G. W. HAGERTY AND P. NAG Received 11 January 2005
Kolmogorov (1949) determined the best possible constant Kn,m for the inequality Mm(f)≤Kn,mM(n0−m)/n(f)Mm/nn (f), 0< m < n, where f is any function withnbounded, piecewise continuous derivative onRand Mk(f)=supx∈R|f(k)(x)|. In this paper, we provide a relatively simple proof for the case of equality.
1. Introduction
While investigating summability methods for infinite series [5], Hardy and Littlewood posed an interesting problem which Kolmogorov solved 28 years later and that is the topic for this paper.
Write f =O(g) if and only if limx→∞f(x)/g(x)<∞. Hardy and Littlewood showed that if f is twice continuously differentiable forx > xo and if f =O(1) and f=O(1), then f=O(1).
More generally, they proved that ifφ,ψare increasing and f(n)is continuous, then for 0< m < n, if f =O(φ) and f(n)=O(ψ), then f(m)=O(φ(n−m)/nψm/n).
These theorems were important due to their applications to Dirichlet’s series—series of the type∞k=1akk−m. In their proof, Hardy and Littlewood show that the quantities
χm(x)=max
y≤x
f(m)(y)
φ(n−m)/n(y)ψm/n(y), (1.1)
are bounded independently ofx.
By lettingφ= f and ψ= f(n) in (1.1) and lettingx→ ∞, one observes that χm is bounded if and only if the inequality
Mm(f)≤Kn,mM0(n−m)/n(f)Mnm/n(f), 0< m < n, (1.2) Mk(f)=sup
x∈R
f(k)(x), (1.3)
Copyright©2005 Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences 2005:11 (2005) 1781–1793 DOI:10.1155/IJMMS.2005.1781
holds for some constantKn,m. Hardy and Littlewood conjectured that a constant Kn,m existed for which the inequality would hold for all functions withnbounded derivatives, and the race was on to find the best constant.
The first breakthrough came in [7]. Motivated partly by the above theorems and partly by his own previous work, Landau was able to show that the valueK2,1=√
2 for func- tions which are twice differentiable. He also considered the related problem on a finite interval, and showed that if f is defined on an interval of sufficient length and if the def- initionMk(f) is modified appropriately, thenK2,1=2. Landau considers the case where the second derivative is continuous separately from the case where it is only assumed to be bounded.
Within the following year, Hadamard [4] extended Landau’s result by proving that Kn,1≤2(n−1)/n.
The best value forKn,mforn <5 andn=5,m=2 was discovered in [1]. Kolmogorov [6] attributes these values to Silov. Silov’s result can be found in a paper written by Bosse [1].
In [3], Gorney obtained an upper bound ofKn,m≤16(2e)m. While Gorney’s value for Kn,1was much larger than the value obtained by Hadamard, Gorney successfully bounded Kn,mfor all values ofmandn, 1< m < n.
Finally, in [6], Kolmogorov observed that the functions used by Bosse could be used to maximize the quantity
γn,m= Mm(f)
M0(n−m)/n(f)Mnm/n(f), (1.4) wheren∈N, 0< m < n. Specifically, Kolmogorov showed that
Kn,m=max
f γn,m(f)=γn,mgn, (1.5) where
gn(x)= 4 π
∞ k=0
sin(2k+ 1)x−nπ/2
(2k+ 1)n+1 (1.6)
is thenth integral of the square function (seeFigure 1.1).
Remark 1.1. In any quarter period where bothgn(x),gn(x)>0, we havegn(x)<0.
