1) and [9, Prop. 2.2]. Similarly, the fourth identity

follows from third one.

Conjecture 1. For a fixed x E (0,1), the functions

sinp (np x/2), tanp (np x/2), sinhp (cp x)

are monotone in p E (1, to). For fixed x > 0, tanhp(x) is increasing in

p E (1, to).

§ 4. Some relations for elementary functions

In this section we give several inequalities involving the elementary functions in form of Lemmas.

Lemma 13. For x E (0,1) and z E (0, to), the following functions f1(k) = sin(xk)1/k , f2 (k) = cos(xk)1/k , fs (k) = tanh(zk)1/k , are increasing in (0, to).

We get

f (k) = (xkcot(xk) log(xk) — log(sin(xk)))sin(xk)1/k/k2, which is positive because

h1 (y) = y cot(y) log(y) — log(sin(y)) > 0 •

For f2 we get

f2 (k) = — (xk tan(xk) log(xk) + log(cos(xk))) cos(xk)1/k/k2, which is positive because the function

h2 (y) = y tan(y) log(y) + log(cos(y)) < 0

For f3 we get

fS(k) = tanhkf " (2zk log(zk)/sinh(2zk) — log(tanh(zk)))•

Let h3(y) = 2y log(y)/sinh(2y) — log(tanh(y)), y = zk E (0, to)^ Clearly hs (y) > 0 for y > 1. For y E (0,1) we see that h3 (y) > 0 if

2y log(y) < 1

sinh(2y) log(tanh(y))

which holds because y > tanh(y). In conclusion, f3 (k) > 0 for all z E (0, to).

Lemma 14. The following inequalities hold

1) Y^arccos(r2)arccos(s2) < arccos(rs) ,r,s E (0,1),

2) arctan(r)arctan(s) < \Jarctan(r2)arctan(s2) < arctan(rs), for r, s E (0,1),

3) Y^arcosh(r2) arcosh(s2) < arcosh(rs); r,s E (1, to) •

For (1) we let f (x) = log(arccos(e_x)) ,x> 0, and get

." ( ) \Je2x — 1 + e2xarccos(e_x) 0

(e2x — 1)3/2 arccos2 (e_x) < ,

hence f is concave, and the inequality follows.

For (2) we define g(x) = log(arcsin(e x)) ,x> 0 and obtain ^ ex ((e2x — 1) tan-1 (e-x) — ex)

g (x) = —------------L2-----r——- < 0 ,

(e2x + 1)2 tan-1 (e-x)2

because y < tan(y/(1 — y2)) for y E (0,1), hence g is concave. Therefore the first inequality of (2) follows and the second one follows from Lemma 13. Finally we define h(x) = log(arcosh(ex)), x > 0 and get

h" ( ) ex(exVe2x — 1 — arcosh(ex)) 0

(e2x — 1)3/2 arcosh2(ex)

This implies the proof of (3).

Lemma 15. For r,s E (0, to), we have

1) cosh(rs) < cosh(r2) cosh(s2) < cosh(r) cosh(s),

here second inequality holds for r, s E (0,1),

2) tanh(r) tanh(s) < \Jtanh(r2) tanh(s2) < \Jtanh(r2 s2) •

For (1) we let g1 (x) = log(cosh(e-x)) and g2(x) = log(cosh(ex)), x > 0, and we get

g1 (x) = e-2x(1/(cosh2(e-x)) + ex tanh(e-x)) > 0,

g’2 (x) = ex (ex/(cosh2 (ex)) + tanh(ex)) > 0,

hence g1 and g2 are convex, and the first inequality of (1) holds, and its second inequality follows from Lemma 13. The firstinequality of (2) follows from Lemma 13. For the second one let h1 (x) = log(tanh(e-x)), x > 0 and get

e-2x (—csch2 (e-x) + 2excsch (2e-x) — sech2 (e-x))

which is negative, hence h1 is concave. Again, let h2 (x) = log(tanh(ex))

and get

—ex (excsch2 (ex) — 2csch (2ex) + exsech2 (ex)) < 0 •

This implies that h2 is also concave, and the second inequality of (2) holds for r, s E (0, to).

