Internat. J. Math. & Math. Sci.
Vol. 9 No. 2 (1986) 331-340
331
IRREGULAR AMALGAMS
JAMES STEWART
Department of Mathematics
McMaster University
Hamilton, Ontario, Canada L8S 4KI
SALEEMWATSON
Department of Mathematics
Pennsylvania State University
Media, Pennsylvania 19063
(Received July 17, 1985)
ABSTRACT.
The amalgam of L
sequence of
LP-norms
p
q
and
consists of those functions for which the
over the intervals
been studied in several recent papers.
{In;neZ
a cover
each of the form
[a,b),
and investigate which properties of
(LP, q)
In particular, we study
These spaces (L p,q) have
[n,n+l)
by
(LP,q) carry over to
how (LP,q) varies as
n
amalgam
1980 MATHEMATICS SUBJECT CLASSIFICATION CODE.
46E30
INTRODUCTION.
The amalgam of L
in L
q.
vary and determine conditions under which translation is continuous.
KEYWORDS AND PHRASES.
[.
is in
of the real line consisting of disjoint half-open intervals I
these irregular amalgams
p, q, and
[n,n+l)
Here we replace the intervals
p
p
q
and
is the space
(LP, q)
of functions f which are locally
and satisfy
f!
LP-norms
p,q
if(x)iPdx]q/p}I/q < =,
[[n+l
-n
intervals [n,n+l] form an q-sequence.
[
(I.I)
n=-
(If either p
way.) Special
cases of amalgams were first introduced by Wiener [I], [2] and Stepanoff [3] in the
1920’s but their first systematic study is due to Holland [4].
p
Notice that for p
q we have (LP, p)
L and indeed some of the properties of
that is, the
over the
or q is infinite, the expression in (I.I) is modified in the usual
Lebesgue spaces can be carried over to amalgams.
norm given by (I.I) makes
I/p + I/p’
(LP, q)
For instance, if
into a Banach space with dual
4
p,q
(LP’,q’),
<
,
the
where
I, and the analogues of Holder’s inequality and Young’s inequality are
also valid.
It is not difficult to establish the following inclusion relations between
amalgams.
If
ql < q2’
then
(LP, ql
c
,
Lp q2 ).
(1.2)
JAMES STEWART and SALEEM WATSON
332
If
P2’
Pl
then (L
P2 ,q)
LPl ,q).
c
(1.3)
For further information about amalgams we refer the reader to [5].
In this paper we propose to replace the cover consisting of .the intervals
In n+l)
in
{I ;neZ}
(I I) by a cover
of the real line consisting of disjoint
[ab),
whose union is the real llne.
n
If
refer
to a cover of this type.
the
cover
use
term
to
Throughout the paper we
half-open intervals I
each of the form
’f!
we define the irregular
[fl n If(x)IPdx]q/p}I/q
{"
P,q,
n
amal.am
(L
p
q)
{f; flp,q,a
[6] on approxima-
Such spaces have arisen in the work of Jakimovski and Russell
tion theory.
Given a complex sequence
(Yn)nEZ
and a linear space S of real-valued
For a given increasing
functions, they considered the interpolation problem:
sequence (=), find a function f in S such that f(an
For certain normed
Yn"
n
sequence spaces E they took S to consist of the functions f with
L
p
an
and showed the existence of optimal solutions to the interpolation problem.
p
then S becomes the irregular amalgam (L q) where
that if we take E
Notice
q
n
An important special case of an irregular amalgam is the dyadic amalgam
(L
p,q)-nwhere
-2
If n
{[n,n+l)}
<
O.
J.W.
is the cover given by
Wells proved that if f
n
Fourier transform f lies in the dyadic amalgam (L
g
o
O,
LP(--,-),
n < 2n
if n
>
O,
2, then its
(Kellogg [7] and Williams
p’,2).
p
[8] proved similar results for the circle group and connected groups, respectively.)
Certain facts about irregular amalgams are straightforward generalizations of
kno results about regular amalgams.
proof.
THEOREM I.
If
whose dual space is (L
THEOREM 2.
<
p,q
p
If f e
,
*
q)
(L
(LP, q)
Thus we state the following theore without
the irregular aIgam (L
p
p, q)
is a Banach space
,q
and g e (L
lfgH
,q
p
), where p,q
glp, ,q
|fl;
P,q,a
I, then
fg e L
and
,
Other aspects of regular amalgams do not always generalize to irregular
amalgams.
For instance, as stated in [5], although the inclusion (1.2) holds for
irregular amalgams, (1.3) does not hold in general.
