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ISSN 2074-1871 Уфимский математический журнал. Том 10. № 3 (2018). С. 121-134.
ON KHABIBULLIN'S CONJECTURE ABOUT PAIR OF INTEGRAL INEQUALITIES
A. BËRDËLLIMA
Abstract. Khabibullin's conjecture is a statement about a pair of integral inequalities, where one inequality implies the other. They depend on two parameters n ^ 2,n £ N, and a £ R+. These inequalities were originally introduced by Khabibullin [6] in his survey regarding Palev problem in Cn and related topics about meromorphic functions. It is possible to express the inequalities in three equivalent forms. The first statement is in terms of logarithmically convex functions, the second statement is in terms of increasing functions, and the third statement is in terms of non-negative functions. In this paper we work solely with the second variant of the hypothesis. It is well established that the conjecture is true whenever 0 ^ a ^ 1/2 for a 11 n. Several proofs exist in the literature among which one is given by the author [2] and it relates the integral inequalities with the general theory of Laplace transform. But it was not known if the statement was true when a > 1/2 until Sharipov [8] showed that the conjecture fails when a = 2, n = 2. However the question of whether this conjecture holds for at least some n ^ 2 and a > 1/2 remained an open problem. In this paper we aim to solve this question. Motivated by Sharipov's approach, we develop a method of constructing a counterexample for the more general case n ^ 2 and a > 1/2. By an explicit counterexample we show that Khabibullin's conjecture does not hold in general.
Keywords: Khabibullin's conjecture, Khabibullin's theorem, Khabibullin's constants, integral inequalities, counterexample, plurisubharmonic function, sharp estimate.
Mathematics Subject Classification: 45A05, 31A05, 31A11). 31B05, 311511)
1. Introduction
A function u with values in [-ro, ro) defined in an open set X c Cn is called plurisubharmonic if
• u is upper semi-continuous
• for arbitrary ^d w £ Cn the function
T M u(z + tw) is subharmonie in the open subset C where it is defined. In 1999 B. Khabibullin [4] conjectured the following inequality for the class of plurisubharmonic functions u(z) in Cra of finite lower order p :
n— 1
^ TT (l + 4) as 0 ^ P ^
sin Mf_AV 2k)
„ M(u,r) hrnrnf ^ « PM={
k= 1
n— 1
П (i+2k) as p> 1, k= 1
(1.1)
A. Berdellima, On Khabibullin's conjecture about pair of integral inequalities. © Berdellima A. 2018. Поступила 24 июня 2017 г.
where M(u,r) = max{«(z) : |z| = r,z E Cn},u+ = max{u(z), 0} and
T(u,r)= u+(r() dsn(()
J&n
represents the Nevanlinna characteristic of «.Here Sn C Cn is the unit sphere, ( E Sn^d dsn is the differential of area on Sn under the normalization f§ dsn = 1, Conjecture (1,1) is the analogue of Paley's problem for entire functions of one variable of order p, which was solved conclusively by Govorov [3], Notice that we say a function u is of finite lower order p if
. flog T(u,r) p = lim inf —--.
v—^^o log r
A sufficient condition for this hypothesis to hold is proving the next statement. (For details on this sufficiency we refer to Khabibullin [5],[6] or Sharipov [7].)
Conjecture 1.1 (Khabibullin's Conjecture). Let a > 1/2. For any nonnegative increasing function h(t) on the interval [0, ro) and for any n ^ 2 if
h(tx)n ^n-1
J 0
then
X
(1 - x)n-1 dx ^ ta, 0 ^ t< ro (1.2)
h(t) dt n/ a
Jo t 1+ < 2 n 0 + d- <L3'
It is also possible to express these inequalities in two other equivalent forms (Khabibullin [5], Baladai, Khabibullin [1]), but in this paper we work solely with the above stated version. For parameter values 0 ^ a ^ 1/2 it has been established that the integral inequalities hold true (see for instance Khabibullin [6], Sharipov [7] or Berdellima [2]). Regarding the same problem, by explicitly constructing a counterexample, Sharipov [8] has demonstrated that the hypothesis doesn't hold for a = 2 and n = 2. However the question of whether this conjecture holds for at least some n ^ 2 and a > 1/2 remained an open problem. In this paper we aim to solve this question.
