condition An Opial-type inequality with an integral

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An Opial-type inequality with an integral boundary condition Richard C Brown and Michael Plum Proc. R. Soc. A 2005 461, 2635-2651 doi: 10.1098/rspa.2005.1449

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Proc. R. Soc. A (2005) 461, 2635–2651 doi:10.1098/rspa.2005.1449 Published online 5 July 2005

An Opial-type inequality with an integral boundary condition B Y R ICHARD C. B ROWN 1

AND

M ICHAEL P LUM 2

1

Department of Mathematics, University of Alabama-Tuscaloosa, AL 35487-0350, USA ([email protected]) 2 Mathematics Institute, University of Karlsruhe, 76128 Karlsruhe, Germany ([email protected]) determine the Ðbest constant K and extremals of the Opial-type inequality ÐWe b b 0 2 0 a jyy jdx % KðbK aÞ a jy j dx, where y is required to satisfy the boundary condition Ðb a y dx Z 0. The techniques employed differ from those utilized recently by Denzler to solve this problem, and also from those used originally to prove the classical inequality; but they also yield a new proof of that inequality. Keywords: Opial inequality; best constant; extremal; homogeneous integral boundary condition

1. Introduction In 1962, Olech (Olech 1960) gave a simplified proof of the following inequality originally due in a slightly less general form to Zdzislaw Opial1 (Opial 1960). Theorem 1.1. If y is a real absolutely Ðcontinuous function on the interval [a,b],KN! a ! b !N, y(a) Z y(b) Z0, and ab ðy 0 Þ2 dx !N, then the best constant K of the inequality ðb

0

jyy jdx % Kðb K aÞ

a

ðb

ðy 0 Þ2 dx

(1.1)

a

is 1/4. Equality holds in (1.1) if and only if

yðsÞ Z

8 > > < cðs K aÞ

if a% s%

> > : cðb K sÞ

if

a Cb ; 2

a Cb ! s% b; 2

where c is an arbitrary constant. 1

Opial required that y 0 be continuous and y O 0. In addition, he did not characterize the extremals.

Received 12 August 2004 Accepted 13 December 2004

2635

q 2005 The Royal Society


2636

R. C. Brown and M. Plum

Embedded in Olech’s proof is the half-interval form of Opial’s inequality, also discovered by Beesack (Beesack 1962), which is satisfied by those y vanishing only at a. Theorem 1.2. If y is a real absolutely continuous function on the interval [a,b], Ð KN!a !b !N, y(a) Z0, and ab ðy 0 Þ2 dx !N, then ðb

jyy 0 jdx %

a

b Ka 2

ðb

ðy 0 Þ2 dx:

(1.2)

a

Equality holds in (1.2) if and only if y Zc(s Ka) for some constant c. Since their discovery, both (1.1) and (1.2) have attracted enormous interest. At least six proofs are known, and a very large number of generalizations have been given. For a survey of the literature on Opial-type inequalities, see the books of Agarwal & Pang (1995) and Mitrinovic´ et al. (1991). For an important, recent result showing that Opial inequalities are equivalent to those of Hardytype (without, however, preserving the constants), see Sinnamon (2004). In addition to their intrinsic interest, Opial-type inequalities have proved essential in developing disconjugacy/stability criteria for differential equations, obtaining sufficient conditions for the positivity of eigenvalues, bounds on the spacing of zeros of a solution, or improving other inequalities, such as the Lyapunov inequality (e.g. Brown & Hinton 1997; Brown et al. 2002; Clark & Hinton 2002). We note that the existence of an inequality of the form (1.1) or (1.2) is quite easy to prove. As noted in Opial (1960), if we apply the Cauchy–Schwarz and the one-dimensional Poincare´ inequality, we find that ðb

0

jyy jdx %

a

ð b

2

1=2 ð b

y dx a

0 2

1=2

ðy Þ dx

a

b Ka % p

ðb

ðy 0 Þ2 dx:

(1.3)

a

The non-trivial part of (1.1) is the determination of the least value of K and the characterization of the extremals, and in all applications this knowledge has been essential. In order to understand the subtleties involved in Opial’s inequality, we sketch an argument for (1.1) which is close to Olech’s. Outline of a proof. Let y be an absolutely continuous function such that Ð y(a)Zy(b)Z0 and ab ðy 0 Þ2 dx !N, and let p 2(a,b) satisfy ðb ðp 0 jy jdx Z jy 0 jdx: a

Define Y ðxÞ Z

p

(Ðx

jy 0 jdx

if x 2½a;p;

Ðb

jy 0 jdx

if x 2ðp;b:

a x

Evidently, Y is absolutely continuous, jyj % jY j, and jy 0 j Z jY 0 j, so that an extremal of (1.1), if any, will be found among the class of appropriate functions y which are non-decreasing on (0, p], non-increasing on (p, 1], and such that y(p)Z 1. From this it follows that the extremal is a linear spline with a unique knot at p, and by varying p we find that the least value of K is 1/4. Through a variation of the Proc. R. Soc. A (2005)


