Proof.
For any l∈{1,…,n}, set
(3.1)al=lim φ(z)→∂𝔻n ∑k=1n|∂φl∂zk(z)|(1-|zk|2)q(1-|φl(z)|2)p.
For the proof of the sufficiency, by Lemma 2.2, we only have to show that al=0 for every l∈{1,2,…,n}. There are two cases to consider.
Case
1. If j∈J with ∥φj∥∞<1, we have
(3.2)lim φ(z)→∂𝔻n ∑k=1n|∂φj∂zk(z)|(1-|zk|2)q≤aj≤lim φ(z)→∂𝔻n ∑k=1n|∂φj∂zk(z)|(1-|zk|2)q(1-∥φj∥∞2)p.
Therefore, for any j∈J, lim φ(z)→∂𝔻n∑k=1n|(∂φj/∂zk)(z)|(1-|zk|2)q=0 is equivalent to aj=0.
Case
2. If i∈I with ∥φi∥∞=1, then for each m≥2, let
(3.3)Am,i={z=(z1,…,zn)∈𝔻n:rm≤|zi|≤rm+1},
where rm=((m-1)/(m-1+2α))1/2.
For each fixed i∈I and every ε>0, there exists a δ0 with 0<δ0<1 such that
(3.4)∑k=1n|∂φi∂zk(z)|(1-|zk|2)q(1-|φi(z)|2)p>ai-ε,
whenever dist (φ(z),∂𝔻n)<δ0.
Since rm→1 as m→∞, we may choose sufficiently large m such that rm>1-δ0. If φ(z)∈Am, i, then
(3.5)rm≤|φi(z)|≤rm+1,1-rm+1<1-|φi(z)|<1-rm<δ0,
thus
(3.6)dist (φi(z), ∂𝔻)<δ0.
There exists wi with |wi|=1 such that
(3.7)dist (φi(z), wi)=dist (φi(z), ∂𝔻)<δ0.
Let w=(φ1(z),…,φi-1(z),wi,φi+1(z),…,φn(z)). Then
(3.8)dist (φ(z), ∂𝔻n)≤dist (φ(z), w)≤dist (φi(z), wi)<δ0.
So we have
(3.9)sup φ(z)∈Am,i ∑k=1n|∂φi∂zk(z)|(1-|zk|2)q(1-|φi(z)|2)p>ai-ε.
Letting ε→0 and by Lemma 2.1, we have
(3.10)ai≤lim m→∞sup φ(z)∈Am,i ∑k=1n|∂φi∂zk(z)|(1-|zk|2)q(1-|φi(z)|2)p≤lim m→∞sup φ(z)∈Am,i ∑k=1n|∂φi∂zk(z)|mp|φi|m-1(1-|zk|2)qmp|φi|m-1(1-|φi(z)|2)p≤lim m→∞sup φ(z)∈Am,i∑k=1n|(∂φi/∂zk)(z)|mp|φi|m-1(1-|zk|2)qlim m→∞inf φ(z)∈Am,imp|φi|m-1(1-|φi(z)|2)p≤(e2p)plim m→∞mp-1∥φim∥q.
From which and (6), we know ai=0. Combining the two cases, thus (2) holds. Note that conditions (5) and (1) are the same, it follows from Lemma 2.2 that Cφ:ℬp→ℬq is compact.
Now we turn to prove the necessity. The result (5) follows by Lemma 2.2. Using Lemma 2.1, we see that
(3.11)lim m→∞mp-1∥zlm∥p=(2pe)p,
for any l∈{1,…,n}.
For any i∈I, we consider the test functions fm(z1,z2,…,zn)=zim/∥zim∥p. It is clear that ∥fm∥p=1 and fm→0 uniformly on compact subsets of 𝔻n as m→∞. If Cφ is compact, then
(3.12)0=lim m→∞∥Cφfm∥q=lim m→∞∥φim∥q∥zim∥p=lim m→∞mp-1∥φim∥qmp-1∥zim∥p=(e2p)plim m→∞mp-1∥φim∥q.
This shows that (6) holds.
For any j∈J, from the discussion of Case 1 in the proof of the sufficiency, it follows that aj=0. And this fact implies that condition (7) holds. Now the proof of the theorem is completed.
Proof.
Suppose first that Cφ:ℬp→ℬq is compact. It is clear that Cφ:ℬp→ℬq is bounded. Taking the test functions fm(z)=zim/∥zim∥p, and using the same arguments as in the proofs of Theorem 3.1, we obtain
(3.13)lim m→∞mp-1∥φim∥q=0,
for any i∈I.
This proves the necessity.
Conversely, by Lemma 2.4, it suffices to show that (4) holds. In fact, for any j∈J with ∥φj∥∞<1, there exists some positive number δ close enough to 1, such that the set Ej={z∈𝔻n: |φj(z)|>δ} is empty. Without loss of generality, we may assume that
(3.14)lim |φj(z)|→1 ∑k=1n|∂φj∂zk(z)|(1-|zk|2)q(1-|φj(z)|2)p=0.
For any i∈I, and m≥2, let
(3.15)Am,i={z=(z1,…,zn)∈𝔻n:rm≤|zi|≤rm+1},
where rm=((m-1)/(m-1+2α))1/2.
Therefore, we have
(3.16)lim |φi(z)|→1 ∑k=1n|∂φi∂zk(z)|(1-|zk|2)q(1-|φi(z)|2)p ≤lim m→∞ sup φ(z)∈Am,i ∑k=1n|∂φi∂zk(z)|(1-|zk|2)q(1-|φi(z)|2)p ≤lim m→∞ sup φ(z)∈Am,i ∑k=1n|∂φi∂zk(z)|mp|φi|m-1(1-|zk|2)qmp|φi|m-1(1-|φi(z)|2)p ≤lim m→∞sup φ(z)∈Am,i∑k=1n|(∂φi/∂zk)(z)|mp|φi|m-1(1-|zk|2)qlim m→∞inf φ(z)∈Am,imp|φi|m-1(1-|φi(z)|2)p ≤(e2p)plim m→∞mp-1∥φim∥q.
This along with condition (9) yields that
(3.17)lim |φi(z)|→1 ∑k=1n|∂φi∂zk(z)|(1-|zk|2)q(1-|φi(z)|2)p=0,
for any i∈I.
Combining the results of the two cases for j∈J and i∈I, we get (4). By Lemma 2.3, we know that Cφ:ℬp→ℬq is compact. This completes the proof of the theorem.