On some properties of a differential operator on the polydisk
On some properties of a differential operator on the polydisk
On some properties of a differential operator on the polydisk
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PROPERTIES OF DIFFERENTIAL OPERATOR ON THE POLYDISK 77<br />
3. Embedding <strong>the</strong>orems for <str<strong>on</strong>g>some</str<strong>on</strong>g> analytic spaces c<strong>on</strong>nected with<br />
R s <str<strong>on</strong>g>operator</str<strong>on</strong>g><br />
In this secti<strong>on</strong> we state <str<strong>on</strong>g>some</str<strong>on</strong>g> new embedding <strong>the</strong>orems for various quasinorms<br />
where <strong>the</strong> R s <str<strong>on</strong>g>operator</str<strong>on</strong>g> is participating, note that practically all results are well<br />
known or obvious in <strong>the</strong> unit disk. Pro<str<strong>on</strong>g>of</str<strong>on</strong>g>s <str<strong>on</strong>g>of</str<strong>on</strong>g> our results are heavily based <strong>on</strong><br />
definiti<strong>on</strong>s and preliminaries from previous secti<strong>on</strong>s.<br />
Theorem 3.1. i) Let 0 < q < ∞, α ∈ [0, ∞), s ∈ N, 1 < p < ∞, f ∈ H(U n ).<br />
Then<br />
( ∫ 1<br />
q/p<br />
(1 − |z|)<br />
∫T α |f(z)| p |z| d|z|) sp dm n (ξ)<br />
n<br />
≤<br />
0<br />
( ∫ 1<br />
q/p<br />
C |R<br />
∫T s f(uξ)| p (1 − u) du) sp+α dm n (ξ).<br />
n<br />
0<br />
ii) Let 0 < q < ∞, α ∈ [0, ∞), s ∈ N, 1 < p < ∞, γ ∈ (−1/p, 1/p ′ ), 1/p ′ + 1/p =<br />
1, f ∈ H(U n ). Then<br />
≤<br />
( ∫ 1<br />
q/p<br />
(1 − r)<br />
∫T γ |f(rξ)| p r dr) p dm n (ξ)<br />
n<br />
0<br />
( ∫ 1<br />
q/p<br />
C |R<br />
∫T s f(rξ)| p (1 − r) dr) sp+γ dm n (ξ) .<br />
n<br />
Pro<str<strong>on</strong>g>of</str<strong>on</strong>g>. Using (1.2) we have<br />
( ∫ 1<br />
M = (1 − |z|) α |f(z)| p |z| sp d|z|<br />
≤<br />
= C<br />
0<br />
( ∫ 1 ( ∫ 1<br />
C<br />
0<br />
∫ 1 ∫ 1<br />
0<br />
where ψ(|z|) ∈ L p′ (d|z|), 1/p ′ + 1/p = 1.<br />
Changing <strong>the</strong> variables we have<br />
∫ 1<br />
0<br />
0<br />
0<br />
0<br />
) 1/p<br />
) p(1 1/p<br />
|R s f(ρz)|(1 − ρ) dρ) s−1 − |z|) α |z| sp d|z|<br />
|R s f(ρz)|(1 − ρ) s−1 (1 − |z|) α/p ψ(|z|)|z| s d|z|dρ ,<br />
|R s f(ρz)|(1 − ρ) s−1 dρ ≤<br />
∫ |z|<br />
Using <strong>the</strong> last inequality and Hölder inequality, we get<br />
M ≤ C<br />
≤<br />
C<br />
∫ 1 ∫ |z|<br />
0<br />
∫ 1<br />
0<br />
( ∫ 1<br />
0<br />
0<br />
|R s s−1 du<br />
f(uξ)|(1 − u)<br />
|z| . s<br />
|R s f(uξ)|(1 − u) s−1 (1 − |z|) α/p ψ(|z|)d|z|du<br />
|R s f(vξ)|(1 − v) s<br />
1 − v<br />
∫ 1<br />
v<br />
) p ′<br />
(1 − |z|) α/p ψ(|z|)d|z|dv<br />
( ∫ 1 1<br />
) 1/p ′( ∫ 1<br />
1/p<br />
≤ C<br />
ψ(|z|)d|z| dv |R s f(uξ)| p (1 − u) du) α+ps .<br />
0 1 − v v<br />
0<br />
From <strong>the</strong> last inequality and Lemma 2.4 we easily get <strong>the</strong> first part <str<strong>on</strong>g>of</str<strong>on</strong>g> <strong>the</strong> <strong>the</strong>orem.