An Opial-type inequality

2637

above argument (Brown et al. 2000), one can show that ð ðb b Ka b 0 2 0 jyy jdx % ðy Þ dx if yðaÞ C yðbÞ Z 0: 4 a a Most of the other generalizations of Opial’s inequality familiar to us involve boundary conditions similar to those of (1.1) or (1.2).2 In this paper we will consider the inequality ðb ðb jyy 0 jdx % Kðb K aÞ ðy 0 Þ2 dx; (1.4) a

a

ðb a

y dx Z 0;

(1.5)

Ð where y again is absolutely continuous and ab ðy 0 Þ2 dx !N. As in the case of (1.1), it is not difficult to show that K!N in (1.4) exists. The argument (1.3) using a form of the Wirtinger inequality (Mitrinovic´ et al. 1991, p. 67) shows that an upper bound for K in (1.4), (1.5) is also 1/p. If we set y Zx K(a Cb)/2, a calculation shows that a lower bound on K is 1/4. We will prove the following result which was conjectured by one of the authors in 2001 and presented as an open problem in the meeting ‘General Inequalities 8’ at Noszvaj, Hungary, in September 2002. Theorem 1.3. The best value of K in (1.4), (1.5) is also 1/4 and all extremals are of the form yc(x)Zc(x K(a Cb)/2) for any constant c. If one assumes that there is a unique extremal for (1.4), (1.5), then it is not hard to show that theorem 1.3 is true (Brown 2003). In an earlier version of the present paper, we showed that the mere existence of an extremal implies theorem 1.3, but could not prove its actual existence. While (1.4), (1.5) is simple in form, it is much harder to handle than (1.1). As in the previous case, the main difficulty is caused by the absolute value signs on the left side, but the technique we utilised to prove (1.1) no longer seems applicable since it is hard to construct a piecewise Ð monotone function y with the properties of Y while preserving the condition ab y dx Z 0. Therefore, we will be forced to utilize a much more complicated technique based on transformation of variables and variational ideas. The proof will be given in §3. In §2, we show that an extremal of (1.4), (1.5) exists in a restricted function set where y 0 is required to have an arbitrary but finite number N of sign changes. This limited existence result turns out to be sufficient preparation for a complete proof of the theorem which will be given in §3. In §4, it will be shown how the argument for (1.4), (1.5) will also work to give yet another proof (perhaps the seventh) of both theorems 1.1 and 1.2. We should point out that our work is now the second proof of theorem 1.3. In 2003, Denzler (2004) found a constructive proof of theorem 1.3 on entirely different lines from our variational method. His basic idea was to modify sequentially an admissible function y using various rearrangements and normalizations so as to 2

However, for a discussion of Opial-type inequalities satisfying the non-homogenous conditions y(a)Z c, y(b)Zd, c, d s 0, see Brown et al. (2000).

Proc. R. Soc. A (2005)


2638


Ð Ð Ð1 0 2 decrease 01 ðy 0 Þdx K 4 01 jyy 0 jdx either Ð 1 by0 decreasing 0 ðy Þ dx while leaving the second term fixed or by increasing 0 jyy jdx while leaving the first term fixed. However, since Denzler’s approach and ours are so different, we feel justified in giving another proof of this problem, especially since the methodology underlying our variational approach might be helpful in dealing with other inequalities (e.g. a generalization of theorem 1.3 to higher dimensions). In fact, additional proofs will probably be found just as in the case of the original Opial inequality. In particular, since both Denzler’s proof and ours are much more complicated than any proof of theorem 1.1, a third, simpler proof would be desirable. We close this section with a few remarks on notation. We denote the Lebesgue space of (equivalence classes of ) real square integrable functions by L2(a,b), and the class of absolutely continuous real functions on [a,b] by AC [a,b]. We set H 1 ða;bÞ dfy 2AC ½a;b : y 0 2L2 ða;bÞg; endowed with its Hilbert space norm ½kyk2L2 ða;bÞ C ky 0 k2L2 ða;bÞ 1=2 : The class of admissible functions for which the inequalities (1.1), (1.2), (1.4) are defined is called D. It is the subspace of H 1(a,b) satisfying the appropriate boundary condition, for example, in the case of theorem 1.3, ðb 1 D d y 2H ða;bÞ : y dx Z 0 : a

Finally, m(S ) denotes the Lebesgue measure of a measurable set S. 2. The existence of an extremal on a restricted function set An essential argument in our proof of theorem 1.3 will be the Euler equation for some suitably chosen variational problem based on transformation of the independent variable. Hence, we (seem to) need the existence of an extremal maximizing Ðb 0 1 a jyy jdx QðyÞ d Ðb 0 2 b K a a ðy Þ dx on D a priori. As mentioned above, we have not been able to find such an a priori existence proof. It turns out, however, that a priori existence of an extremal on a restricted set involving artificial compactness (instead of the full set D) is sufficient for our argument. Let N 2 N be fixed. We say that a function f 2L2(a,b) has at most N sign changes on [a,b] if t1, ., tN 2 [a,b] exist such that a et0 % t1 % t2 %.% tN % tNC1db and, for j Z1, ., N C1, f R0

a:e: on ½tjK1 ;tj or f % 0

a:e: on ½tjK1 ; tj

(which is clearly always satisfied if tj Kl Z tj). We define the restricted function set (still for N fixed) DN dfy 2D : y 0 has at most N sign changes on ½a;bg: Proc. R. Soc. A (2005)