Lemma 16. For y E (0,1), we have

2 y cot (f) log y < log (sin (f)) , (4)

y coth (y) log y < log (sinh (y)) , (5)

log (tan (f)) > log(y)csc( f) sec (f). (6)

Let f (y) = 2 y cot (ny) log y — log (sin (ny)) • We get

f' (y) = 2 cot (n2y ) log y — 4 yn2 csc2 ( n2y ) log y =

n .(ny\ i 1 2 2(ny

2coH Wlog y — 4 yn csM^

n log(y-1) (ny______1_______cos(ny/2A

2 V 2 sin2(ny/2) sin(ny/2) J

n log(y-1) (ny • (ny) (ny

— sin —— cos

2 sin2 (ny/2) ( 2 V 2 J V 2

n log(y-1) (ny sin(ny)

2 sin (ny/2) \ 2 2

This is positive because x > sin x for x E (0, 2n), and f (1) = 0 and this completes the proof. Next, let

g(y) = ycoth (y) log y — log (sinh (y)) •

We get

g (y) = (y — sinh(y) cosh(y)) < 0,

log(1/y)

sinh2(y)

because sinh x > x/ cosh x for x > 0. Moreover, g tends to zero when y tends to zero and this implies the proof of (5). Next, let

h(y) = log (tan (y)) — 2 y log(y) csc (?)sec (?

We see that

h’(y) = — ylog(y)sec2 (Y) + 1 ^ylog(y)csc2 (Y)

— ^csc (f) sec( f) =

= n log I — ) csc2 (ny)(sin(ny) — ny cos(ny)) < 0, y

because x < tanx for x E (0,1). Hence h is increasing and tends to log( n/2) when y tends to zero and this implies the proof.

Lemma 17.

1) The function

H (y) = r-log (yy \cot (?)— log (csc ( t ))

is decreasing from (0,1) onto (0, log(n/2)).

2) The function

G(y) = log (<»sh(y)) — 2ny log(y) tan^n2y)

is increasing from (0,1) onto (0, n log(cosh(n/2))/2).

We get

H' (y) = — Jcsc2 ( y) (n log (y-y) + log(y) sin(ny)) =

TT / ny \

= — 4csc VY)( ny log(1/y) — sin(ny)log(1/y)),

which is positive. Next,

G (y) = — ^ log(y) tan^n2y) — 4n2y log(y)sech^^ > 0 , and the limiting values follow easily.

Lemma 18. The following function is increasing from (0,1) onto

(0, n(log( n/2))/2)

g(x) = x 2 log (1) — arcsin(x) log ( —) •

V1 — x2 \x/ V arcsin(x) J

In particular,

xx/VI-x < arcsin(x)arcsin(x) < (2)n/2 xx/VI-x •

We get

/ x2 log(x) log(x) log (arcsin(x))

g (x) = — — — x =

log(1/x) — (1 — x2) log(1/arcsin(x))

= (1 — x2)3/2 =

log(arcsin(x)(1' x2) /x)

= (1 — x2)3/2 ,

which is clearly positive, and g tends to zero when x tends to zero and 1. Lemma 19. For x E (0,1), the following functions

f (k) = sin (2 x^ , g(k) = tan ^2 x^ , h(k) = sinh (xk)1/k ,

are decreasing in (0, to) . In particular, for k > 1

k

^tan(| x^ < tan (2 ^ < tan (2 k ,

^sinh (xk) < sinh (x) < sinh (px)k •

We get

k \ / nxk log(x)cot ^ log ^sin ^ njr'j'j

f (^=(/^(2^) I ---2k V ^ 2

1 fc/ • /" nxk

' sin —— I X

2 k2

x ^ n k x log(1/x) cot ^ ——j — 2log^1/sin which is negative by Lemma 17(1). Next, we get

kk xk xk

g (k) = Jtan —- x

xk

/nxk log(x)csc fsec flog (tan fnr))

X 1 2k k2 1 < 0

by (6). Finally,

h (k)=xk (*k)—iog(sikhx»

= Jsinh (xk) (xk log(xk) coth (xk) — log (sinh (xk))) (1/k2), which is negative by inequality (5), and this completes the proof. Lemma 20. The following functions

f (k) = cos ^2x1/k) ,x e (0,1),

g(k) = cosh (xk)1/k , x E (0,1),

'n kp 1/k

2 ^

are decreasing in (0, to) . In particular, for k > 1

( n k\ 1

h(k) = arcosh 2 x J , x E (1, to) ,

x

cos(2 ^ < cos(2 ^ ^cos(2 x^j ,

Jcosh (xk) < cosh (x) < cosh (px)k ,

Jarcosh ^2 xk^j < arcosh ^-2 xj < arcosh ^-2 i/x^J •

f (x) = cosk ^ 2 nx1/^ X

k,g(x)tan (nx1/k/2) + log (cos (nx1/k^ < 0

and proof of g follows from Lemma 14(1). Finally, for y > n/2, let

y log(2y/n)