The first few values ofKn,mare [6]
K2,1=√
2, K3,1=
√3
9
2 , K3,2=√3
3. (1.7)
Kolmogorov’s proof, although elementary, was very complicated. In this paper we will give a modified proof of Kolmogorov’s theorem. Our techniques give us the insight
1 0
−1 g0(x)
−8 −6 −4 −2 0 2 4 6 8
1 0
−1 g1(x)
−8 −6 −4 −2 0 2 4 6 8
1 0
−1 g3(x)
−8 −6 −4 −2 0 2 4 6 8
1 0
−1 g4(x)
−8 −6 −4 −2 0 2 4 6 8
Figure 1.1. Plot of comparison functions.
needed to characterize all functions for which equality holds in (1.2) withKn,m=γn,m(gn).
We note that for everyn,gnhas a discontinuousnth derivative, and in fact we will show that all functions for which equality holds have discontinuousnth derivatives.
Boor and Schoenberg [9] proved that the case of equality was true only for the com- parison functions whenn≥3 and true for a class of functions which were a modifica- tion of the comparison function forn=2. The proof however is quite complicated and technical. In [8] Schoenberg discusses the results forn=2 and 3 using concepts from elementary differential and integral calculus. However, in this article Schoenberg points out that though the underlying ideas for proving the result forn≥4 are simple as the cases n=2 or 3, the elementary approach does not work because the tools necessary to establish them becomes quite involved and complicated. Finally, Cavaretta [2] proves Kolmogorov’s theorem for all values ofnusing Rolle’s theorem and the Leibnitz formula for differentiation of a product.
The modification that is made in this article significantly modifies the case of equality for all values ofn.
2. Comparison functions
Forn∈N, letᏮndenote the class of all bounded (n−1) times differentiable functions whosenth derivative is continuous almost everywhere and bounded.
Definition 2.1below is a modification of a definition of Kolmogorov and is the key for simplifying the proof in the case of equality.
Definition 2.1. Supposen∈N,f ∈Ꮾn. We say thatφnis a comparison function of order nof f if and only if
φn(x)=agn(bx+c), (2.1)
wheregnare the functions defined in (1.6) and the constantsaandbare chosen such that M0(f)≤M0
φn
, Mn(f)=Mn
φn
. (2.2)
We say thatφnis a comparison function of f atx0if in addition we have fx0>φn
x0. (2.3)
Note that for any f ∈Ꮾn a comparison function of orderncan be constructed by lettinga=M0(f)/M0(gn) andabn=Mn(f). Furthermore, sinceφntakes all values be- tween±M0(φn), we can choosecso thatφnis a comparison function atx0, provided that
f(x0)=0.
Also, note thatγn,m(φn)=γn,m(gn)=Kn,mfor all choices ofa,bandc.
One advantage of the new definition is that ifφnis a comparison function of f atx0, then it is also a comparison function at all pointsxin some interval containingx0.
Comparison functions possess the following remarkable property.
Theorem2.2. Letn≥2, f ∈Ꮾn. Ifφnis a comparison function of f of ordernatx0, then
fx0<φnx0. (2.4)
The proof will be given later. For now, we will assumeTheorem 2.2to be true and prove some important consequences.
Corollary2.3. Supposen≥2, f ∈Ꮾn, and supposeφnis a comparison function of or- dern. Thenφ(m)n (x)is a comparison function of f(m) of order(n−m)for 0< m < n. In particular,Mm(f)≤Mm(φn).
Proof. We provem=1 only, since the other cases follow inductively.
Notice that ifφn(x)=agn(bx+c), then φn(x)=abgn−1(bx+c) and thatMn(φn)= Mn−1(φn). Thus, sinceM0(f)=M1(f),M0(φn)=M1(φn), to finish the proof it suffices to prove thatM1(f)≤M1(φn).
Choosex0such that
fx0=M1(f). (2.5)
If f(x0)=0, then we can translateφnto be a comparison function atx0. Consequently, byTheorem 2.2we have
M1(f)=fx0<φnx0≤M1
φn
. (2.6)
If f(x0)=0, then we may assume that there exist pointsx1 arbitrarily close tox0 such that f(x1)=0. ByTheorem 2.2, we have
fx1<φnx1≤M1
φn. (2.7)
By lettingx1→x0and using continuity of f, we obtain the result.