(2.1)


2639


Proposition 2.1. There exists y 2DN\{0} which maximizes Q on DN. Proof. Let KN d

sup

QðyÞ;

y2DN nf0g

and choose some sequence (yn) in DN such that ðb ðb 0 2 ðyn Þ dx Z 1 ðn 2NÞ; jyn yn0 jdx / ðb K aÞKN a

ðn/NÞ:

(2.2)

a

ðnÞ

ðnÞ

Since, for each n; yn0 has at most N sign changes, there exist a et0 % t1 % . % such that, for j Z1, ., N C1, h i h i ðnÞ ðnÞ ðnÞ ðnÞ or yn0 % 0 a:e: on tjK1 ;tj : (2.3) yn0 R 0 a:e: on tjK1 ;tj Ð Since ab yn dx Z 0 and therefore yn has at least one zero, (2.2) implies that (yn) is bounded in H 1(a,b). Hence, a subsequence (denoted again by (yn)) can be chosen such that yn . y ðweaklyÞ in H 1 ða;bÞ (2.4) ðnÞ ðnÞ tN % tNC1 db

for some y 2H 1(a,b), and, in addition, ðnÞ

tj / tj

ð j Z 1; .; N ; n/NÞ;

(2.5)

where aet0 % t1 % t2 % . % tN % tNC1db. The weak convergence (2.4) implies, by the Sobolev–Kondrachev–Rellich embedding theorem, that yn / y uniformly on ½a;b

(2.6)

yn0 . y 0 ðweaklyÞ in L2 ða;bÞ:

(2.7)

and

The zero integral condition for yn and (2.6) imply that y has zero integral, i.e. y 2 D. To prove y 2 DN , we show that, for each j Z1, ., N C1, y0 R 0

a:e: on ½tjK1 ;tj or y 0 % 0 a:e: on ½tjK1 ;tj :

(2.8)

Assuming the contrary, we obtain, for some j 2{1, ., N C1}, subsets U C,U K3[tj K1,tj ], which both have positive measure, such that y 0 O0 on U C and y 0 ! 0 on U K. Possibly after reducing U C and U K (but still keeping their measures positive), we may assume that U C,U K3[tj K1Cd,tj Kd] for some d O0, whence (2.3) and (2.5) imply yn0 R 0 a:e: on U Cg U K or yn0 % 0 a:e: on U Cg U K (2.9) for n sufficiently large. On the other hand, with cC denoting the characteristic function of U C, (2.7) implies ðb ðb ð ð yn0 dx Z yn0 cC dx / y 0 cC dx Z y 0 dx O 0; a

UC

Ð

and, analogously, UK yn0 dx / (2.8). In particular, y 2DN. Proc. R. Soc. A (2005)

a

Ð

UC UK y

0

dx ! 0. This contradicts (2.9), and thus proves


2640


Furthermore, by (2.3), ðb N C1ð tjðnÞ X 0 0 jyn yn jdx Z jy jy dx ðnÞ n n a jZ1 tjK1 Z

C1 1 NX ðnÞ ðnÞ jðyn jyn jÞðtj Þ K ðyn jyn jÞðtjK1 Þj 2 jZ1

C1 1 NX ðnÞ ðnÞ jðyjyjÞðtj Þ K ðyjyjÞðtjK1 Þj 2 jZ1 NC1 X ð tjðnÞ Z ðN C 1Þkyn jyn j K yjyjkN C ðnÞ jyjy 0 dx t jK1 jZ1

%ðN C 1Þkyn jyn j K yjyjkN C

%ðN C 1Þkyn jyn j K yjyjkN C

ðb

jyy 0 jdx;

a

whence (2.6), implying ynjynj/yjyj uniformly, and (2.2) give ðb jyy 0 jdx R ðb K aÞKN : a

Ð In particular, y s0. Furthermore, ab ðy 0 Þ2 dx % 1 by (2.2) and (2.7). Thus, Q(y)RKN, but also Q(y)%KN since y 2DN. Hence, y is the maximizer we are looking for. & Remark 2.2. The proof shows that the statement of the proposition remains true when the zero integral condition is replaced by any set of conditions fi ½y Z 0

for i 2I ;

(2.10)

with some index set I, and with fi (i 2I ) denoting some bounded linear functionals on H 1(a,b) which are such that D:Z{y 2H 1(a,b): y satisfies (2.10)} and DN defined by ((2.1) using this D) contain non-zero elements, and such that the Poincare´ inequality kykL2 ða;bÞ % Cky 0 kL2 ða;bÞ is true for y 2DN. Choosing, for example, f1[y]Zy(a) and f2[y]Zy(b), or just f1[y]Zy(a), one obtains the a priori statement of proposition 2.1 for the situations underlying theorem 1.1 or theorem 1.2, respectively.