2. Preliminary discussion A change of variable in inequality (1.2) tx M- x transforms it into the following form
T - x)n-1 dx ^ t»+n-1, 0 ^ t< ro. (2.1)
Jo x
The integral inequality (2.1) can be regarded as an inequality of two functions f (t) ^ g(t) for all t E [0, ro), where f (t) denotes the integral and g(t) := ta+n-1. On the other hand, inequality (1.3) is an inequality of quantities that are fixed for any given parameters a and n. Therefore, it is reasonable to think of getting from (2.1) to (1.3) through some integration over [0, ro) with some appropriate real valued function v(t). If so, then there should exist some relation of v(t) with the function <fr(t) := (1 + t2a)-1, which appears in (1.3). To find an appropriately related function v, first we should study 0 and in particular, its derivatives. In general, for 0 we have the following representation for its derivatives
dn P (ta)
T-^) =-r^(/ ^ + , n E N, (2.2)
dtn^K ' tn • [Q(ta)]n+1' ' v ;
where Q(ta) = 1 + t2a and P(ta) is a polynomial function in ta of degree 2n. One can show (2.2) by using an argument of mathematical induction on n. An important implication of equation
(2.2) is the growth order of 0(n\t) like 0(t n 2a) as t ^ x1. In particular, this fact together with the inequality (2.1) imply the following vanishing limits
lim I4>(k)(t)l r h(x)(t-x)k dx = 0, VkG N. (2.3)
t^ Jo x
If additionally the function h(t) satisfies a restriction of the form h(t) = o(t2a) as t ^ x, then
(2.3) holds as well for k = 0. Another key characteristic of 0(t) is its sign variation over the interval [0, x). It is possible, though technical, to show that when 0 < a ^ 1/2, the derivatives 0(n\t) are of constant sign for every n G N. In fact, they satisfy the following equation2
sign0(n)(t) = (—1)ra, V > 0, Vn G N, (2.4)
where sign(-) is the usual sign function which takes the value 1 on the positive numbers, —1 on the negative numbers and 0 at the origin. Since successive integrations bring ^^(t) to 0(t), it is justifiable to assume that there exists some relation of our sought function v(t) with the derivatives of 0(t). But integrating both sides of (2.1) with v(t) over [0, x) is valid for our problem only if v(t) ^ 0 for all t ^ 0 because otherwise the order of inequality in (2.1) would reverse. At least, as 0 < a ^ 1/2, we can set v(t) := (—1)ra0(ra)(i). By (2.4) it is immediate that v(t) ^ 0. Then multiplying both sides of (2.1) by v and integrating over [0, x) yields
/ v(t) f(t)dt ^ v(t)g(t)dt. (2.5)
Jo Jo
Using the vanishing limit conditions in (2.3) and the assumption that h(t) = o(t2a) as t ^ x, we get from integrating n times by parts in (2.5) the following inequality
h( )
i \n±f.\ i. ^ / 1 \nr(a + n)
"OO
Jo t r(a) Jo
where we use definitions of f and g and the fact that 0 can be obtained from v by integrating successively n times. After proper simplifications in the last inequality we arrive at
f ~ htt),.., r(a + n) [ ~ „ k r(a + n) k ( a \
—6(t)dt ^ ^ ; ta-16(t)dt = — ^————= — TT (1 +—— I, Jo t T(a)T(n)Jo 0() 2a r(a)r(n) 2 J^V k
which is exactly the inequality (1.3). This is but a special case of Khabibullin's theorem. One then might want to apply this method for a > 1/2. But in this case the derivatives of our function 0(t) are not guaranteed to be of constant sign over [0, x) and we cannot follow a similar argument. In particular, for n = 2 it is easy to see that 0"(t) changes sign on [0, x). Explicit calculations show that
2at2a-2((2a + 1)t2" — (2a — 1))
0 (t) =-aT*^--(2.6)
and evidently 0'(t) < 0 for t G [0, r) and 0"(t) > 0 for t > r, where r is the nonzero solution of (t) = 0 given by the formula
2a - 1 - 1
^> = 120+1) * <2-7)
While on one hand the failure of derivatives of 0 to be of constant sign over [0, x) invalidates the integration by parts technique, on the other hand, it provides us with an opportunity to build a counterexample. We use this sign variation of derivatives of 0 to successfully set up
1With slight abuse of notation 0 and 4>(n)(t) will be used interchangeably to mean the n-th order derivative of a function ^ with respect to a real variable t.
2For rigorous derivations of (2.2),(2.3), and (2.4) please refer to Appendix.
oo
sufficient conditions for a counterexample to work and then we explicitly show the existence of such a counterexample.
3. Khabibullin's conjecture
Before solving the problem we develop some definitions and establish some preliminary results that will be crucial in reaching the conclusion of this work.
3.1. Preliminary analysis. Studying the monotonicity of 0(n\t) over R+ is very important in our task of constructing a suitable h that would violate the conjecture. In the case n = 2 it is relatively easy to characterize monotonicity of <fr"(t). However, as n > 2 getting such explicit expressions as in (2.7) becomes very difficult. By nature of 0(i), its higher order derivatives become more intricate the higher the order. In what follows for t G R+ and a> 1/2 we define a new function
1
n+l
in(z) := Pn(z) • with z(t) = ta. (3-1)
It is immediate from (2.2) and (3.1) that tnt{n)(t) = 1n(z(t)). Proposition 3.1. The followings are true:
1. The function 1n(z) satisfies the recursive ordinary differential equation:
n^n(z) + 1n+1(z) = az^'n(z) for all z E R+, n E N.
