Proof of Theorem 2. We shall divide the proof into 8 steps. In Step 1, by Lemma 13 we obtain that (b) implies (c) and (c) implies (d).
In Step 2, we prove (a) implies (b). We firstly consider that f∈Ap(·)(Bn) such that fLp(·)(Bn)=1. We follow the idea of the proof of Theorem 2.16 in [5] and make some crucial modifications where needed. Fix r∈(0,1). It follows from Lemma 2.4 in [5] that for fixed β>α=0 there exists a constant C>0 such that(39)∇g0≤C∫Dz,rgwdνβw, for all holomorphic g in Bn.
Now, for each z∈Bn, let φz be the biholomorphic mapping of Bn which interchanges 0 to z, and let g=f∘φz. Then by making an obvious change of variables according to Lemma 12, we obtain (40)∇~fz≤C1-z2n+1+β∫Dz,rfw1-z,w2n+1+βdνβw=C1-z2n+1+β∫Dz,rfwCβ1-w2β1-z,w2n+1+βdνw≤C1-z2n+1∫Dz,rfwdνw.Since the volume of D(z,r) is comparable to (1-z2)n+1, then we have (41)∇~fz≤C⨍ Dz,rfwdνw. Then by Lemma 4 we have (42)∇~fzpz≲⨍ Dz,rfwpwdνw+1≲∫Dz,rfwpw1-z2n+1+β1-z,w2n+1+βdνβw+1.
Integrating both sides of the above inequality over Bn with respect to dν(z) and using Fubini’s theorem, we have that (43)∫Bn∇~fzpzdνz≲∫Bn∫Dz,rfwpw1-z2n+1+β1-z,w2n+1+βdνβwdνz+1≲∫Bnfwpwdνβw∫Bn1-z2n+1+β1-z,w2n+1+βdνz+1≲∫Bnfwpw1-w2βdνw∫Bn1-z2n+1+β1-z,w2n+1+βdνz+1≲∫Bnfwpw1-w2βdνw∫Bn1-z2n+1+β1-z,w2n+1+βdνz+1.By Lemma 14 we have that(44)∫Bn1-z2n+1+β1-z,w2n+1+βdνz~1-w2-β. Thus using Lemma 3 and the above result we obtain that (45)∫Bn∇~fzpzdνz≲∫Bnfwpw1-w2β1-w2-βdνw+1≲∫Bnfwpwdνw+1. Therefore ∇~f∈Lp(·)(Bn,dν).
Finally, for a general 0≠f∈Ap(·)(Bn), define h=ffLp(·)(Bn)-1, and we apply the previous result for h and conclude that ∇~f∈Lp(·)(Bn,dν).
In Step 3, to prove (d) implies (a), we assume that f is a holomorphic function in Bn such that the function 1-z2Rfz∈Lp(·)(Bn,dν). Let β be a sufficiently large positive constant. Then by the proof of Theorem 2.16 in [5] we have that (46)fz-f0≤∫BnRfw1-z,wn+βdνβw=Cβ∫BnRfw1-w2β1-z,wn+βdνw≲∫BnRfw1-w21-z,wn+1dνw≲P1-·2Rf·z, where P is the Bergman projection operator. By Lemma 16 we know that f(·)-f(0)∈Lp(·)(Bn,dv). Therefore f∈Ap(·)(Bn).
In Step 4, we prove (a) implies (e). If f∈Ap(·)(Bn) and z∈Bn, we have (47)fz-f0=∫01∑k=1nzk∂f∂zktzdt. Fix r∈(0,1). It follows that, for ρ(z,0)<r, we have (48)fz-f0=∫01∑k=1nzk∂f∂zktzdt≤∫01∑k=1nzk∂f∂zktzdt≤∫01∑k=1nzk21/2∑k=1n∂f∂zktz21/2dt≤∫01z∇ftzdt=z∫01∇ftzdt≤zsup∇fw:w∈D0,r. Since in the relatively compact set D(0,r) the Euclidean metric is comparable to the pseudohyperbolic metric, then ∇fw is comparable to ∇~fw in the relatively compact set D(0,r). Thus, there is a constant C>0, which depends only on r, such that (49)fz-f0≤Cρz,0sup∇~fw:w∈D0,r, for all z∈D(0,r). Replacing f by f∘φw, and z by φw(z), respectively, and using the Möbius invariance of the pseudohyperbolic metric and the invariant gradient, we obtain (50)fz-fw≤Cρz,wsup∇~fu:u∈Dz,r, for all z and w in Bn with ρ(z,w)<r.
