j=jo such that

We put

where the infimum is taken over all sequences {a.,} satisfying (4) and (5). (iii) If b = 0, then Ae<q(\ogA)b = Ae,q.

Standard arguments show that Ae!g(logyl)6 is a Banach space in all cases of b G R. It is independent of jo (equivalence of norms). If we replace the real method by the complex interpolation method, that is to say, if we replace in (i) A„jtq by (A0, ^i)^] and in (ii) we put (A),^i)[Aj] instead of A\j<q, then we obtain the spaces studied by Edmunds and Triebel [15]. They are different from the complex scale and complement it.

In our case spaces Ag^log-A);, are also different from the real interpolation spaces, but they coincide with the spaces generated by the function parameters te(l + | log £|)—6 :

Theorem 3.1. Let 1 < q < oo, 0 < 9 < 1, and b € R. Let ge,b(t) = t9 (1 + | log t\)~b,t > 0. Then we have, with equivalent norms,

Proof. Sketch of the case b < 0. It is sufficient to prove that the respective norms are equivalent to each other. First note that

Using that A0 <—► Ai, one can check that the last expression is equivalent i to (]Cm=i 2~maiqKq(2m,a))q, moreover constants in equivalences can be chosen independent of aj and 6. Similarly oo r^i

YJ2~meqmbqKq{2m,a).

For the norm in the other space, we have m—1

To work with the last sum in (6) let j = [logm], where [•] is the greatest integer function. Then jb - m2~j ~ b [logm] +bk - 2~k. We obtain oo oo jbq-m2-jq ^ mbq ^ 2kbq2~q2'k .

Now, using that b < 0, the last sum can be estimated from above and from below by positive constants which are independent of m. This yields that the norms of the two spaces are equivalent. □

The proof for the case b > 0 is more involved. Details can be found in [10]. Let me only mention that it is based on the description of real interpolation spaces in terms of the J—functional.

Theorem 3.1 allows us to use the results known on interpolation with function parameter to study spaces A#)9(log A)b- A first consequence is that Ao is dense in Ae,q{\ogA)b if q < oo. Concerning duality, if A0 is dense in Ai, then we have (Ai)' <-> (Ao)', so we can consider logarithmic spaces generated by the couple {(Ai)', (Ao)'}. Call them

If q < oo, since Ao is dense in As,q(logA)b, we get

and we can compare the spaces (Ae,g(log A)b)' with the spaces A^g(log A')&. Using Theorem 3.1 and the duality formula for the real method with a parameter function we derive for 1 < q < oo, 1/q + 1/q' = 1 and 0 < 9 < 1 that

4. Concrete Logarithmic Interpolation Spaces

In this final section we apply Theorem 3.1 to concrete situations. We start with the couple (Loo(Çl),Li(iî)). We get:

Corollary 4.1. Let Q be a domain in K" with finite Lebesgue measure. Let 1 < p < oo, 1 < q < oo and let j0 = jo(p) € N such that for all je N with j > jo,

(i) Let b < 0. Then LPtq(\ogL)b(0.) is the set of all measurable functions f : Çl —> C such that

(equivalent norms).

(ii) Let b > 0. Then LPig(log L)b(ft) is the set of all measurable functions f : il —> C which can be represented as oo f-T.fi with e (7)

Furthermore, the infimum over the expression in (8) with respect to all representations (7), (8), is an equivalent norm in Lpq(\og L)b(£l).

For p = q we derive a representation theorem for Zygmund spaces Lp(logL)b(il) in terms of Lorentz spaces LrjS(Q). Note that Theorem 2.1 requires only the simpler Lebesgue spaces. However, the next lemma yields that Theorem 2.1 follows also from our approach.

Lemma 4.1. Let A0, A\ be Banach spaces with Ao <-> Ai. Let 0 < 0 < 1, be R and let jo = jo(0) £ N such that, for all j > jo,

(i) Let b < 0. The norm of Aei/e (log A) b is equivalent to

OO a

3=30

(ii) Let b > 0. Then ^0,1/0(log A)b is formed by all those a £ A\ which

Was this article helpful?

## Post a comment