2. Pn(z) = z2Rn(z) with Rn(z) = cn.n-iz2{n-l) + Cn,n-2Z2{n-2) + ... + Cn,lZ2 + Cn,0.
3. cn,k = c„,,k(a) for k = 0,1,... ,n — 1 are polynomials in a. In particular,
n- l n- l
Cn,n-i = (— 1)n II(2a + k) and cnfl = — JJ(2a — k). k=0 k=0
Proof. 1. We differentiate both sides of the equation in0(n)(i) = ^n(z(t)) with respect to t:
ntn-1t(n)(t) + tnt(n+1)(t) = 1'n(z) ^
This is the same as
ntn(t>(n) (t) + tn+1t(n+1)(t) = 1'n(z) ^t. Using definition of 1n and tdz/dt = ata = az gives
nipn(z) + 1n+1(z) = az^'n(z).
2. For n = 1 we have
,„ , 2 at2a 2az2 , . .
^ ® = — = — (T^^ = №)
with P1(z) = —2az2 and R1(z) = —2a. Suppose the statement is true for n = k. Let n = k +1. From the differential equation for 1n(z) we obtain
, ^ „, , , , , , az[Pk(z)Q(z) — (k + 1)Pk(z)Q'(z)] — kPk(z)Q(z) ^k+1(z) = az^k(z) — kik(z) = 1 ^ ^ '- [Q(;g)]k+2 --.
By induction hypothesis Pk (z) = z2Rk (z) and together with Q(z) = 1 + z2 this yields
z2[((2a — k)(1 + z2) — 2 a( k + 1) z2)Rk (z) + az(1 + z2)R'k (z)]
Ik+1(Z) = [Q^p2 . (3.2)
By induction hypothesis we also have
Rk (z) = Ck,k-1Z2(k-1) + Ck,k-2Z2(k-2) + ... + Ck,1Z2 + Ck,0,
which upon termwise differentiation gives
R'k(z) = 2(k - 1)ck>k-iz2k-3 + 2(k - 2)ck,k-2z2k-5 + ... + 2cMz.
Substituting this in the numerator in (3.2) yields
[(2a - fc)(1 + z2) - 2a(k + 1)z2\Rk(z) + az(1 + z2)Rk(z)
k-1 k-1 = [(2a - fc)(1 + z2) - 2 a(k + 1)z2] ^ ck>iz2i + 2 az(1 + z2) ^ ick>iz2i-1
i=0 i=1
k-1 k-1 = ^(2 a(i + 1) - k)Ck,iz2t + ^ -(2a(k - i) + k)ck>iz2t+2 i=0 i=0
= (2a - k)Ck,o + (-1)(2 a + k)Ck,k-iz2k k-1
+ ^[(2a(i + 1) - k)ck>i - 2 a(k + 1 - i)ck>i-1]z2* =1
k-1
=Ck+1,0 + Ck+1,kz2k + ^ Ck+1,iz2% := Rk+1(z).
=1
where we have defined
{(2a - k)ck,0 as i = 0,
(-1)(2 a + k) Ck,k-1 as i = k, (3.3)
(2a(z + 1) - k) ck,i - 2a( k + 1 - i ) ckti-1 as % = 1, 2,... ,k - 1.
This completes the proof of (ii).
3. The identities for ck,0 and ck, k-1 follow directly the definition of ck,i for i = 0 and i = k- 1 with initial condition c1t0 = -2a. Also using this initial condition, one can show by induction that ck,i = ck,i(a) are polynomials in a for all k E N and i = 0,1,... ,k - 1. This is implied by (3.3) as polynomials form a ring closed under addition and multiplication. □
Definition 3.1. Let T C R. A real valued function f : T ^ R is called proper onT if f is identically different from zero on every compact subset S C T.
Notice that for any n ^ 2 the function 4>(n)(t) is proper on R+. Otherwise there would exist some compact set S C R+ with P(ta) = 0 on S. Since P is a polynomial function then it would follow that P is identically equal to zero on all R. But this contradicts (2.2), which states that P is a nonconstant polynomial function.
Definition 3.2. Let T C R and f : T ^ R be a continuous real valued proper function. The sign variation set of f, denoted by 7}, is defined as
Tf := {tET : 3r > 0, E (0, r] f(t + 5) = 0& sign(/(i + 5)) = - sign(/(i - 6)) }.
In simple words, Tf is the set of points where the function f changes sign. Another way to think about the sign variation set of a continuous proper function is in terms of its zeros. Notice that if r E Tf then f(r) = 0 by continuity of f. In fact 7} coincides with the set of zeros of f of odd order (multiplicity). We are interested now in the sign variation set of the function 4>(n\t) over R+ that we denote by . When n = 2, $"(t) < 0 for all i <t and $"(t) > 0 for all t E (r, to) with r given by (2.7). This implies 7^» = {r} = 0. For n > 2 the higher order derivatives of <fr(t) become more complicated and it is not clear whether T^(n) is nonempty or not. The next statement remedies this issue.