3. Proof of theorem 1.3 For fixed N 2 N, we define ~ dDN either ðiÞ D

~ dD; or ðiiÞ D

(3.1)

and K d sup QðyÞ: ~ y2Dnf0g


(3.2)



2641

We will prove, for both alternatives in (3.1), that ~ ~ i:e: QðyÞ Z K; if y 2 Dnf0g is a maximizer of Q on D; 1 a Cb then K Z and yðxÞ Z c x K for some c 2R: 4 2

(3.3)

The proof of theorem 1.3 is then straightforward: using alternative (i) in (3.1), and proposition 2.1, we obtain from (3.3) that KZKN Z1/4. This holds for every 1 N 2 N. Moreover, gN NZ0 DN is dense in D with respect to the H (a,b)-norm, since 2 for any given y 2D, the density of C [a,b] in L (a,b) in combination with Weierstrass’ approximation theorem gives a sequence (Qn) of polynomials converging to y 0 in L2(a,b), whence defining ð b ð t ðx 1 Pn ðxÞ d Qn ðsÞds K Q ðsÞds dt ðx 2½a;b; n 2NÞ; b Ka a a n a Ð and noting that ab yðxÞdx Z 0 implies ðx ð b ð t 1 0 0 yðxÞ Z y ðsÞds K y ðsÞds dt ðx 2½a;bÞ; b Ka a a a we obtain Pn /y in H 1(a,b), and Pn 2gN NZ0 DN for each n 2 N. This density result proves that sup{Q(y):y 2 D\{0}} is also 1/4, and thus the first part of theorem 1.3. The second part about the form of extremals follows immediately from (3.3) when using alternative (ii) in (3.1). To prove (3.3) we proceed in a series of lemmas. We restrict ourselves to the case a Z0, b Z1; the general case then follows by a change of variables. ~ ~ We normalize y by Thus, let y 2 Dnf0g be a maximizer of Q on D. ð1 jy 0 j2 dx Z 1; (3.4) 0

so that (3.2) gives 1 K

ð1 0

jyy 0 jdx Z 1:

(3.5)

Let MC dfx 2½0;1 : yðxÞO 0g;

MK dfx 2½0;1 : yðxÞ! 0g:

Since y has zero integral, we have ð ð y dx Z K y dx O 0: MC

(3.6)

MK

With no loss of generality (possibly after replacing y by Ky), we may assume that mðMKÞ% mðMCÞ; (3.7) whence, in particular, 1 0! mðMKÞ% : 2 Proc. R. Soc. A (2005)

(3.8)


2642


Lemma 3.1. Let 1 gd Ð MC y dx 1 hÐ MK y dx

ð

1 K

1 K

0

jyy jdx K

ð

0 2

jy j dx

MC

MC

ð

ð

0

jyy jdx K

MK

0 2

(3.9)

jy j dx

MK

where the last equality follows from (3.4) to (3.6). Then, jy 0 j2 Z 1 C 2gy

a:e: on ½0;1:

(3.10)

Proof. Let 42C [0,1] be fixed, and choose 30 O0 such that, for 32(K30,30), ð ð minð1 C 34ÞO 0; ð1 C 34Þy dx O 0; ð1 C 34Þy dx ! 0 ½0;1

MC

MK

(note (3.6)). Define for x 2[0,1] Ðx ð1 C 34Þdt ; F3 ðxÞ d Ð01 0 ð1 C 34Þdt which are both increasing C 1-mappings of 8 C l yðJ3 ðxÞÞ > < 3 y3 ðxÞ d lK 3 yðJ3 ðxÞÞ > : 0

J3 dFK1 3 ;

[0,1] onto itself. Set if x 2F3 ðMCÞ; if x 2F3 ðMKÞ;

(3.11)

otherwise;

where lC 3 dÐ

1 ; MC ð1 C 34Þy dx

lK 3 dK Ð

1 : MK ð1 C 34Þy dx

(3.12)

Since y is continuous and thus vanishes on vMCh(0,1) and on vMKh(0,1), y3 vanishes on vF3(MC)h(0,1) and on vF3(MK)h(0,1), and is therefore in H 1(0,1). Furthermore, MCh(0,1) and MKh(0,1) are both disjoint unions of open intervals, whence we can utilize the transformation t ZJ3(x) to obtain ð1 ð ð K y3 ðxÞdx Z lC yðJ ðxÞÞdx C l yðJ3 ðxÞÞdx 3 3 3 0

F3 ðMCÞ

ð

Z lC 3

yðtÞF30 ðtÞdt C lK 3

MC

Z Ð1 0

ð

F3 ðMKÞ

yðtÞF30 ðtÞdt

MK

1 ð1 C 34Þdt

ð ð C K ð1 C 34Þy dt C l3 ð1 C 34Þy dt Z 0 l3 MC

MK

~ in the case of alternative (ii) in (3.1). In by (3.12), whence y32 D, that is, y3 2 D the case of alternative (i), there exist a et0 % t1 %.% tN % tNC1db, such that y 0 R0 a.e. on [tj K1,tj ] or y 0 % 0 a.e. on [tj K1,tj ], for each j Z1, ., N C1. Hence, K 0 0 (3.11) and the positivity of lC 3 ; l3 and J3 show that, for j Z1, ., N C1, y3 R 0 Proc. R. Soc. A (2005)


