Radial variation in some function spaces

In a previous paper [8] we considered properties of the radial variation of analytic functions in a class of Besov spaces Apq , s > 0. Here we wish to extend these results to certain related spaces. These are the Lipschitz classes Λs and the mean Lipschitz classes Λp,s where p ≥ 1, 0 < s < 1. We also consider Apq , where s = 0, although the results obtained for these are not as good as when s > 0.


Introduction
If f is analytic in the disc, the radial variation function of f is the function defined on the disc by ( 1) and F (r, t) is a majorant for f .The function F (r, t) represents the length of the image of the radius vector [0, re it ] under the mapping f .It is clear from the definition, that the boundary function F (t) = lim r→1 F (r, t) exists, finite or infinite, for all t ∈ [0, 2π].It is known as the radial or total variation.An immediate property of F is that if F (t) < ∞, then lim r→1 f (re it ) exists.We saw in [8] that the property that f ∈ A s pq , 0 < s < 1, 1 ≤ p, q < ∞, translated into meaningful results for F , in particular that F (r, t) satisfies an analogous condition on the disc.In Section 1 we are led naturally to consider the case s = 0 when we ask for a condition under which F (t) is an integrable function on the circle.It follows immediately that F ∈ L 1 if and only if f ∈ A 0 11 .We then show that F (r, t) satisfies a corresponding condition to that by f in the disc .This result extends to the general case f ∈ A 0 pq .In Section 3 we suppose that f belongs to a Lipschitz space or a mean Lipschitz space and show that both F (r, t) and F (t) exhibit the expected behaviour.
1.1 Preliminaries.Let D denote the unit disc, T the unit circle in the complex plane and L p = L p (T ) the usual Lebesgue space when 0 < p < ∞.For p ≥ 1 we denote the norm of a function f ∈ L p by ||f || p .For convenience we shall let m denote normalised Lebesgue measure on the circle T .
denote the integral mean of f of order p.It is well known that M p (f, r) is an increasing function of r on [0, 1) and that the class of functions f for which sup r<1 M p (f, r) < ∞, is the familiar Hardy space H p [2].For 1 ≤ p, q < ∞, s > 0 , and an arbitrary integer m > s, we define the Besov space B s pq by It is well known that the definition is independent of m.For a discussion of these spaces see [1], [3], [4], [6], [7].When s passes through a positive integer value, the working definition of the Besov space B s pq may require a change as indicated above.
The previous definition is no longer valid when s ≤ 0 ; for these cases another description is required.For n ≥ 1 we let W n be the polynomial on T whose Fourier coefficients satisfy Ŵn ( 2 It is known that this description is equivalent to the previous one for s > 0, but for s = 0 in particular, only the second definition is valid.See [4] Appendix 2, [1].In fact when q > p there exist f ∈ B 0 pq such that f / ∈ L p .Let A s pq denote the subspace of B s pq consisting of analytic functions.The space A s pq for s > 0 , may be characterized as follows: for an arbitrary integer m > s the analytic function f ∈ A s pq if and only if Once again the definition is independent of m for m > s.For s = 0 this definition is easily modified.This is because of the property that f ∈ A 0 pq if and only if If ∈ A 1 pq where I is the integration operator.Therefore and with m = 3 , if and only if We shall need both of these representations.In particular with p = q = 1 we have f ∈ A 0 11 if and only if

Integrability of F
The function F (t) = F (1, t) is given from (1) by We now ask what is a sufficient condition that F ∈ L 1 ?Since F ∈ L 1 if and only if 2π 0 1 0 |f (re it )|r dr dm < ∞, the answer is immediate from the definition: It may be observed here that if f ∈ A 0 11 then its boundary function f (e it ) exists a.e.; in fact f ∈ H 1 .This follows by integrating the obvious inequality |f We can equally express the relationship in terms of the A-norm of F (r, t).For this purpose we introduce the gradient of If the integral is finite then it follows very simply that f ∈ A 0 11 and that f A ≤ |f (0)| + F A .The proof in the other direction has already been done in essence in [8] where we considered only s > 0 .In fact we can state a more general result which follows from Theorem 1 there, and which works without any changes for our situation.
Proof.The proof in [8] goes through word for word with s = 0.In the case p = q = 1 it is simpler since the use of Hölder's inequality is not needed.We do make use of the alternative definitions of A 0 pq mentioned above.
If the double integral for F (r, t) is finite then as noted already it is clear that f ∈ A 0 pq .The question when F ∈ L p , p > 1 , does not have so neat an answer.A reasonable sufficient condition is given by Proof.By Minkowski's Inequality in continuous form Remark.The condition f ∈ A 0 p1 implies that f ∈ H p for all p ≥ 1.To see this we note that for r < 1 On using Minkowski's Inequality again we obtain and the result is immediate.
In [8] it was shown that if f ∈ A s pq , 0 < s < 1 , then the boundary function F ∈ B s pq .We do not know whether this is true for the case s = 0 since the proof given there is no longer valid.

The Lipschitz spaces
The Lipschitz space Λ s , 0 < s < 1 , may be regarded as the Besov space B s ∞∞ .It is well known that for an analytic function f on the disc, f ∈ Λ s if and only if there exists M such that This property has its counterpart for the function F (r, t).
Proof.Suppose f ∈ Λ s and let M be the number noted above.First we show that F (t) is bounded.
In the second inequality above we used Theorem 5.5 of [2].The result follows.
There is a corresponding result for F (t).
Proof.We have shown that F is bounded.We write With the aid of this, similar results to those of the last two theorems can be shown to hold and the proofs are straightforward.
Whether a particular type of continuity for f implies the same holds for F is uncertain.The boundary function f (e it ) is absolutely continuous if and only if f ∈ H 1 .We dont know that this implies that F (t) is absolutely continuous but it does imply that F is continuous.In [8] it was seen that if we assume slightly more, namely if f ∈ A 1 11 , then F ∈ B 1  11 which implies that F is absolutely continuous.However mere continuity of f on the circle does not even imply that F is bounded.In fact Walter Rudin [5] has shown that there exists an analytic function f continuous in the closed disc, such that F (t) = ∞ almost everywhere.
on using the previous theorem.If we now choose 1 − r = |x − t| we get |F (r, x) − F (r, t)| ≤ M |t − x| s and F (t) ∈ Λ s .The mean Lipschitz classes Λ p,s (T ), 1 ≤ p , 0 < s < 1 , are indentical with the Besov spaces B s p∞ .They satisfy the condition: A function g ∈ L p (T ) belongs to Λ p,s if