Proposition 3.2. For any n ^ 2 and n E N the sign variation set T^(n) is nonempty. Moreover, T^(n) is finite.
Proof. Suppose that there exists n G N such that 7^(n) = 0 and without loss of generality let it be the smallest such n. Clearly, n > 2 as we know by construction that 7^» = 0 as n = 2. Now for any t G 7^(n-i) we have
^^(r) = o ^ Tn_i^(n_i) (r) = 0 *) = 0 with z* := ra.
By proposition 3.1 (1) it follows that
az*tfn_i( z*) = n'tpn_i( z*) + z*) = (z*).
Since T4>(n) = 0, this yields that 0(n)(t) ^ 0 or 0(n)(t) ^ 0 for all t G R+. Without loss of generality we assume that 4>(n)(t) ^ 0 for all t G R+. This is equivalent to ^n(z) ^ 0 for all z ^ 0 by definition of in (3.1). In particular, for t = t we have <j)(n\T) ^ 0 and this implies z*) ^ 0. Therefore, we have
*^n-i(z*) = Mz*) ^ 0 ^ ^'n_i(z*) ^ 0. (3.4)
On the other hand, Q(z*) = 1 + (z*)2 > 0. Since ^n_i(z*) = 0 then it follows that Pn_i(z*) = 0. Explicit differentiation of ipn_i(z) and its evaluation at z* yields
,, ( *) = K_i(z*)Q(Z*) -nPn_i(z*)Q(z*) = P^_i(z*)Q(z*) (3 5)
Vn_i(Z ) [Q(z*)]"+i [Q(z*)]ra+i ( )
and (3.4) and (3.5) imply Pln_i(z*) ^ 0. Note that Pn_i(z) is a polynomial function and hence is continuously differentiable everywhere on R. Now suppose that there exists z** > z* with Pn_i(z**) = 0, then by Rolle's theoremi there exists w G (z*, z**) such that P'n_i (w) = 0. Moreover, by the fact that P^_i(z*) ^ 0 it must be the case that Pn_i(w) > 0, which in turn implies ^n_i(w) > 0 and ^'n_i(w) < 0. From the differential equation we then get
0 > aw^'n_i(w) = mpn_i(w) + ^n(w) > ^n(w) ^ ^n(w) < 0.
However, this contradicts our assumption that ^n(z) ^ 0 everywhere on R+. Therefore, an arbitrarily chosen r is the only element of 7^(n-i). Since Pn_i(z*) ^ 0, this implies
iim %riy >0 ^iim >0 ^ >0 ^ (-1)™-1 TT(2«+k) > o. (3.6)
Z—X Z2(n-1) Z—X Z2\n-2) -L-L
k=0
On the other hand, by assumption we have 4>(n)(t) ^ 0 for all t G R+. This in turn implies ^ 0 for all z > 0. In particular, we have lim^^ r^n(z) ^ 0, which then yields
Lz—yx n\
n— 1
lim ^T > 0 ^ lim > 0 ^ ^ 0 ^ (-1)" TT(2a + k) > 0. (3.7)
k=0
However, (3.6) and (3.7) cannot hold simultaneously. Therefore, we arrive at a contradiction. This establishes 7^(n) = 0 for any n ^ 2 and n G N. The set 7^(n) is finite for any n ^ 2 as it is implied by the fact that each element r G 7^(n) is vis-a-vis related to a zero z* of odd multiplicity of the polynomial Rn(z) through the relation z* = ra. By the fundamental theorem of algebra2 Rn(z) has at most deg(Rra) zeros, not necessarily distinct. Therefore, |7^(n) | ^ deg(Rn) < x. □
1Rolle's theorem: If f : [a, 6] ^ R is differentiable on (a,b) and continuous on [a, 6] with f (a) = f (6) then there exists c g (a, b) such that f '(c) = 0.
2The fundamental theorem of algebra: Every nonzero, single variable, polynomial of degree n ^ 0 with complex coefficients has at most n complex roots (not necessarily distinct).
3.2. Sufficient conditions. Now let 0 < ^ < t2 < ... < rm < to be an ordering of the elements of 7^(n) for some n ^ 2 and m = |7^(n) |. We partition R+ into disjoint intervals:
R+ = (0, n] U (n, T2] U ... U (Tm-1, rm\ U (rm, to).
Define the following sets: 7n+ := {t E R+ : sign(0n(i)) = (-1)n+1} and the index set T+ := {tE N : sign (0n(i)) = (-1)n+1 , Vt E (n-1, n]}. Since sign(0(n) (t)) = sign(cn,n-1) = (-1) n by the proof of the last proposition for all t > rm we have (rm, to) E T"+ and so we get
X+ = U(^^ Ti].
iei.