j(y) = arcosh (y) log (arcosh (y)) —

vV — 1 ’

and

■' ( ) log(2y/n) , log (arcosh(y)K 0

j (y) = ^ + VW—1 > 0

and

j(n/2) = arcosh(n/2) log(arcosh(n/2)) = 0^0235 • With z = xk we get

/ arcosh (nz/2)1/k

h (x) = k2 arcosh (nz/2) X

x | nz log(z) — arcosh (nz/2) log (arcosh (^)^ •

V(nz/2)2 — 1 V2 1 J \2J)

This is negative, because j(y) > 0 for y > n/2.

Proof of Theorem 3. The proof for arcosh follows from Lemma 20, and the rest of proof follows from the Lemma 3, if we take p = 2. □

Lemma 21. The following relations hold

1) sin(r) sin(s) < y/sin(r2) sin(s2), r, s E (0,1),

2) cos(r) cos(s) < y/cos(r2) cos(s2) < cos(r s),

3) tan(r) tan(s) > yjtan(r2) tan(s2) > tan(rs),

the first inequalities in (2) and (3) hold for r,s E (0, y/n/2), and second ones for r, s E (0,1).

Clearly (1) and the fist inequality of (2) follwos from Lemmas 13 and 20, respectively. Let g(x) = log(cos(n e-x/2)), x > 0, we get

2x n -2x_ n2 e M -x

g"(x) = —- e-2x sec2 -------------e-x tan

4 V 2 J 2 V 2

7r

= — 4 e

2x

sec2 ^ —2—^ (ex sin (e xn) + n) < 0,

and the second inequality of (2) follows.

For (3), we define h(x) = log(tan(ne-x/2)), x > 0, and we get

h"(x) = e-xn (1 — e-xn cot (e-xn)) csc (e-xn) > 0,

hence h is convex, and the second inequality follows easily, and the first one follows from Lemma 19.

Lemma 22. For a fixed x E (0,1), the function g(k) = (cos kx+ + sin kx)1/k is decreasing in (0,1).

Differentiation yields

g (k) = (sin(kx) + cos(kx))J x

/kx(cos(kx) — sin(kx)) n , . ,, , ,, NN\ /j2

x I ------—— --------——:----log(sin(kx) + cos(kx)) I /k •

V sin(kx) + cos(kx) )

To prove that this is positive, we let z = kx ,y = cos z + sin z < 1 • 1442

h(z) = (cos z + sin z) log(cos z + sin z) — z(cos z — sin z),

and observe that

h (z) = z cos z + (cos z — sin z) log(cos z + sin z) + z sin z =

= zy + log ycos 2 — log ysin 2 > 0,

because e2y > ysin 2. This implies that g (k) > 0.

§ 5. Appendix

In the following tables we give the values of p-analogue functions for some specific values of its domain with p = 3 computed with Mathemati-ca® • For instance, we can define [11]

arcsinp[p_, x_]:= x *Hypergeometric2F1[1/p, 1/p, 1 + 1/p, x~p] sinp[p_, y_]: = x /.FindRoot[arcsinp[p, x] == y, {x, 0.5 }]

x arcsinp (x) arccosp(x) arctanp (x) arsinhp (x) artanhp (x)

0-00000 0-25000 0^50000 0^75000 L00000 0^00000 0^25033 0^50547 0^78196 L20920 L20920 1 • 17782 L07974 0^88660 0^00000 0^00000 0^24903 0^48540 0^68570 0^83565 0^00000 0^24968 0^49502 0^72710 0^93771 0^00000 0^25099 0^51685 0^85661 to

x sinp (x) cosp(x) tanp(x) sinhp (x) tanhp(x)

0^00000 0^25000 0^50000 0^75000 L00000 0^00000 0^24967 0^49476 0^72304 0^91139 L00000 0^99478 0^95788 0^85362 0^62399 0^00000 0^25098 0^51652 0^84704 L46058 0^00000 0^25033 0^50518 0^77588 L08009 0^00000 0^24903 0^48517 0^68283 0^82304

With a normalization different from ours, some eigenvalue problems of the p-Laplacian have been studied in [20].

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The work is received on May 1, 2013.

University of Turku,

Department of Mathematics and Statistics,

FI-20014 Finland.

E-mail: [email protected], [email protected]