2643


a.e. on [F3(tj K1),F3(tj)] or y30 % 0 a.e. on [F3(tjK1),F3(tj)]. This gives y32 DN, ~ also in this case. i.e. y3 2 D ~ Q½y3 % Q½y for 32ðK30 ; 30 Þ: Moreover, Therefore, since y maximizes Q on D; K C lC 0 Z l0 by (3.6) and (3.12), implying that y0 Z l0 y and hence Q[y0]Z Q[y]. Consequently, d (3.13) Q½y3 j3Z0 Z 0; d3 provided that the derivative exists, which however will follow from the calculations below. The next step is to compute Q[y3] and ðd=d3ÞQ½y3 j3Z0 . By (3.11), ð1 ð 0 C 2 jy3 y3 jdx Z ðl3 Þ jyðJ3 ðxÞÞy 0 ðJ3 ðxÞÞjJ30 ðxÞdx 0

F3 ðMCÞ

ð

2 CðlK 3 Þ

2 Z ðlC 3 Þ

ð

jyðJ3 ðxÞÞy 0 ðJ3 ðxÞÞjJ30 ðxÞdx

F3 ðMKÞ 2 jyðtÞy 0 ðtÞjdt C ðlK 3 Þ

MC

ð1 0

ef ð3Þ: Moreover, ðy30 Þ2

dx

ð

jyðtÞy 0 ðtÞjdt

MK

ð3:14Þ ð

2 Z ðlC 3 Þ

y 0 ðJ3 ðxÞÞ2 J30 ðxÞ2 dx

F3 ðMCð Þ K 2 Cðl3 Þ

ð

y 0 ðJ3 ðxÞÞ2 J30 ðxÞ2 dx

ð 1 1 K 2 y ðtÞ 0 dt C ðl3 Þ y 0 ðtÞ2 0 dt F3 ðtÞ F3 ðtÞ MK ð ð ð 1 MC jy 0 j2 jy 0 j2 C 2 K 2 dt C ðl3 Þ dt Z ð1 C 34Þdt ðl3 Þ 1 C 34 1 C 34 0

Z ðlC 3 Þ

F3 ðMKÞ 0 2

MC

MK

ð3:15Þ

egð3Þ: By (3.12), dlC 3 2 Z KðlC 3 Þ d3

ð 4y dx;

dlK 3 2 Z ðlK 3 Þ d3

MC

ð 4y dx; MK

l0 , give whence (3.14) and (3.15), and the fact that ð ð ð ð 0 K 3 0 0 C 3 f ð0Þ Z K2ðl0 Þ jyy jdx 4y dx C 2ðl0 Þ jyy jdx 4y dx K lC 0 Z l0 e

Z 2l30

MK MK ðMC ðMC ð ð 0 0 K jyy jdx 4y dx C jyy jdx 4y dx ; MC


MC

MK

MK


2644


and 0

g ð0Þ Z

ð1 4 dx 0

2 ðlC 0Þ

ð

3 K2ðlC 0Þ

ð

0 2

jy j dx

MC

jy j dx

MC

2 KðlC 0 Þ

Z l20

ð

MC

2 4jy 0 j2 dx K ðlK 0Þ

MC

ð 1

ð1

4 dx K 0

C2l30

4y dx

4jy 0 j2 dx

0 2

jy j dx

MK

ð

0 2

ð

2 C ðlK 0Þ

ð

3 C 2ðlK 0Þ

0 2

jy j dx

MK

ð

ð 4y dx MK

4jy 0 j2 dx

MK

0

ð

0 2

jy j dx

K MC

ð

ð 4y dx C

MC

0 2

jy j dx

MK

ð 4y dx MK

(note also (3.4)). Since Q½y3 Z f ð3Þ=gð3Þ and f ð0Þ=gð0ÞZ K, condition (3.13) amounts to g 0 ð0ÞK K1 f 0 ð0ÞZ 0, that is, ( ð ð ð ð1 1 0 2 0 0 2 4½1 K jy j dx C 2l0 jyy jdx K jy j dx 4y dx K 0 MC

MC

MC

) (3.16) ð ð ð 1 jyy 0 jdx K jy 0 j2 dx 4y dx Z 0: K K MK

MK

MK

Using (3.12), (3.6) and (3.9), we find that (3.16) is equivalent to ð1 4½1 K jy 0 j2 C 2gydx Z 0; 0

which implies (3.10) since 4 2C [0,1] is arbitrary.