Kolmogorov’s inequality is an immediate consequence ofCorollary 2.3.
Theorem2.4 ([6]). Supposen≥2,f ∈Ꮾn. Then
Mm(f)≤Kn,mM0(n−m)/n(f)Mnm/n(f), 0< m < n, (2.8) whereKn·m=γn,m(gn).
Proof. Choose a comparison functionφnsuch thatM0(f)=M0(φn). Then byCorollary 2.3and (1.4) and (1.5), we have
Mm(f)≤Mm φn
=Kn,mM0(n−m)/nφn
Mnm/nφn
. (2.9)
Therefore, we obtain
Mm(f)≤Kn,mM0(n−m)/n(f)Mnm/n(f), (2.10)
whereKn·m=γn,m(φn).
It is also interesting to note thatTheorem 2.4impliesCorollary 2.3.
Theorem2.5. IfTheorem 2.4is true, thenCorollary 2.3is true.
Proof. Supposen≥2, f ∈Ꮾnand thatφnis a comparison function of ordernof f. Then sinceM0(f)≤M0(φn) andMn(f)=Mn(φn), we have byTheorem 2.4,
Mm(f)≤Kn,mM0(n−m)/n(f)Mnm/n(f)
≤Kn,mM0(n−m)/n
φn
Mnm/nφn
=Mm
φn
. (2.11)
Therefore,M0(fm)≤M0(φn(m)). SinceMn−m(fm)=Mn−m(φ(m)n ) we conclude thatφ(m)n is
a comparison function f(m)of order (n−m).
3. Proof ofTheorem 2.2
We will prove Theorem 2.2 by an inductive process involving both Theorem 2.2 and Theorem 2.4. The proof follows the same strategy that Kolmogorov used, but with sim- plification afforded by our modified definition of comparison functions. We will prove the Theorem by proving the following lemmas.
Lemma3.1. Theorem 2.2is true forn=2.
Lemma3.2. IfTheorem 2.2is true forn=k≥2, thenTheorem 2.4is true forn=k+ 1and m=1.
Lemma3.3. IfTheorem 2.2is true forn=k≥2, andTheorem 2.4is true forn=k+ 1and m=1, thenTheorem 2.2is true forn=k+ 1.
Proof ofLemma 3.1. SupposeTheorem 2.2is not true forn=2. Then there exists a func- tion f ∈Ꮾ2, a pointx0, and a comparison functionφ2of f atx0such that
fx0>φ2
x0, fx0≥φ2x0. (3.1) Without loss of generality, we may assume that f(x0)>0 and f(x0)≥0. If not, we can replace f with±f(±x). We can also assume thatφ2(x0),φ2(x0)≥0 by changing the sign ofaand shifting if necessary.
Since M0(f)≤M0(φ2), it follows thatφ2(x0)=M0(φ2). Letx1 be the first point to the right such thatφ2(x1)=M0(φ2). Note that we will haveφ2(x)<0 for allx∈(x0,x1).
Furthermore, since f(x0)> φ2(x0), f(x0)≥φ2(x0), and f(x1)≤φ2(x1), we have x1
x0
f(x)x1−xdx= fx1
−fx0
−fx0
x1−x0
< φ2
x1
−φ2
x0
−φ2x1−x0
= x1
x0
φ2x1−xdx.
(3.2)
Therefore there existsx2∈(x0,x1) such that fx2
< φx2
. (3.3)
Sinceφ2(x2)<0 andφ2 is the square wave function, we obtain fx2>φ2x2=M2
φ2
, (3.4)
contradictingM2(f)=M2(φ2). This completes the proof ofLemma 3.1.
Proof ofLemma 3.2. Choosex0such that|f(x0)| =M1(f). Without loss of generality, we may assume that f(x0)>0. Letφkbe a comparison function of fof orderksuch that φk(x0)=M0(φk)=M0(f). Letx1be the first point to the left ofx0such thatφk(x1)=0.