Fix any n ^ 2, n E N and a > 1/2. Next we need to construct a pair of functions {f, h} with h being continuously differentiable on R+. This pair of functions should satisfy the following relation for all t E R+
rt h( x) 1 r]n
№ = J h(--)(t - x)n-1 dx or equivalently h(t) = —iW (3.8)
Let {Tii}ieI+ be a family of functions rji : (ri-1, Tj\ ^ [0,1] such that rji(t) = 0 for all t E (U-1, Ti] and r/i is sufficiently smooth (derivatives of orders at least n + 1 exist) on (n-1, Ti) for all i E T+. Define function in the following way
i ta+n-1 (1 -e E VO) as tE%+
№= \ (3.9)
[ ta+n-1 as tE R+ \T+
Then f(t) satisfies the inequality 0 ^ f ^ ta+n-1 whenever 0 ^ e ^ 1. The family of functions {Vi}iei+ should satisfy conditions derived from the properties of h. Differentiability of h and relation (3.8) imply that must have continuous derivatives at least to the order n + 1. This leads us to the following limits
dk dk lim — [ta+n-1ill(t)}=0, lim — [ta+n-1Vl(t)\ = 0, k=1, 2,... ,n + 1, ieZ+. (3.10)
T- dtk t^t+_1 dtk
Condition (3.10) illustrates the fact that the left and the right limits at t = ^ for i E T+ of derivatives of up to order n + 1 should coincide. The increasing and nonnegativity property of h lead us to the following conditions
1 dn
f^dnJ(i) > 'E T'+ (3-11)
1 dn _ 1 dn+1
w \ 1 f(t) + TTTTf(t) > 0, fE X+. (3.12)
r(n) dtn r( n)dtn+1jy) n v 7
We consider only T,+ because (3.11) and (3.12) trivially hold on R+ \ T,+ by definition of f in (3.9). From the same definition and (3.8) one can get also an explicit piecewise formulation of h:
{F dn
C(n,a)ta - r-rn-tY, IT" lta+"-1 "¡Ml as t E T+
C (n,a) ta as tE R+ \X+,
__r(n+a)
where C(n,a) := ra)Tan). We consider the integral
J£(a) := / ^mdt, J R+ 1
Using the piecewise representation of h(t) in (3.13), this follows that
c C r]n r
Js («) = -7^/ Y.^-a [ta+n_i rn(tMt)dt + C(n,a) ta_i<P(t)dt r(n) J]n JR+
= E - r(n) I £" "WI^Wdt + L ¥<«<>dt
where hn(t) = C(n,a)ta. We integrate n-times by parts in the integral expression inside the sum. Notice that from conditions in (3.10) its integrand vanishes on the boundary dT,+ = 7^(n). Therefore, we arrive at the following expression
Js(«) = (-1)^ £ -n / m(t)d^m dt + f dt> i ^m dt
— r(n) JT+ Mn Jr+ t Jr+ t
whenever
dn
M)^^(i) > 0, tGTn+. (3.14)
This inequality is implied by the definition of T+. Therefore, the second integral inequality would violate the conjectured upper estimate. Naturally, one raises the question of whether a family of functions {rit}t£i+ satisfying (3.10)-(3.12) exists. This will be shown in the next section.
3.3. A general counterexample. Let n 2 2. Consider the collection of functions {r]t}tex+ given by
/ nt \ fnt\ i2n .
, cos - cos — as t G (rt-1, rt\
Vi(t)={1 ^2Ti_J V2rj\ K tU tJ (3.15)
as tG (n-1, rt],
with the convention that if 1 G Z+ as r/^t) = cos2ra ^t^J for all t G (0, Ti] and zero otherwise. By definition in (3.15), the function rjt is smooth (infinitely differentiable) on (rt-1, rt) for all i G Z+. On the other hand, by Leibniz's formula we have for k = 0,1, ..,n +1
lim fr[ta+n~1 m(t)] = Vf k 1 lim d-ta+n~1 cos2n (—) lim ^ cos2n (—) = 0, --idt"[ /t( )] Jt->m-idP V2 Tjt-yrt idtk_ V2rt_ J
lim [ta+n_1 Vi(t)] = V PI lim d-ta+n_1 cos2n (—) lim cos2n (—) =0.
t-r-dtk [ /t()] Jt-T-dP V2Tt_J t-T-dtk_ \2rj
Hence, the limit conditions in (3.10) are fulfilled. Next we need to check (3.11), which by piecewise representation of h is equivalent to
f dn
C(n,a)r 2 r~jt £ ddf [r+n_1 t G V-. (3.16)
( ) tei+
Another application of Lebniz's formula on the last expression yields
d /.\n X ^ I n\ d . d ^
3=0 VJ/
j=0Q\j )v(a + n - jf dt-3nt(t).
