Taylor & Francis
J. oflnequal. & Appl., 2002, Vol. 7(5), pp. 663-672
Taylor & Francis Group
A Landau-Kolmogorov Inequality for
Orlicz Spaces
HA HUY BANG* and MAI THI THU
Institute of Mathematics, National Center for Sciences and Technologies, P.O.
Box 631, Bo Ho, 10000 Hanoi, Vietnam
(Received on 12 December 2000; In final form: 27 February 2001)
In this paper we prove that the Landau-Kolmogorov inequality for functions on the half line
holds for any Orlicz space with the constants, which are best possible for L-space.
Keywords: Landau-Kolmogorov inequality; Inequality for derivatives; Orlicz spaces.
Classification: 2000 AMS Subject Classification. 26D 10.
1
INTRODUCTION
The Landau-Kolmogorov inequality
-
f(k)I1 _< K(k, n)ll fll f(n)I1,
(1)
where 0 < k < n, is well known and has many interesting applications
and generalizations (see [1-6, 15, 18-21]). Its study was initiated by
Landau [11] and Hadamard [7] (the case n 2). For functions on the
whole real line [, Kolmogorov [9] succeeded in finding in explicit
form the best possible constants K(k, n) Ck,. in (1), and Stein proved
in [20] that inequality (1) still holds for Lp-norm, 1 < p < c, with
these constants (the same situation also happens for an arbitrary Orlicz
*Corresponding author. E-mail: hhbang@thevinh.ncst.ac.vn
This work was supported by the Natural Science Council of Vietnam.
ISSN 1025-5834 print; ISSN 1029-242X. (C) 2002 Taylor & Francis Ltd
DOI: 10.1080/1025583021000022441
H.H. BANG AND M. T. THU
664
Q,+,,, for the half line [+ [0, cxz) are not
known in explicit form except for n 2, 3, 4 (see 11, 13]), but an algorithm exists for their computation (Schoenberg and Cavaretta [17]). In
this paper, essentially developing the Stein method [20], we prove
that, for the half line, inequality (1) still holds for an arbitrary Orlicz
norm with the constants Ck+,,.
norm [1 ]). The best constants
2 RESULTS
"
Let G [, [+ or [a, b],
[0, +cx) --+ [0, +c] be an arbitrary
Young function [10, 12-14], i.e., (0)= 0, (t)> 0, (t) 0 and
is convex. Denote by
(t)
sup ts
(s)
s>0
the Young function conjugate to
functions u such that
I<u, v)l-
and L,(G)-the space of measurable
U(X)V(x)dx
for all v with p(v, ) < cx, where
(lv(x)l)dx.
p(v, )
G
Then L,(G) is a Banach space with respect to the Orlicz norm
IlulI,,G
sup
p(v,)_<
which is equivalent to the Luxemburg norm
Ilfll(’)
inf
{2 IG
>0
O(If(x)I/2)dx<_
I}
<co.
A LANDAU-KOLMOGOROV INEQUALITY FOR ORLICZ SPACES
665
Recall that II" II(,G
II(G) where (I)(t)= p with _< p < c,
and
[[c(6) when (I)(t) 0 for 0 _< t _< and (I)(t) o
[[(,6)
fort> 1.
We have the following results [13-14]"
LEMMA
Let u E L)(G) and v
’GlU(x)v(x)ldx
LEMMA 2 Let u
L(G).
Then
< IlulI.GIIVlI(,G.
L@() and v Ll ().
Then
LEMMA 3 [5, p. 37] Let n > 1. Iff L,toc([+) has a generalized n-th
derivative g E L,toc(+), then f can be redefined on a set of measure
zero so that f (n-) is absolutely continuous and f (n) g a.e. on ff+.
THEOREM
derivative
be an arbitrary Young fitnction, f and its generalized
be in L,(+). Then f(k)L(+) for all k E
and
Let
f(n)
{1,...,n- 1}
n-k
k
+
f(n) II,,+.
iIf(,)l.I,+ < C),,n
fll,i,,+
(2)
Proof We divide our proof into two steps.
Step 1 We begin to prove (2) with the assumption that f(k)6
n.
La>([+), k O, 1
Fix O<k<n. Let e>O be given. We choose a function
v, L([+), p(v, (I)) _< such that
()(x)v(x)dx
> f)ll,+
e.
(3)
H.H. BANG AND M. T. THU
666
Put
F.(x)
Then F,:(x)
6
f(x + y)v.(y)dy.
Loo([+) by virtue of Lemma 1, and it is easy to check that
Ix)
f(") (x + y)v,,. ( y)dy, r
O,
in the 79’(0, cxa) sense.
Since p(v., (I)) < 1, [[v.[[(,u+) < 1. So, for all x
(4)
n
+, clearly,
IF")(x)l IIf(r)(x + ")ll,,+llv,:llt,+)
F
Now we prove the continuity of ") on E+. We show this for r 0 by
contradiction: Assume that for some 6 > 0, a point x and a sequence
{tin} in [ with x + tm .>_ 0 and t,,, --+ 0 we have
(f(x
+ tm + y) -f(x + y))v,:(y)dy
> 6, rn
.
(5)
Since f 6 L,(+) we easily get f L,to(+). Then f(x+
Therefore, there exists
t,,, + .) --+ f(x + .) in L[0,j] for any j 1,2
such
denoted
a subsequenee,
again by {tin},
thatf(x + tm + y) -’+ f(x + y)
a.e. in [0,j]. So, there exists a subsequence (for simplicity of notation we
assume that it coincides with {tin}) such that f(x +tm +y)--+ f(x+ y)
a.e. in [0, cxa).