We claim that
f(x)≥φk(x), ∀x∈ x1,x0
. (3.5)
If not, then choose x2∈(x1,x0) such that 0< f(x2)< φk(x2). Let φkc(x)=φk(x+c) wherec <0 is chosen such thatφkcis increasing on [x2,x0],
φkcx2
=fx2
, φkcx0
< fx0
. (3.6)
Letx3be the first point to the left ofx0(seeFigure 3.1) such thatφkc(x3)= f(x3). Then 0< φkc(x)< f(x), for allx∈(x3,x0) and
x0
x3
φkc (x)dx=φkcx0
−φkcx3
< fx0
−fx3
= x0
x3
f(x)dx. (3.7)
−1
−0.5 0 0.5 1 1.5
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8
df /dx φk
φkc
x2 x3 x0
Figure 3.1. Construction for the proof ofLemma 3.2.
Therefore, there existsx4∈(x3,x0) such that 0< φkc
x4
< fx4
, 0< φkc x4
< fx4
. (3.8)
This contradictsTheorem 2.2and proves the claim.
Similarly, choosex1the first point to the right ofx0such thatφk(x1)=0. By the same argument as above, we obtain
f(x)≥φk(x)≥0, ∀x∈
x0,x1. (3.9)
Combining (3.5) and (3.9), we obtain 2M0(f)≥ fx1−fx1
= x1
x1
f(x)dx≥ x1
x1
φk(x)dx. (3.10) Now note thatφk(x)=agk(bx+c) is the derivative ofφk+1(x)=ab−1gk+1(bx+c). Since the pointsx1andx1are zeros ofφk(x), then we have
x1
x1
φk(x)dx=2M0
φk+1
. (3.11)
Therefore we have
M0(f)≥M0 φk+1
. (3.12)
Finally, sinceM0(φk)=M1(φk+1), we obtain M1(f)=Kk+1,1M0k/(k+1)φk+1
Mk+11/(k+1)φk+1
≤Kk+1,1M0k/(k+1)(f)M1/(k+1)k+1(f). (3.13)
This completes the proof ofLemma 3.2.
−1
−0.5 0 0.5 1 1.5
−0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 f
φk+1,shifted
φk+1
x0 x1
Figure 3.2. Construction for the proof ofLemma 3.3.
Proof ofLemma 3.3. SupposeTheorem 2.2is not true forn=k+ 1. Then, for an arbitrary functionf ∈Ꮾk+1and a pointx0, there exists a comparison functionφk+1of f atx0such that
fx0>φk+1x0, fx0≥φk+1x0. (3.14)
Since M0(f)≤M0(φk+1), the pointx0 cannot be a maximum forφk+1. Consequently, φk+1(x0)=0, which implies f(x0)=0.
Without loss of generality, we may assume thatf(x0)>0,f(x0)>0,φk+1(x0)≥0, and φk+1(x0)>0. Furthermore, by shiftingφk+1slightly to the left if necessary, we can replace
≥in the inequality (3.14) with>(seeFigure 3.2).
We now have
fx0
> φk+1
x0
>0, fx0
> φk+1x0
>0. (3.15)
Now letx1be the maximum ofφk+1which is closest tox0on the right, such that fx1
≤M0(f)≤M0
φk+1
=φk+1
x1
. (3.16)
We have x1
x0
f(x)dx=fx1
−fx0
< φk+1x1
−φk+1x0
= x1
x0
φk+1(x)dx. (3.17) Consequently, there existsx2∈(x0,x1) such that
fx2
< φk+1 x2
. (3.18)
Since we also have f(x0)> φk+1 (x0), there exists anx3to the left ofx2such that fx3
=φk+1x3 , f(x)> φk+1(x)>0, ∀x∈
x0,x3
. (3.19)
Thus, x3
x0
f(x)dx=fx3
−fx0
< φk+1x3
−φk+1x0
= x3
x0
φk+1(x)dx. (3.20) Therefore, there exists a pointx4∈(x0,x3) such that
fx4
> φk+1x4
>0, fx4
< φx4
<0. (3.21)
On the other hand, byTheorem 2.5, whenn=k+ 1,φk+1is a comparison function of orderkfor the function f. This concludes the proof ofLemma 3.3.