(3.17)
a—
Relabelling j m n - j in (3.17) and using the identity ( ) = ( ), we rewrite (3.16) as
n -
C (n, a) ^ e
5 § 0
i£l+ 3=°
n\ r(a +n) d?
j) r(n)r(a +j)tdtirh(t) ytEln
for some suitably chosen e E [0,1]. In fact, using the definition of C(a, n) one can simplify even further the last inequality into the following equivalent form
1 > e
§(n)
^ ^ tel?.
\jJ r(a + i) dP
i€L+ 3=° XJ/ K J>
(3.18)
Similarly using the piecewise representation of h, we rewrite condition (3.12) as
d
dn+1
aC<n,a)ta-1 > ^-sY: ^[r+n-1 *<*)] + r<n)t E ^ [r+n-1 f G <3.19)
r(n) ^ dt iZL,
r<n) ^ dtn+1 ieL+
Again by Leibniz's formula we have
^ i*—»«)) = g(n+^
dtn+11
dn+1-j
dtr dtn+1-3Vl()
,a-2
n+1
£
3=° n+1
n + ^ r(a + n) tn+- dn+1-[l^
j JT(a + n - j)
dtn+1-i
(3.20)
,a-2
zrt'Uia+n)„
3=0
j J r(a +j- 1) dp-
Using (3.20) in conjunction with (3.17) we can write (3.19) as follows
aC(n, a)ta-1 >eta-1\Y ■ ^ , ■
( , ) L4+ ^VJ) r(n)r(a + j) dP
%eLri J
n\ r(a + n) dj
+
££("+0
i = 0 \ J /
n + 1\ r(a + n) • dj 3 J r(n)r(a + j- 1) ' dP
t G T„+.
iex+ j=o
Since a > 0, after proper simplifications we have the inequality in its final form
n+1
5 \s(n)
^ £
^ a ^ L ^ \j) r(a + j) dP
iEL+ 3=0 v J>
n\ r(a) , d3 . . v+ /n + 1
£(n+0
r(a) , d3 . . _+
—-Trr Vi (t) , teTn+
j jT(a + j- 1) dP
(3.21)
For each z E Z+, i]i(t) is smooth on the bounded interval (n-1, Ti), and the supremum of each derivative of rji exists on this interval and is finite. For brevity denote the supremum of the j-th derivative of rji, j = 0,1,..., n, by
d1
vh := suP { -itM^) j -
We get the estimates:
'n\ r(a)
5 §C)
ieL+ 3=°
jjT(a +j) dP
<
5 S©
n\ r(a) _j *
ieL+ 3=°
jj r(a +j)
t G T„
A r(a\^W +
^ E (J T(q + jY dtj
iex+ j=0 v v J>
" + Orr ) 1) (j ^W
j /r(Q + j - 1) dP
<
<
5 EC)
n\ T(q)
ieirt j=0
jJ r(Q + j)
j
j
+ ')
j=o v J / + £ (" +1)
j=0 v J 7
r(Q)
5 EC)
n\ T(q)
jjr( Q +j)
j * |
j) n i]i,j + i€l+ ^^^ ■ j) j=0
j yr(Q + j- 1) r(Q) j * -
j
dj . . dP^(t)
j JT(q + j- 1)
< oo, te T„ .
In order to prove (3.18) and (3.21), it is sufficient to show the following system of inequalities hold:
£ A < 1, where A :=
5 SG)
n\ r(Q)
j*
i
iei+ j=0
jjT(Q +j) i
(3.22)
£B < 1, where B := —
E[EQ
iex+ j=0 KJ/
n\ r(Q)
j
Q— l — \jJr(Q + j)ili,j + j ;r(Q +j- 1)
>,-1 /n + A r(Q) j *
Z^ y j y TV™ /1_r/i'3
(3.23)
If max{A, B} ^ 1, then each e ^ 1 would work. Otherwise if max{A,B} > 1, we choose e = (max{A, B})-1 < 1. In any case we are guaranteed of the existence of such 0 ^ £ ^ 1. We state these observations in the following theorem:
Theorem 3.1. There exists no q > 1/2 and n ^ 2 such that Khabibullin's conjecture is true.
The proof follows the analysis in Sections 3.1-3.3.