In fact we show in Section 2
that the analogue of (1.3) holds if and only if the lengths of the intervals I
bounded above.
.
are
n
In Section 3 we discuss inclusion relations between irregular amalgams when p
and q are fixed and the sequence
We write
varies.
intersect boundedly many of the intervals of
then
(LP’q)c (LP’q)B with
reversed.
B
=.
strictness if
It then follows that (L
p
q)
(L
p
B
if the intervals I
and show that if q
Eq)
If
B
q
p and
n
=
E
B,
p, the inclusion is
if and only if
=
B
and
B
.
333
IRREGULAR AMALGAMS
B and
We write the latter conditions as
this defines an equivalence relation on
the set of all amalgams.
In Section 4 we investigate conditions under which translation is a continuous
operator on (L p,gq)
In particular we show that all the translation operators are
continuous, with uniform bound on their norms, if and only if
On
{[n,n+l)}
=
p, where
is the cover that gives the regular amalgam.
Finally, in Section 5, we discuss generalizations to functions on measure
spaces and groups.
2.
THE VARIATION OF (L p,gq)
q
WITlt p
In this section we consider the irregular amalgam (L p,q) defined by a fixed
cover a
{I and investigate how it varies when p and q vary.
n
The variation with q is an immediate consequence of
Jensen’s inequality [9,
p. 28]:
I/q
I/q2
lanl 2]
q
lanl I] I,
q
[
Taking
an Ill n If(x)IPdx] I/p,
THEOREM 3.
If
[.
flfH
and
(LP, q)
(]I n I;nEZ}
(2.1)
ql < q2
(LP, q2
fif
P,q2 ,a
P,ql ,a
varies with p, the truth of the corresponding inclusion will
depend on the lengths
THEOREM 4.
<
then
(LP, ql
When
0
we have the following inclusion.
q2’
ql
<
llnl
Let
of the intervals
<
Pl
Then (L
P2"
In
E a"
P2 ,gq)
c
(L Pl ,q)
if and only if the set
is bounded above.
PROOF:
Let f
n
Suppose that the set
be the function which agrees with f on I
{[I n l;nZ}
n
and is 0 elsewhere.
It follows from Holder’s
is bounded above.
inequality that
and so
gfflq
Pl
’q’=
.
n
Hf n
gf
n
flq <
Pl
< If n
Pl
I
n
flf n Iq
where K is an upper bound for the set
(L P2 ,q)
c
ilnl
P2
P2
II n
{II n
I/PI-I/P 2
q(I/p
-I/P2
< K If q
p2,q,
q(I/PI-I/P 2)
nZ}.
It follows that
Pl
(L ,q)
Now assume that
may assume that
{llnl;nZ}
llnl lln+ll.
Without loss of generality we
(If not, we take a subsequence, reorder if necesis not bounded above.
sary, and assume the function constructed below is zero on intervals not in this
then
c on I
say f(x)
subsequence.) Observe that if f is constant on I
n
n
n
llf
c
I/p.
n II n
In order to construct f
n p
choose the c’s in such a way that
n
E
(L
P2 q)
with f
(L Pl
q)
we shall
JAMES STEWART and SALEEM WATSON
334
P2
’q’=
If n p
n
If n
.
q
Pl
I
P2 n [CnlInl
I
n
while
If
q
’q’
Hf|q
q
I [Cnllnl
n
1/P2 ]q <
I/p
.
(2.2)
l]q
To do this we use the following fact due to Stieltjes [I0, p. 41]:
there is a sequence (d
n)
of positive numbers such that
If we choose
converges.
d
If a
x
diverges while
n
.
(2.3)
0, then
a d
n n
q(l/P2-1/P
,.
then a
II n -.
0 since
n
a d
n n
<
d
and
.
n
By the result of Stieltjes there exist numbers d n with
Let
d
1/q
n
1/Pl
n
Then (2.2) and (2.3) give
I cqn
n
llflq
P2 ’=
’q
Pl ’q’
3.
P2
,q)
THE VARIATION OF (L
q/P2
n
(L Pl
is not contained in
p
q)
.
I a nd n
I cqn lln lq/pl
Iflq
This shows that (L
]I n
d
<
n
,gq).
WITH
In this section we fix p and q and consider how the irregular amalgam (L
varies when the cover
=
For that purpose we need the inequalities provided
varies.
.
by the following lemmas.
LEMMA I.
If a
0 and p
n
f]En,
Suppose f n
of R and f
I, then
p
llf
n=l
If
a
n=l
n
where
El,
...,
E
N
are disjoint measurable subsets
f
N
p, then lfll
q
.