&

1

Lemma 3.2. Let f be any function in H (0,1) such that f O0 on MC, f !0 on MK, and jf 0 j %1 a.e. on [0,1]. Then ð 1 f dx % mðMCÞ2 ; (3.17) 2 MC

with equality holding if and only if MC is an interval and f (x) Zjx K xj on MC, with x denoting one of the endpoints of MC. Correspondingly, ð 1 K f dx % mðMKÞ2 ; 2 MK

with equality holding if and only if MK is an interval and f(x) ZKjx K hj on MK, with h denoting one of the endpoints of MK. Proc. R. Soc. A (2005)


2645


Proof. Since y is continuous, MCh(0,1) is the disjoint union of finitely or countably many open intervals Ii . For each fixed i, the function f vanishes at least at one endpoint xi of Ii, since f is continuous and at least one endpoint of Ii does not belong to MC. Thus, since jf 0 j %1 a.e. on [0,1], f ðxÞ% jx K xi j which implies

ð

on Ii ;

(3.18)

ð 1 1 f dx % jx K xi jdx Z mðIi Þ2 % mðMCÞmðIi Þ; 2 2

Ii

(3.19)

Ii

with equality holding (everywhere in this chain) if and only if equality holds in (3.18), and m(Ii)Zm(MC). Summation over i in (3.19) yields (3.17), with equality holding if and only if there is just one interval Ii and equality holds in (3.18). The corresponding statement on MK follows analogously. & Lemma 3.3. gR0, and ð nm o ; 1 K 2m ; G dg y dx % min 2

(3.20)

MK

where mdm(MK) (see (3.8)). Proof. We utilize again (as in the proof of lemma 3.2) the subdivision of K MCh(0,1) and of MKh(0,1) into disjoint open intervals fI C i g and fI j g, respectively, and the fact that for each i and j, y vanishes at least at one endpoint K of IC i and I j , respectively. Using the half interval inequality (1.2) in theorem 1.2, we obtain for each i that ð ð ð 1 1 0 C 0 2 jyy jdx % mðIi Þ jy j dx % mðMCÞ jy 0 j2 dx; 2 2 IiC

IiC

IiC

whence summation over i gives ð ð 1 0 jyy jdx % mðMCÞ jy 0 j2 dx: 2 MC

(3.21)

MC

Since K R 1=4 (which follows from (3.2) and Q½~ yZ 1=4 for y~ðxÞZ x K 1=2), (3.21) further implies that ð ð ð 1 0 0 2 jyy jdx % 2mðMCÞ jy j dx % 2ð1 K mÞ jy 0 j2 dx: K MC

Analogously, 1 K

MC

ð MK


0

jyy jdx % 2mðMKÞ

MC

ð MK

0 2

jy j dx Z 2m

ð MK

jy 0 j2 dx:


2646


By (3.9), we therefore obtain ð ð g y dx % ð1 K 2mÞ jy 0 j2 dx % 1 K 2m MC

(cf. (3.8)), as well as

(3.22)

MC

ð

g

y dx % ð2m K 1Þ

MK

ð

jy 0 j2 dx % 0

MK

by (3.8), whence (3.6) proves that gR 0. Moreover, (3.10) implies ð ð 1 m 0 2 g y dx Z jy j dx K mðMKÞ RK ; 2 2 MK

MK

which, together with (3.6) and (3.22), gives (3.20).

&

Now, we want to ‘solve’ the differential equation (3.10). Formally (i.e. without pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 regarding zeroes of 1C 2gy), (3.10) yields g ð 1C 2gyÞ 0 Z 1 if g s0, that is, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 1 g 1C 2gy Z zðCconst:Þ for some function z such that jz j h1. The ‘natural’ specification for the constant is 1=g, since 1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi zZ 1 C 2gy K 1 g gives z hy in the limit g/ 0, whence jz 0 j h 1 in every case. For a rigorous proof, we have to circumvent differentiability problems (and possible multiple solutions of initial-value problems) when 1C 2gy has zeroes. For this purpose, we ‘regularize’ the above function z by a parameter 3O0, which leads to the definition (3.23) below. Lemma 3.4. mðMCÞZ mðMKÞZ 1=2; gZ 0; and K Z 1=4. Proof. Let 3O 0, and define z 2H 1(0,1) by 8 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi 1 > > 1 C 3 C 2gy K 1 C 3 y > : pffiffiffiffiffiffiffiffiffiffiffi 1 C3

if g s0; (3.23) if g Z 0:

(Note p ffiffiffiffiffiffiffiffiffiffi that 1C2gy R 0 by (3.10).) Therefore, in both cases, 1C 3 C gz, implying pffiffiffiffiffiffiffiffiffiffiffi gz RK 1 C 3 on ½0;1;

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1C 3C 2gy Z (3.24)

and pffiffiffiffiffiffiffiffiffiffiffi 1 y Zz 1 C 3 C gz : 2 Proc. R. Soc. A (2005)

(3.25)



2647

pffiffiffiffiffiffiffiffiffiffi Inequality (3.24) shows in particular that 1C 3 C gz=2O 0 on [0,1], whence (3.25) gives z O0 on MC and z! 0 on MK. Moreover, (3.23) and (3.10) imply sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0 jy j 1 C 2gy % 1 a:e: on ½0;1; (3.26) jz 0 j Z pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Z 1 C 3 C 2gy 1 C 3 C 2gy so that by lemma 3.2

ð 2 z dx % mðMKÞ2 Z m2 :

(3.27)

