1.1.6. Suppose that e+, e- e R, r 0, and let E e S(e+,e_,r).
The space Y(E) consists of all distributions / G S', which have a representation
(1.1.7), converging in 5', satisfying (1.1.8) and
(1.1.10) supp^/o C 5(0,2),
(1.1.11) suppTji C 5(0,2
i + 1
) \ 5(0,2
i _ 1
), i e N.
If £ =
with A e R, 0 p,0 oo, or E = Lp(lg) with
0 p o o , 0 # o o , then Definition 1.1.6 is a classical definition of the Besov
spaces Bpe and Lizorkin-Triebel spaces F£e, respectively. See e.g. the books by
J. Peetre [42] and H. Triebel [53], [54]. The spaces obtained in Definition 1.1.5
are in this case denoted BL^e and FL^e. They have been less studied (in the
cases when the two definitions do not give the same spaces; see Proposition 1.1.12
below), but see Netrusov [33], [34].
It is easily seen that if ||/||yz,(£;) is defined by
where the infimum is taken over all representations of / as in Definition 1.1.5,
||/||Y(# )
is defined analogously, the spaces YL(E) and Y(E) are quasi-normed
spaces, and normed spaces HE is.
We shall prove that the spaces YL(E) and Y(E) are complete, i.e., quasi-
Banach or Banach spaces.
For the proof we need the following lemma of so called Polya-Plancherel type,
which plays an important role for the whole theory. It is a slight modification of
Theorem 1.3.1 in Triebel [53].
LEMMA 1.1.7. Let f e Sf, suppTf c B(0,R), r 0, d 0, and M e N
Then there is a constant C, independent of R, such that for all multi-indices a with
\a\ M,
R-M\D°f(x + z)\ Y J _ f \f(x + y)\r N
S?" {l + R\z\)»/*+* -CZPo\a»JB{0,a)(l + R\y\y*C
PROOF. For the reader's convenience we give the proof, essentially following
Triebel [53]. The idea of the proof is due to Peetre [41]. For r 1 a simpler proof
using Holder's inequality is possible.
We first prove that for any multi-index a and any A 0 there is C so that
n i 1
+ z)\ ^ \f(x + z)\
(1.1.13) SUP —! ' ... C SUP -y- „. , ' .
We can assume that R = 1, i.e., that suppTf C 5(0,1); otherwise replace f{x) by
f(x/R). It is also no restriction to assume x = 0.
Let (pbea function in S such that J-^{i) = 1 on 5(0,1). Then / = / * y, and
= / * D . For any A 0 there is C such that
(1 + M)-
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