On some properties of a differential operator on the polydisk

PROPERTIES OF DIFFERENTIAL OPERATOR ON THE POLYDISK 71
also establish connections between holomorphic spaces with quasinorms in the
subframe and polydisk.
2. Some observations concerning R s and D s <strong>differential</strong> <strong>operator</strong>s
and pro<strong>of</strong>s <strong>of</strong> preliminaries
When we look at R s <strong>operator</strong>s then we have the following natural problem:
Is it possible to reduce the study <strong>of</strong> R s <strong>operator</strong>s to the study <strong>of</strong> D s <strong>operator</strong>s?
which was studied by many authors (see for example [4] and references there).
Then we will be able to use known <strong>properties</strong> <strong>of</strong> D s to get new results for R s
<strong>operator</strong>s. The <strong>differential</strong> <strong>operator</strong> D s is much more convenient at least because
<strong>of</strong> the following property <strong>of</strong> g function. Let
We have
and
R s g(z) =
g(z) = 1
1 − z = ∑
∑
k 1 ,··· ,k n≥0
k 1 ,··· ,k n≥0
z k 1
1 · · · z kn
n .
(k 1 + · · · + k n + 1) s z k 1
1 · · · z kn
D s g(z) = ∑ 1 + 1)
k 1 ≥0(k s z k 1
1 · · · ∑
(k n + 1) s z kn
For D s we reduce things to one dimensional <strong>differential</strong> <strong>operator</strong>s for one function
in the unit disk, that why we will find ways to reduce the study <strong>of</strong> R s to D s .
Let
˜R s f(z) =
∑
k 1 ,··· ,k n≥0
k n≥0
n
n .
(k 1 + · · · + k n ) s a k1 ,··· ,k n
z k 1
1 · · · z kn
The following lemma is playing an important role in the study <strong>of</strong> R s <strong>operator</strong>. We
use <strong>some</strong> known facts in the pro<strong>of</strong> <strong>of</strong> Lemma 2.1 about action <strong>of</strong> one dimensional
D α <strong>operator</strong>, for example the following estimate (see [2])
∫
∫
|D α g(τξ)| p dξ ≤ C |D β g(τξ)| p dξ (2.1)
T
which can be transferred by induction to polydisk.
T
n .
(τ ∈ (0, 1), 0 < p < ∞, α ≤ β, g ∈ H(U))
Lemma 2.1. Let w = |w|ξ, w, z, ∈ U n , 1 − wz = ∏ n
k=1 (1 − w kz k ), s ∈ {0} ∪ N,
β > 0, p ∈ (0, ∞). Then we have
∫ ∣ ∣∣R s 1
∣ ∣∣
p
dmn (ξ)
T (1 − ξ|w|z) n β
≤
C ∑ α j ≥0
P αj =s
( n
∏
k=1
1
(1 − |w k ||z k |) p(α k+β)−1
)
, p >
1
min k α k + β .