MK

ð

Furthermore, using (3.25), ð ð ðð 1 1 yzdxK ydx zdxZ m 2m

MK

MK

MK

½yðxÞKyð~ x Þ½zðxÞKzð~ x Þdxd~ x

MK !MK

ðð

1 Z 2m

pffiffiffiffiffiffiffiffiffiffi 1 ½zðxÞKzð~ x Þ x ÞÞ dxd~ x; 1C3C gðzðxÞCzð~ 2 2

MK !MK

which is non-negative by (3.24). Thus ð ð ð 1 yzdx R ydx zdx : m MK

MK

(3.28)

MK

pffiffiffiffiffiffiffiffiffiffi In ffiaddition, by (3.25) and lemma 3.3, yR 1C3 z on [0,1], whence jyj% pffiffiffiffiffiffiffiffiffi 1C3jzj on MK, and thus ð ð 1 z 2 dx R pffiffiffiffiffiffiffiffiffiffi yzdx: (3.29) 1C3 MK

MK

Moreover, with x 02MC chosen such that z (x 0)Z max{z (x):x 2MC}ez max, (3.26) implies zðxÞRz max KjxKx 0 j for x2½0;1:

(3.30)

Since z has at least one zero, (3.30) shows that at least one of the two intervals (x 0Kz max,x 0) and (x 0,x 0 Cz max) is completely contained in MC. Thus, with I denoting this interval, (3.30) gives ð ð ð 1 zðxÞdxR zðxÞdxR ½z max KjxKx 0 jdxZ z 2max : (3.31) 2 MC

I

I

Ð1 Ð1 Finally, Ð 0 (bylemma 3.3) and 0 ydxZ0, (3.25) implies that 0 zdx%0, Ð since gR that is MC zdx% MK zdx , whence (3.31) gives 1=2 ð z max % 2 zdx : (3.32) MK



2648


Now, using (3.5), (3.25), (3.24), (3.26) and (3.6), we obtain ð1 ð1 pffiffiffiffiffiffiffiffiffiffi KZ jyy 0 jdxZ jyjjz 0 jð 1C3CgzÞdx 0

0

ð1 jyjdxCg jyjzdx 0 ð 0 ð ð pffiffiffiffiffiffiffiffiffiffi ZK2 1C3 ydxCg K yzdxC yzdx

pffiffiffiffiffiffiffiffiffiffi % 1C3

ð1

MK

MK ð MC ð ð pffiffiffiffiffiffiffiffiffiffi 2 zdxCg K 1C3 z dxK yzdxC yzdx :

ð ZK2ð1C3Þ

MK

MK

MK

MC

Thus, estimating according to (3.29), (3.28), (3.32), and using gR 0 and G defined in lemma 3.3, we obtain ð ð ð K %K2ð1C3Þ z dx Cg K2 yz dx Cz max y dx MK

ð %K2ð1C3Þ

MK

MC

ð ð 1=2 ð ð 2 z dx Cg K y dx z dx C 2 z dx y dx m

MK

MK

MK

MK

MK

1=2 ð ð 1 Zð1C3K GÞ2 z dx CG 2 z dx : m MK

MK

Since 1C3 K(1/m)GR 0 due to (3.20), we then utilize (3.27) to obtain 1 K %ð1C3K GÞm2 CGm Zð1C3Þm2 : m This holds for every 3 O0, whence K% m2. On the other hand, K R1=4. Therefore, by (3.8) we obtain KZ

1 1 and m ZmðMKÞ Z : 4 2

Now, (3.7) shows that also mðMCÞZ1=2, and finally (3.20) gives GZ 0, and thus gZ0. & It is now easy to complete the proof of (3.3). Lemma 3.4 and (3.10) give jy 0 j Z 1 a:e: on ½0;1;

(3.33)

so lemma 3.2 provides ð MC


1 y dx % ; 8

ð K MK

1 y dx % : 8

(3.34)


2649


Moreover, by (3.5), (3.33) and lemma 3.4, ð ð ð1 1 Z K Z jyjdx Z y dx K y dx: 4 0 MC

MK

Therefore, equality must hold in both inequalities in (3.34). Thus, lemma 3.2 implies that both MC and MK must be intervals (of length 1/2), that is

(3.35) MCh ð0; 1Þ Z 0; 12 ; MKh ð0; 1Þ Z 12 ; 1 or vice versa, and that yðxÞ Z x

or

1 2

Kx

on

0; 12 ;

1 2

yðxÞ Z x K 1

or

Kx

on

yðxÞ Z 1 K x

or x K 12

on

1 2

;1

in the case of (3.35), and yðxÞ Z Kx

or x K 12

on

0; 12 ;

in the opposite case. Since y is continuous, we obtain yðxÞZ x K 12 on [0,1].