IfH q.
p
q
n p
N
q
P
If
n=l
N
(c)
n
Jensen’s inequality (2.1).
LEMMA 2.
(b)
I aP]n I/p
n=l
The left inequality follows from Holder’s inequality and the right one
PROOF:
If q
N
N
a
n=l
(a)
.
I, then
N
N_I/p,
is just
p, q)
If p,q
PROOF:
l, then
(a)
N-q/P’( I
n=l
Since
q/p
If n
q.
n p
q)
p
Ifl
q
p
N
Nq/q’
I
n=l
If n
q)
p
I, Jensen’s inequality gives
q/p
q
If lq
f
p
[I
IP]
n p
I
(b)
Here q/p
and the result again follows from
(c)
From Lemma
we have
If n p
Jensen’s inequality.
IRREGULAR AMALGAMS
Iflq
[I
If n
N-q/P’II
IP]q/P
p
335
N-q/P’II Ifnlq
)q
If
n p
p
The inequality on the right follows from Minkowski’s inequality and Lemma I:
p
Ifl
=
Let
DEFINITION.
q
p
Ifnl p )q Nq/q’[l If n q]
p
be
of
the
covers
B
{Jn}
(I
{Inl
and
We say that
described in Section 1.
N of the J ’s.
[ntersects at most
has index N in
B
The notation
m
B and
real llne of the type
"N
write
B
if each I
n
K N 8 for
will mean that
sore N.
Suppose a has index N in B.
THEOREM 5.
(i)
>
If q
(LP,q)Bc (LP,q)=
p
I, then
q
p
p, then (L
and
P,q,B
P,q,
(ii) If
<
q)
NfH
q)
N
P,q,B
and
I/q
NfN
I Nfll n Nqp nI Nq/q’ mI Nfll n NJ m
NfNq
p,q,=
n
Nq/q’
I llflJ m qp
N
m
IfN
Thus
Nf
>
N
P,q,
I
l’q’
NfN
q
p
q/q’ NfN q
P,q,
P,q,B
and 2 we write
(ii) Using Theorems
But q’
P,q,
Using parts (c) and (a) of Lemma 2, we have
(i)
PROOF:
(L p
c
Np,q, B
sup
{lffgl; Nglp, ,q, ,B
sup
{NfN P,q,
NgN
;
P’,q
P ,q
; NgNp, ,q
,B
p’ and so, by part (i),
Ifllp,q,
sup
{fl P,q,
Igll
P ,q
igi
, N I/q}
N llq Ifl
constant N
P,q,
and 2), but the
We note that part (ii) could also be proved directly (using Lemmas
I/q would be replaced by N I/p’.
THEOREM 6.
(i)
PROOF:
If B
If B
,
,
there is an interval, say
then the inclusions in Theorem 5 are strict.
there are two cases.
First we consider the case where
which intersects infinitely many I
Jl’
Define f to be 0 outside J
n
’s.
Let I’n
n
and
fl I’n
c
n
[nlI’n I] -lIp
Then
n
slnce q
>
p, but
If
If no
Jm
"fl l’nq-I cqn II’1n q/p= I nqp <
n p
n
NPP
q g
fj Ifl p: nI cpn li’l
n
intersects infinitely many
such that J
m
In’S
17
then we may assume there exist
intersects at least m of the I
’s, say I m.
n
.-., I m
n
Jl’ J2’
and the intervals
JAMES STEWART and SALEEM WATSON
336
l’m. Im.
Let
are all disjoint.
Ik
fll’m.
c.
j
. .
.
.
3
U J
0 off J
and f
2
Ifq
P’q’a
U
k
IfllklqP
I flfl I’k fiqp
n k=l
k
,q
y. ’flJ mp
m
q
Pq,8
3
cqn I’m q/p=
n k=l
nfn
-I/p
Then
n
and
Jm and define
n-(I/p+I/q)ll’m.
N
m
cp.
I
m
3
m
n
n-q/p-l=
nk=l
n-q/P<
n
Ifll’mj ,P]q/P
p
m
j=l
I’m.
I] q/p
.
m
I I m-l-p/q] q/p
m j
m
Thus, in both cases,
(LP,q)
(LP,q) s
(ii) We consider the same two cases as in (i).
In the first case we define
I’n -I/p
-I/q
n
n
Then
’Ifq
p,q,a
I cqn I’n q/p nI
n
whlle
I cpn I’1n
’fliPp,q,13
since q
<
I n-P/q
n
In the second case we take
p.