The inductive process proves thatTheorem 2.2holds forn≥2, and thatTheorem 2.4 holds forn≥3. It was proved earlier thatTheorem 2.4in the casen=2 follows directly fromCorollary 2.3.
4. The case of equality
Theorem4.1. Supposen≥2, f ∈Ꮾn, and suppose that for somem,0< m < n,
Mm(f)=Kn,mM0(n−m)/n(f)Mnm/n(f). (4.1) Then there exists constanta,b, andcsuch that
(a)forn=2,f(x)=φ2(x)=ag2(bx+c)forx∈[x0,x1]is a half period ofφ2for which
|φ2(x0)| = |φ2(x1)| =M0(φ2)=M0(f);
(b)forn≥3, f(x)=φn(x)=agn(bx+c)for allx∈R. Proof. We will do the proof in three steps.
Step 1. If (4.1) is true for n≥2 andm=1, then there existsφn(x)=agn(bx+c) such that f(x)=φn(x) forx∈[x0,x1] is a half-period ofφnfor which|φn(x0)| = |φn(x1)| = M0(φn)=M0(f).
To prove this, suppose that (4.1) holds for f. Choose a comparison function off such thatM0(f)=M0(φn) andMn(f)=Mn(φn). By (4.1) we haveM1(f)=M1(φn).
Letx2be a point such that|f(x2)| =M1(f).
If f(x2)=0, then there exists c such thatφn is a comparison function of f at x2; by Theorem 2.2, |f(x2)|<|φn(x2)|. But this contradictsM1(f)=M1(φn). Therefore
f(x2)=0.
Without loss of generality, we may assumef andφnare increasing atx2andφn(x2)=0.
Choose [x0,x1] centered atx2, a half period ofφn.
−1
−0.5 0 0.5 1 1.5
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8
φn
f
x2 x4 x3 x1
Figure 4.1. Construction for the proof ofTheorem 4.1.
We now claim that
f(x)=φn(x) ∀x∈ x2,x1
. (4.2)
Assume otherwise and suppose that there existsx3∈(x2,x1) such that f(x3)> φn(x3). Let x4be the first point to the left ofx3such that f(x4)=φn(x4) (seeFigure 4.1).
Thenx4∈[x2,x3) and
f(x)> φn(x)>0 ∀x∈ x4,x3
. (4.3)
Furthermore, x3
x4
f(x)dx= fx3
−fx4
> φn x3
−φn x4
= x3
x4
φn(x)dx. (4.4) Hence there existsx5∈(x4,x3) such thatf(x5)> φn(x5)>0, which along with (4.3) con- tradictsTheorem 2.2. Thus
f(x)≤φn(x), x∈ x2,x1
. (4.5)
To prove the inequality in the other direction, assume that there exists anx3∈(x2,x1) such that f(x3)< φn(x3). Then, sincef(x2)=φn(x2)=0, we have
x3
x2
f(x)dx=fx3
< φnx3
= x3
x2
φn(x)dx. (4.6)
Then there existsx4∈(x2,x3) such thatφn(x4)> f(x4).
We can assume that f(x4)≥0; if it were negative, then (since by assumption f(x2)>
0) we can choose a new pointx4where f(x4)=0, in which caseφn(x4)> f(x4) would hold trivially.