1 2

;1

or

1 2

Kx

4. A new proof of theorems 1.1 and 1.2 The following is by no means the easiest or shortest proof of Opial’s inequality or its half interval version.3 However, it is a new argument and we include it since the technique may be useful in the proof of other inequalities. Again, it suffices to consider the case a Z0, b Z1. Using remark 2.2 after proposition 2.1, we see that it is sufficient to prove a statement corresponding to (3.3), with the obvious changes concerning the value ~ defined correspondingly, let of K and the form of extremals. Thus, with D ~ y 2 Dnf0g be a maximizer of Ð1 0 jzz j dx Q½z d Ð01 0 2 0 ðz Þ dx ~ normalized by (3.4) and (3.5), with K given by (3.2). Let 42C[0,1] be on D, fixed, and choose 30 O 0 such that, for 3 2(K30,30), minð1 C 34ÞO 0; ½0;1

and define again

Ðx

F3 ðxÞ d Ð01

ð1 C 34Þdt

0 ð1 C 34Þdt

on ½0;1;

J3 dFK1 3 ;

and this time y3 ðxÞ dyðJ3 ðxÞÞ 3

on ½0;1:

This distinction probably belongs to Mallows, whose proof of theorem 1.2 takes two lines. The proofs of Olech (1960) and Levinson (1962) are also short (see Agarwal & Pang 1995, pp. 5–9).



2650


~ in both theorems 1.1 and 1.2 (by Since J3(0)Z 0 and J3(1) Z 1, y3 lies in D the same argument as given in the proof of lemma 3.1), whence d Q½y3 j3Z0 Z 0: d3

(4.1)

In addition, because ð1 ð1 ð1 0 0 0 jy3 y3 jdx Z jyðJ3 ðxÞÞy ðJ3 ðxÞÞjJ3 ðxÞdx Z jyy 0 jdt; 0

ð1 0

0

ðy30 Þ2 dx Z

ð1 0

ð1 Z

0

y 0 ðJ3 ðxÞÞ2 J30 ðxÞ2 dx Z

ð1 C 34Þdt

ð1

0

0

ð1

y 0 ðtÞ2

0

1 F30 ðtÞ

dt

ðy 0 Þ2 dt; 1 C 34

(4.1) amounts to ð ð1 ð1 ð1 d 1 ðy 0 Þ2 0Z ð1 C 34Þdt dt Z 4 dt K 4ðy 0 Þ2 dt d3 0 0 1 C 34 0 0 3Z0 (note (3.4)). Since 4 2C[0,1] is arbitrary, we obtain that jy 0 j Z 1 a:e: on ½0;1;

(4.2)

and (3.5) gives ð1 KZ

jyjdx:

(4.3)

0

Now, to distinguish between the two theorems we wish to prove, we set y~ dminfx;1K xg for theorem 1.1 and y~ dx for theorem 1.2. Property (4.2) and the condition y(0)Zy(l)Z0 for theorem 1.1, or y(0)Z 0 for theorem 1.2, imply that jyj% y~ on [0,1], whence ð1 ð1 jyjdx % y~ dx Z Q½~ y % K: (4.4) 0

0

Now, (4.3) shows that equality must hold everywhere in the inequality chain (4.4), so that K Z Q½~ yZ 1=2 in theorem 1.2 (Z1/4 in theorem 1.1), and jyjZ y~ on ½0;1, whence, by continuity, yZ y~ or yZK~ y. & The authors are grateful to Jochen Denzler and to two anonymous referees for useful suggestions.

References Agarwal, R. P. & Pang, P. Y. 1995 Opial inequalities with applications in differential and difference equations. Dordrecht: Kluwer. Beesack, P. R. 1962 On an integral inequality of Z. Opial. Trans. Am. Math. Soc. 104, 470–475. Brown, R. C. 2003 An Opial-type inequality with an integral side condition. J. Inequal. Pure Appl. Math. (electronic) 4, 37–42. Proc. R. Soc. A (2005)



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Brown, R. C. & Hinton, D. B. 1997 Opial’s inequality and oscillation of 2nd order equations. Proc. Am. Math. Soc. 125, 1123–1129. Brown, R. C., Fink, A. M. & Hinton, D. B. 2000 Some Opial, Lyapunov, and de la Valee-Poussin inequalities with non-homogeneous boundary conditions. J. Inequal. Appl. 5, 11–37. Brown, R. C., Clark, S. & Hinton, D. B. 2002 Some function inequalities and their application to problems in differential equations. In Analysis and applications (ed. H. P. Dikshit & P. K. Jain), pp. 19–42. New Delhi, India: Narosa Publishing House. Clark, S. & Hinton, D. B. 2002 Positive eigenvalues of second order boundary value problems and a theorem of S.G. Krein. Proc. Am. Math. Soc. 130, 3005–3015. Denzler, J. 2004 Opial’s inequality for zero-area constraint. Math. Inequal. Appl. 7, 337–354. Levinson, N. 1962 On an inequality of Opial and Beesack. Proc. Am. Math. Soc. 15, 61–63. Mitrinovic´, D. S., Pecˇaric´, J. E. & Fink, A. M. 1991 Inequalities involving functions and their integrals and derivatives. Dordrecht: Kluwer. Olech, C. 1960 A simple proof of a certain result of Z. Opial. Ann. Polon. Math. 8, 61–63. Opial, Z. 1960 Sur une ine´galite´. Ann. Polon. Math. 8, 29–32. Sinnamon, G. 2004 Reduction of Opial-type inequalities to norm inequalities. Proc. Am. Math. Soc. 132, 375–379.