I’
.
f
so that
lf
lq
p,q,a
c-
m.
n
-2/q
3
n
I cqn
k=l
n
<
n
I’m q/p
-I/p
.
I’
m.
n
I
n
I n-I
-2
n k=l
k
n
while
"f’q
p,q,
m
m
q/p= re(q/p-2)
Im I =P. I’m. I] q/p= mY j=lI m-2p/q]
m
3
3
j
<
This completes the proof.
By combining Theorems 5 and 6 we obtain the following.
p
c (L
(i) If q > p
I, then (L p
q)B
THEOREM 7.
(ii)
If
p,q)Cp (L p,q) if
q) (L q) if and
<
p, then (L
p
q, then (L
q
p,q
I, p *
Furthermore, the norms are equivalent.
DEFINITION.
We write a
if a
From Theorem 7 (iii) we note that
B
and
q) if
and only if
=
only if
K
.
=
and only If
and
(iii)
If
K
< a.
is an equivalence relation on the set of
all irregular amalgams.
(r)
{[an,an+i)
For example, consider the class of amalgams given by
where
r
n
n > O. (In these and the following examples we assume that
a
and
0
n
o
(r)
p q)
(s whenever 0 < r < s and
.) Then
is
O, where (L
-n
n
p
the regular amalgam (L q) discussed in Section 1. Thus we have infinitely many
=-
(I)
p,
.
IRREGULAR AMALGAMS
p and
mutually inequlvalent amalgams both
(LP,q)
with
n
2n
a(r)
satisfies
>
337
Clearly the dyadic amalgam
p.
{[n’n+l )I
where
but
3n
n
Another amalgam of some interest is defined by taking intervals I
llmn_l
llmn
I,
such that
n
This amalgam is equivalent to the regular
1, 2, ..-.
I/n, n
We shall see in the next
amalgam but has intervals of arbitrarily small length.
section, however, that no amalgam with arbitrarily large intervals is equivalent to
the regular amalgam.
TRANSLATION.
4.
For the case of regular amalgams, Holland [4] showed that the translation
operator is continuous on (L p,q) and used this fact to investigate the Fourier
Here we determine when the translation
transform of functions in this space.
In order to state the first result
operator is continuous on an irregular amalgam.
we use the notation a
t
f(x-t).
ft(x)
THEOREM 8.
on
(LP, q)a
PROOF:
(i)
If q
>
(LP,q) a
(i)
If a
a
(ii)
t.
if and only if a
"N at
>
and q
p
we have
at,
(LP,q)=t
(ii)
then by Theorem 6, (L p
If a
"<N at
q
and
<
Thus
at,
such that f
<
ft
is continuous
T_t
p, the operator
is
t
I, then, by Theorem 5 (i) applied
N
to
ft’
I/q’
p,q, a
t
p
q)a and so there exists
f_t (Lp’ q)a and (f_t)t (Lp’ q)"
f!
f
that is
t
Nl/qf p,q,a
P’q’at
(LP,q)at
(LP,q) a
then by Theorem 6,
(LP,q)t
q
p, then Theorem 5 (ii) gives
f -t p,q,a
If a
a
q)at q (L
(LP’q)a"
such that f
If
Nl/q’lftl p,q,a
Ift Wp,q,a
=
-
I, the translation operator T t f
p
if and only if a
continuous on
If
units to the right and
to mean the cover a shifted
(Lp
(L p
so there exists f
q)a
q)a
REMARK.
It is possible for the translation operator T to be continuous for
t
some values of t but not for other values. For example, consider the amalgam
obtained by splitting each unit interval [n,n+l) at the points n+2 -k k
2 3
-..
By Theorem 8, T t is continuous for integer values of t but for no other value.
We also note that it is possible for all translation operators T to be
t
continuous, but without a uniform bound on their norms.
the dyadic amalgam has this property.
THEOREM 9.
{TtW
t
R} is bounded if and only if
If there exists N such that a
"N at
conversely.
II
.
PROOF. If a
we claim that sup
n <
given any positive integer k and k adjacent intervals
{Inl
interval I
N
such that
INI
>
.
-.., 4, 3, 2, I, I/2, I/3, I/4,
intervals of successive lengths
show in Theorem I0 that
It is not hard to see that
Another example is obtained by taking
lnl
+
+
Inkl"
In fact we
a
for all t, then a
For if sup
Inl,...,Ink
I
n
and
-,
then
we can choose an
Thus there exists t such that
JAMES STEWART and SALEEM WATSON
338
(InlU
Ulnk)tC IN
"I"
Since k is arbitrary, this contradicts our hypothesis.
lnl
Therefore we can write sup
interval
In
M where M is an integer.
meets at most M +
Suppose now that
Then there are intervals in 0 which contain
a.
arbitrar1y many l.’s.