Letφnc(x)=φn(x+c) wherec <0 is chosen such that φncx4
= fx4
. (4.7)
Note that since f(x2)=M1(f), we will haveφnc(x2)< f(x2). Letx5be the first point to the right ofx2such thatφnc(x5)= f(x5). Hence
f(x)> φnc(x)≥0, x∈ x2,x5
, x5
x2
f(x)dx=fx5
−fx2
< φncx5
−φncx2
= x5
x2
φnc(x)dx. (4.8) This implies that there existsx6∈(x2,x5) such that
fx6
> φncx6
≥0, fx6
< φncx6
<0. (4.9)
Ifn=2, this contradictsM2(f)=M2(φnc).
Ifn≥3, then byCorollary 2.3,φnis a comparison function of fof order≥2, which implies thatφncis also a comparison function of fof order≥2. In this case (4.9) con- tradictsTheorem 2.2. Therefore f(x)=φn(x) for allx∈[x2,x1].
A similar argument shows that f(x)=φn(x) for allx∈[x0,x2]. This completes the proof ofStep 1.
Letting f be defined by
f(x)=
−M0
g2
ifx≤ −π 2, g2(x) if−π
2 < x≤π 2, M0
g2
ifx >π 2,
(4.10)
we see that this is the best possible result forn=2.
Step 2. If the hypothesis inStep 1is true forn≥3, then f(x)=φn(x) for allx∈R. To prove this, note that fromStep 1andCorollary 2.3,φnis a comparison functionf, φn(x)=f(x) for allx∈[x0,x1], and since fis continuous,
M2
φn
=φnx0=fx0. (4.11) Using this last expression and the definition of a comparison function, we find
M2
φn
=M2(f). (4.12)
Therefore (4.1) is true for the function f.
Sinceφn(x0)= f(x0)=0, then byStep 1, we can extend the equalityφn(x)=f(x) to the left ofx0by a quarter period to a pointx0. Similarly, we can extend the equality to the right ofx1by a quarter period to a pointx1.
We now have
φn
x0=φn
x1=fx0=fx1=0. (4.13) Hence, we can extend the equality in both directions another quarter period.
By continuing this process of going back and forth between the original functions and their first derivatives, we can extend the equality so thatφn(x)= f(x) for allx∈R. This completes the proof ofStep 2.
Step 3. If (4.1) is true for anyn≥2 and at least onemsuch that 2≤m < n, then there exists a comparison functionφn(x) such that f(x)=φn(x) for allx∈R.
To prove this, choose φn a comparison function of f such that Mn(f)=Mn(φn), M0(f)
=M0(φn). We will show thatM1(f)=M1(φn), such that the conclusion will follow from Step 2.
Suppose thatM1(f)< M1(φn). Choose a comparison functionψn−1of ordern−1 for fsuch thatM1(f)=M0(f)=M0(ψn−1). Then we have
M0
ψn−1
< M0
φn. (4.14)
Now, we can write
ψn−1(x)=a1gn−1
b1x+c1
, φn(x)=a2gn−1
b2x+c2
, (4.15)
where we assumea1,a2,b1, andb2 are nonnegative real numbers. From (4.14) we have a1< a2. FromMn−1(ψn−1)=Mn−1(f)=Mn(f)=Mn(φn)=Mn−1(φn), we havea1bn1−1= a2bn2−1. It follows that
Mm−1
ψn−1
=a1bm1−1< a2bm2−1=Mm−1
φn, 2≤m < n. (4.16) On the other hand, byCorollary 2.3,
Mm−1
f≤Mm−1
ψn−1
. (4.17)
Taken together, (4.16) and (4.17) contradictMm(f)=Mm(φn).
This provesStep 3, which completes the proof ofTheorem 4.1.
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G. W. Hagerty: Department of Mathematics, Black Hills State University, Spearfish, SD 57799- 9115, USA
E-mail address:[email protected]
P. Nag: Department of Mathematics, Black Hills State University, Spearfish, SD 57799-9127, USA E-mail address:[email protected]