Let (51)gz≔fzr+hz, where h(z)=C sup{∇~fu:u∈D(z,r)}. Then it is easy to see that the function g is continuous on Bn and (10) holds. Now we only need to prove that g∈Lp(·)(Bn,dν). Since f is already in Lp(·)(Bn,dν), the remainder is to show that h∈Lp(·)(Bn,dν).
Recall that if r and R satisfy (8), then D(z,r)=E(z,R). So we can choose r′∈(0,1) such that D(z,r′)=E(z,2R) for all z∈D. For any u∈E(z,R), any w1∈E(u,R), we have β(w1,u)<R. Then we have β(u,z)<R due to u∈E(z,R). By the triangle inequality, we have (52)βw1,z≤βw1,u+βu,z<R+R<2R. So w1∈E(z,2R). That is, E(u,R)⊂E(z,2R). Equivalently, we have D(u,r)⊂D(z,r′) whenever u∈D(z,r).
By Lemmas 17 and 3 we obtain that (53)hz≲11-z2n+1∫Dz,rhwdνw≲11-z2n+1∫Dz,rC sup∇~fu:u∈Dw,rdνw≲11-z2n+1∫Dz,r′∇~fwdνw≲⨍ Dz,r′∇~fwdνw, where we used that the volume of D(z,r′) is comparable to (1-z2)n+1. Without loss of generality, we assume ∇~fLp(·)(Bn,dv)=1. Therefore, using Lemma 4 we obtain that (54)hzpz≲⨍Dz,r′∇~fwpwdνw+1.≲11-z2n+1∫Dz,r′∇~fwpwdνw+1. Integrating both sides of the above inequality over Bn with respect to dν(z) and using Fubini’s theorem, we have that (55)∫Bnhzpzdνz≲∫Bn11-z2n+1∫Dz,r′∇~fwpwdνwdνz+1≲∫Bn∇~fwpwdνw∫Dw,r′11-z2n+1dνz+1≲∫Bn∇~fwpwdνw+1. Due to f∈Ap(·)(Bn), by Step 2 we have that ∇~f·∈Lp(·)(Bn,dν). Thus h∈Lp(·)(Bn,dν).
In Step 5, it is easy to see that (e) implies (f), according to ρ(z,w)≤β(z,w), for all z,w∈Bn.
In Step 6, we prove that (f) implies (a). Assume that there exists a continuous function g∈Lp(·)(Bn,dν) such that (11) holds for all z,w∈Bn. We fix z∈Bn and let w=tz, where t is a scalar. Then(56)fz-fwz-w≤βz,wz-wgz+gw, for all z≠w in Bn. Let t approach 1 and apply Lemma 18; we obtain that 1-z2Rfz≤2g(z) for all z∈Bn. Since g∈Lp(·)(Bn,dν), then 1-·2Rf·∈Lp(·)(Bn,dν), and by Step 3 we have that f∈Ap(·)(Bn).
In Step 7, we prove that (g) implies (a). If there exists a continuous function g such that (1-·2)g in Lp(·)(Bn,dν) and for all z,w∈Bn,(57)fz-fw≤z-wgz+gw, then(58)fz-fwz-w≤gz+gw. Let z=(z1,…,zk,…,n) and fix z. For each k∈{1,…,n}, let w=(z1,⋯,wk,…,zn). Let wk tend to zk, and we obtain(59)∂f∂zkz≤2gz. Therefore for all z∈Bn, we obtain(60)∇fz≤2ngz. Thus(61)1-z2∇fzpz≤2n1-z2gzpz. Since (1-·2)g∈Lp(·)(Bn,dν), thus 1-·2∇f∈Lp(·)(Bn,dν). Therefore, we have f∈Ap(·)(Bn) by the equivalency of (c) and (a) which we have proved.
In Step 8, we prove that (e) implies (g). If Condition (e) holds in this theorem, Lemma 1.2 in [5] says that(62)1-φzw2=1-z21-w21-z,w2. Thus(63)ρz,w2z-w2=z-w2+z,w2-z2w2z-w21-z,w2. By the Cauchy inequality, z,w2≤z2w2, we obtain that, for all z,w∈Bn,(64)ρz,w≤z-w1-z,w. By condition (e), there exists a continuous function h in Lp(·)(Bn,dν) such that, for all z,w∈Bn,(65)fz-fw≤ρz,whz+hw. Therefore,(66)fz-fw≤z-w1-z,whz+hw. Since (67)1-z,w≥1-z,1-z,w≥1-w, we have (68)fz-fw≤z-whz1-z+hw1-w. Write g(z)=h(z)/(1-z), then (69)fz-fw≤z-wgz+gw. Since z∈Bn, we have (70)11-z≤21-z2,so (71)gz=hz1-z≤2hz1-z2. Since h∈Lp(·)(Bn,dν), we have (1-·2)g∈Lp(·)(Bn,dν).