4. Concluding Remarks
In this work we constructively showed that for any q > 1/2 and n ^ 2, n G N, there exists a nonnegative increasing function h violating the conjecture. The method of construction relied strongly on the monotonicity of the higher order derivatives of the function 0 over R+. In fact, it is the nonemptiness of the set 7^(n) for any n ^ 2,n G N that opens up the way of defining a pair of functions {f, h} that fulfils the task. Another key result implied by Theorem 3.1 relates to the question in Berdellima [2] of whether the inverse Laplace transform of the function 0(z) is everywhere nonnegative over R+ for some q > 1/2 and 2q g N. Notice that here 0 is extended to the complex plane, 0(z) = (1 + z2a)-1 for z G C. If $(i) := (£-10(z))(t) denotes the inverse Laplace transform of 0, then one can show that
= i
rj+ix
<f>(z) eztdz = ^
' j-ix
keZ2a
exp( zk t)
JJ (Zk - Zm) k=m
(4.1)
where Zk = exp((2 k + /2q) and Z2a = {0,1,..., 2q — 1}. Indeed, formula in (4.1) follows the Cauchy residue theorem applied on a halfdisk contour around the origin enclosing all poles (singularities) of 0(z) in the complex plane. The line integral in (4.1) would work for any 7 ^ 1 since all poles of 0(z) lie on the unit circle and Zk = 1 for any k G Z2a. In fact, we can establish a more general result that $(i) in (4.1) is not of constant sign on R+. If for some 2q g N we had $(i) ^ 0 for all t G R+ (or $(i) ^ 0 for all i G R+), then by theorem 6 in Berdellima [2] it would follow that Khabibullin's conjecture would be valid for all nonnegative increasing functions h. However, since q > 1/2, this contradicts Theorem 3.1 of this paper which always guarantees a counterexample. Therefore, (4.1) cannot be of constant sign everywhere on R+
for any 2 a E N. It must be pointed out that finding sharp estimates when n ^ 2 and a> 1/2 is still an open problem. Sharipov [8] showed that for each n ^ 2 and a > 0 there exists a positive constant Cxh(n,a), which is a sharp upper bound for the integral inequality (1.3). Note that as 0 ^ a ^ 1/2, we have
n-1
CKh(n,a) = 2 n 0 + D k=1
It is possible to numerically bound Khabibullin's constants Cxh(n, a) but finding them explicitly as a > 1/2 is yet unsolved. It would certainly be desirable to get, if not in a closed form, at least some explicit formula for Cxh(n, a). These constants have implications in estimating the growth rate of a plurisubharmonic function.
A. Technical details
Lemma A.1. Let 6(t) =-—, then for all m ^ 1,m E Z we have
1 +12a J
dmA P(ta)
dtm tm • [Q(ta)]m+1'
where Q(ta) = 1 + t2a and P is a polynomial function in ta of degree 2m.
Proof. We prove this lemma using mathematical induction. For m = 1 we have
, -2at2a _ P (ta)
A = t(1 + t2a)2 = t • Q(ta)]2 ,
where P(ta) = -2at2a as a polynomial function in ta is of degree 2. Let the statement be true for m = n, that is
A>(n)(t) = _p(-)_
A () tn • [Q(¿a)]n+1 ,
where P is a polynomial function in a of degree 2 n. Differentiating both sides of the last equation with respect to , we obtain
d An)m d \ P(fa) ^ a,(n+1) (a ataP'Q - nPQ - (n + 1)taPQ'
ltn • [Q(ta)]n+^J r w tn+1[Q(ta)]n+2
Denote by deg P the degree of the polynomial P. By the inductive hypothesis, since deg P = 2n, we obtain deg P' = 2n-1. We observe that deg ta = 1 and by definition of Q we have deg Q = 2. Setting P0 = ataP'Q - nPQ - (n + 1)taPQ' and using deg PQ = degP + degQ for any two polynomials P and Q, then the degree of polynomial P0 is
deg(Po) = max{deg(taP'Q), deg(PQ), deg(iaPQ')} = max{1 + 2n - 1 + 2, 2n + 2,1 + 2n +1} = 2n + 2 = 2(n + 1)
□
Lemma A.2. Let h(t) = o(t2a) as t m to and suppose h(t) satisfies the inequality (2.1). Then the following limits hold
lim IA(k)(T)| [T h(x)(T -x)k dx = 0, kE Z+ U{0}. l0 x
Proof. First we show for k = 0. We observe that both 0(T) and f^ dx are differentiate functions. It is sufficient to prove that the limit vanishes, when the improper integral f^^^dx diverges. We get an indeterminate form x/x. On the other hand,since
>0 x
hit) = o(t2a) as t ^ x, we obtain
lim
T—»00
_d_ dT
,-T
h(x)
x
dx
lim hill
2 a t^x T2a
By L'Hospital's Rule [9] we get
cT
lim $(T) [T hix) dx
lim
h( x)
x
T ^X
x
dx ——
v dT lim
T ^X
cT
h(x)
x
x
d
T^x 1 + T2a
Let k ^ 1, then from the first integral inequality we obtain
0 ^ I h(x)(T — xf iv ,
If (1 + T
0 CT<oo.
■1 lim 'hi?!
2 a t^x T2»
0.
0
x
On the other hand, by Lemma 1 we have that 0(T) = 0(T k 2a) as t ^ x. Therefore,
|0(fc)(T)| [T hix)(f-x)kdx ^ 0(T-k-2a)Ta+k = 0(T-a) ^ 0 as T ^ x.
0 x
This completes the proof.