For simplicity of notations we consider only the case when x ---0.
Because inequality (2) holds for f if and only if it holds for f/C,
where C is an arbitrary positive number, without loss of generality we
may assume that p(2f, ) < cxz. By the Young inequality we get
I.f(tm + Y) f(y)llv.(y)l
_< I)(I f(tm + y) -f(Y)l) + ((Iv,,,(y)l)
< 21-(2lf(y)l) + 1/2dO(2lf(tm +Y)I) + (Iv,(y)l).
(6)
A LANDAU-KOLMOGOROV INEQUALITY FOR ORLICZ SPACES
- ---
Since (2lf[), (Iv,l) LI([+) and tm
bers M and h such that for all m
>t
-
O, there are positive num-
((21 f(Y)l) + (21 f(tm "+ Y)I) + (Iv(y)l))dy <
and
t(21f(y)l)dy <
667
d(21f(tm +y)l)dy < g,
(7)
(]v(y)l)]dy <
(8)
if B C N+, mes(B) < h. On the other hand, by the Egorov theorem,
there is a set A C [0, M], mes(A) < h such thatf(tm + y)v,(y) uniformly
converges tof(y)v,(y) on [0, M]\A. Therefore, applying (6) and (8), we
have
lmi.rno
f(t + y) -f(Y)l Iv,(y)ldy
< lim
m---> cx:
If(t,, + y)
[0,MI\A
f
6
6 6 6
lina JA If(tm + y) -f(y)l[v(y)ldy Z + + 5"
(9)
Combining (7), (9) and using (6), we get for sufficiently large m
l(f(tm
+ y)-f(y))v(y)ldy <
,
which contradicts (5). The cases 1 < r < n are proved similarly. The
continuity of r) has been proved.
F
H.H. BANG AND M. T. THU
668
+.
F
The functions ") are continuous and botmded on
Therefore, it
follows from the Landau-Kolmogorov inequality and (3)-(4) that
(llJm)ll.,+- e,)" IF)(0)l" IIFk)ll
< C+
(lO)
On the other hand,
IIf(x + ")11..+ IIv,(’)ll,+) Ilfll,+,
IIF,llo
IIF,l")llo < IIf(")(x + ")ll,,+llv,,(’)ll(,+)
(11)
(12)
Combining (10)-(12), we get
(11 f) I1,,+
e,)"
+k,n II.zCll "-a,,+ f")
< C
By letting e -+ 0 we have (2).
Step 2 To complete the proof, it remains to show that f(’)6
if f,ft") L.(R+). By Lemma 3 we can
n
L.(+), Yk 6
assume thatf,f’
f("-) are continuous on andf("-!) is absolutely
continuous on
+
+.
We define for k
0,
f)(x)
Let
J 6 C(O, cx)),
put t/C(x)
n,
f(t)(x),
o,
>_ O, tp(x) 0 for x >_
1/2.(x/2), 2 > 0 and J). -fo)*
.
and
Ju t/J(x)dx
1. We
A LANDAU-KOLMOGOROV INEQUALITY FOR ORLICZ SPACES
669
Fix b>0. Then YqgC(b,o) we have for 0<2<b and
n
k--1
So, we have proved that for 0 < 2 < b and k
=),
in the
"D’(b, c)
n
,
(13)
sense. Therefore, for 0 < 2 < b we have
II(fio) * ’z)(’) IIq>,i,) lift,) * ll,,t,)
_< fn) * q’, I1, --< Jn)I1,
J,,)II,i,,+ f(n)I1,+.
(14)
On the other hand, using ( fo) * P)(’) fo) * P’) L([),
Yk O,
n and the proved in Step
Landau-Kolmogorov inn l,
equality for functions on [b, cx)), we get for k 1
i1,,_:
fj(n) ik
H.H. BANG AND M. T. THU
670
Hence, combining (13), (14) we obtain for all 0<2<b, k=
n-l,
(15)
On the other hand, because f,) is continuous on I+, we easily get
lim fk)
20
* $),(x) =fo(x) =f()(x), ’v’x > O.
(16)
Indeed, for 2 < x we have from the continuity offk) at x that
For each function v L-[b, oo), p(v, )< and 0 < 2 < b, by (15)
and the definition of the Orlicz norm we get
I(fk) * q),)(x)v(x)ldx
<
Q,+,,,llfll "-k
*,to,) f(")
Therefore, using Fatou’s lemma, (15) and (16) we obtain
k
*,Io,)"
A LANDAU-KOLMOGOROV INEQUALITY FOR ORLICZ SPACES
So, by the definition of the Orlicz norm we have
"- f(n) Ila,,[0,o <
f(k) ll,t,o C+,.ll fll,,t0,ll
671
.
On the other hand, it follows from the continuity off(k) on [0, ) that
f(k) 6 Lo[O, b] for any b > O. Therefore,
f(k)II.,t0,)
f(k)I[,[0,bl -+- f(k)Ila,t,) <
.
The proof is complete.
Remark 1 To obtain Theorem 1 we have developed the Stein method
because, for example, the property [g(x + h)- g(x)]/h g’(x) in the
Lp mean (1 < p < ), which is used in [16], holds for L, only if
satisfies the A2-condition (see 12, 14]).
REMARK 2 By the representation [14]
ull,)
sup
Ilvll._<1
it is easy to see that Theorem 1 still holds for any Luxemburg norm.
Acknowledgements
In conclusion the authors would like to thank Professor Dinh Dung for
the valuable discussions.
The second author would like to thank Hanoi Institute of Mathematics for a research grant
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