It follows that any
intervals from p, i.e.,
It follows from the pigeon-hole principle that there are
interval of R of any given size which contain arbitrarily many
l.’s.
In
lj s
particular, a given interval in a can be translated to cover arbitrarily many
so a
.
at"
N
To prove the converse we assume that a
there is no N such that a
for all t.
a
This means
N t
in a containing arbitrarily many I
llnl
that we
K
Then sup
<
.
Suppose
can find an interval
II
+t’s for some t. But, since
K, it follows
n
n
(as above) that there is a unit interval which intersects arbitrarily many I +t’s
n
and therefore a unit interval [k,k+l) meeting arbitrarily many I ’s. This is a
n
contradiction.
THEOREM I0
Tt,
The translation operators
with uniform bound on their norms, if and only if a
sup
sup
PROOF.
;ITt
if q
p
l;Tt
I/q
N
if
q
N at
q)
In this case we have
O.
I/q’
N
where N [s the smallest integer such that a
(L p
R, are continuous on
t
p
for all t.
This follows from the two preceding theorems together with the proof of
Theorem 4.1.
REMARK.
8].
Continuity of translation is an essential ingredient in the proof that
(L q
p,q
2, then the Fourier transform f
[4, Theorem
,P
(L p,q), I
if f
Thus, in view of Theorem I0, it seems unlikely that any such result will hold
for amalgams other than those equivalent to the regular amalgam.
Translation also arises in the consideration of convolution and
{In}
If
inequality.
~N(a
we write y
B
+ B) if y
N
state the following version of
Z},
{Jm }, and y are covers and a + B {I n
(a + B) and (a + B) N Y" With this notation we can
Young’s inequality for irregular amalgams.
I/Pi
+
I/qi
~N r (a+B).
r
Suppose y
I, then f*g
(L
f*gI
Theorem
|I
2
The proof
[II].
is similar to that of Theorem 4.2 in
THEOREM II.
Young’s
+ Jm; n,m
(L Pl
If f
)y,
where i/r
N f
rl,r2,Y
i
Pl,P2 ,a
P2
I/Pi
and g e (L
+
I/qi
q2
ql
),
where
and
gq l,q2,B
holds for amalgams on groups (see the next section for definitions)
but on R the condition y
~N
(a + B) holds if and only if a,
,
and y are all
equivalent to the regular amalgam, as we now show.
THEOREM 2.
PROOF.
Then
5.
a
If y
If y
(a + ), then a,
,
y
(a + B), then clearly y
O by Theorem 9.
Similarly
B
p.
a
for all t and so a
t
0 and it follows that y
a
t
for all t.
also.
AMALGAMS ON MORE GENERAL SPACES.
In this section we discuss the extent to which our results can be generalized
from functions on R to more general functions.
sense on any measure space.
The amalgam spaces themselves make
339
IRREGULAR AMALGAMS
{E;
Let (X,v) be a measure space and E
J} any covering of X by disjoint
%
measurable sets of finite measure:
u(Ek) <
kJ EX’
X
p
In terms of this decomposition E of X we define the amalgam (L
q)
E
to consist of
functions f such that
llfH
I
p q E
]I/q
lflq
p
L (E
J
<
)
Then all of the results of the first three sections extend to this setting
follows from a general result [12, p. 359]
particular, the extension of Theorem
concerning the dual of the space
a Banach space, and llxN
[
In
q(B%)
of nets x
llxnUq]I/q < .
(xk),
Taking
where
B-- LP(Ek),
x
B%,
each
B is
and using this
result, we have
(LP’q)*=E
q(LP(Ex))* q’(LP’(Ex)) (Lp’’q’)E
Theorems 3, 4, 5, 6, and 7 have verbatim proofs in the general context of measure
spaces
For translation, of course, we need algebraic structure and so we assume that G
is a locally compact abelian group (with Haar
s
valid for amalgams on G.
10 for groups
measure) and easily see that Theorem 8
The question then arises as to an analogue of Theorem
Regular amalgams on groups have been defined and studied in [13],
[14], and [II].
Nonetheless we do not see how to extend Theorem I0 to groups
without imposing severe restrictions on the shape of the sets E k, even for groups as
R
simple as G
2.
We thank John Fournier for useful suggestions
ACKNOWLEDGEMENTS.
This research was
supported by NSERC Grant A4833.
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