Lemma A.3. For all 0 ^ a ^ 1/2 and t > 0 the identity holds
□
(-1)ra as 0 < a ^ 1/2, n ^ 0,
sign(^\t)) = Proof. We begin with the identity
0
as a as a
0, n ^ 1, 0, n = 0.
dn
dtn
(1 + t2a) = 2 a(2a - 1)(2a - 2)... (2a - m + 1)i
2a—m
This implies
sign
d"
'(-1)m-1 as 0 < a < 1/2, m ^ 1, as a G {0,1/2}, m ^ 2,
idt r>
-(1 + t2a)
as a = 0, as a = 1/2, as a ^ 0,
0 and a
(A.1)
m = 1 , m = 1 , m = 0,
1/2 separately. Assume
for all t > 0 and m G Z. We will consider the cases q that 0 < q < 1/2. This reduces to showing that
sign(^(i)) = (—1)"
for all n ^ 0 and t > 0. We prove this statement by a strong mathematical induction. The base case would be for n = 0. By the definition of 0(t) we have
(1+t2a)m = 1 ^ sign((1+12a)<p(t)) = sign(1)
Since 1 + t2a > 0 for all t > 0, then sign(1 + t2a) = 1. Thus, we obtain
sign(0(i)) = 1 = (—1)0
0
0
0
0
Therefore, the base case is true. Assume that the equation sign(A(n)(t)) = (- 1)n is valid for all n =1, 2,.., k. Now let n = k + 1. Differentiating (1 + t2a)<fi(t) successively k + 1 times and using Leibniz's formula, we get:
k+1
dk+l dt k+l
(1 + 12")<(t)
h + 1 \ rfi Jk+l-i
4(1+ t2a) TJ+^W). dtiV Jdtk+l-iYK '
i=0 v 7
Since (1 +1 )<(t) = 1 for all t ^ 0, then we obtain
±ß +1) ^ (1+ ^ <(t) = 0
i—n \ /
k+1-
By the latter equation we arrive at
sign
1
dk+l 1 r t+i f k + 1\ di dk+
a + <2")i+M = (-1)sign [£ * + M£o + **)^m
(A.2)
For 1 ^k + 1 the sign of each summand in (A.2) can be computed as follows:
\dk+1-1 ,, ; ^A(t)
sign
By (A.1) we have
Ji Jk+l-i
4 (1+t2a) i——<(t)
VdtiK 'dtk+l-iYK '
= sign
di
à <> +j2° )
sign
Idt k+l~
sign
i
i <1+(2">
(-1)
l
for all = for i = 1, 2,.
Therefore,
1,2,...,k + 1 and t > 0. On the other side, applying strong induction . ,k on sign(A(k+1-l\t)), we get
sign
k+
l
Idt k+l-i
<(t) =(-1)
k+l-i
sign
i k+l-
4(1+t2a) i k+l-
(-1)i-l(-1)
k+l-
(-1)k
for all i = 1, 2,... ,k and t > 0. For i = k + 1 we have
k+
sign
k+
j (1+ t2")
^k+i-(k+i) dtk+l-(k+l)
< ( )
rdk+l
sign
Idt k+l
T(1 + t2a) • <(t) = (-1)k+l-l(-1)0 = (-1)k.
This shows that all terms (A.2) are of the same sign (- 1)k for all i = 1, 2,...,k +1 and therefore, we are led to the conclusion that
sign
k+1
(1 + ^M+W
r dk+1 1
(-1) • (-1)k ^ sign l^+m] = (-1)
k+
This proves the statement for all 0 < a < 1/2.
Now let a = 0. Then A(t) = 1 and this implies sign(A(i)) = 1 and sign(A(n)(i)) = 0 for all n ^ 1 (since A(t) in this case is a constant function, its derivatives of all orders vanish). As a = 1/2, we need to prove again that sign(A(n)(i)) = (- 1)n for all n ^ 0. We need only to use the mathematical induction. For the base case n = 0 we have
(1 + t)A(t) = 1 ^ sign(A(t)) = 1 = (-1)0
for all > 0. Suppose the statement is true for n = , that is sign( A(k)(t)) n = k + 1. Differentiating (1 + t)A(t) successively k + 1 times, we obtain
(-1)k. Let
dk+l dtk+1
(1 + ) < ( )
kt{k + 0 £ ^ ^<(t) = (1+t) ^ +
(
k + ^dk 1 ) dfk<(t).
Since (1 + t)((t) = 1, by definition of ((t), we necessarily have
Jk+1 /u +1\ v rtk+i i r dk 1
(1 + i) = — ( 1 ) - sig4s-r^} = (—= H0«
where in the last step we have used the inductive hypothesis. This completes the proof. □
I am grateful to Prof. Dr. R. Luke and our scientific group for making it possible to organize a seminar at our Institute on this mathematical problem. Also I would like to thank the anonymous referee for his/her useful comments in improving the exposition of the paper.
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Arian Berdellima,
Institute for numerical and applied mathematics, University of Gottingen, 37083 Gottingen, Germany
E-mail: arian.berdellima@mathematik.uni-goettingen.de berdellima@gmail.com
