Composition Operators fronm Weak to Strong Vector-Valued Hardy Spaces
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Complex Analysis and Operator Theory
Composition Operators fronm Weak to Strong Vector-Valued Hardy Spaces Oscar Blasco1 Received: 31 October 2019 / Accepted: 10 September 2020 © Springer Nature Switzerland AG 2020
Abstract Let φ be an analytic map from the unit disk into itself, 1 < p < 2 and 1 ≤ q ≤ p. It is shown that the composition operator Cφ ( f ) = f ◦ φ is bounded from p Hweak (D, L q (μ)) into H p (D, L q (μ)) if and only if Cφ is a 2-summing operator p from H p (D) into H p (D). Here Hweak (D, X ) and H p (D, X ) are the weak and strong formulation of X -valued Hardy spaces on the unit disc. Keywords Composition operator · Vector-valued Hardy space · p-summing operator Mathematics Subject Classification Primary 47B33; Secondary 46E40 · 47B10
1 Introduction and Notation Throughout the paper X is a complex Banach space, 1 ≤ p < ∞ and we write H(D, X ) for the space of X -valued holomorphic functions and H p (D, X ) for the space of functions F ∈ H(D, X ) such that F H p (D,X ) = sup
0 0 is called the p-summing norm of u and is denoted by π p (u). Thus, u ∈ p (X , Y ) if and only if uˆ ∈ L(wp (X ), p (Y )). A basic observation that plays an important role in the sequel is that if Cφ : φ , H p (D) → H p (D) and X = q for 1 ≤ q < ∞ then CφX = C Indeed, if F ∈ H(D, q ) then it can be seen either as a Taylor series F(z) =
∞
xn z n , xn = xn (k) k∈N ∈ q
n=0
where the convergence is uniform on compact sets of D or as a sequence of holomorphic functions F(z) = f k (z) with f k ∈ H(D) for k ∈ N. Of course F(z) =
∞ k=1
f k (z)ek
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and f k (z) =
∞
xn (k)z n , k ∈ N, z ∈ D.
n=0
Therefore CφX (F) =
∞
φ ( f k ) . f k (φ(z))ek = Cφ ( f k ) = C
k=1
Let us now see that Hweak (D, q ) = qw (H p (D)) for 1 ≤ p, q < ∞. p
Lemma 2.1 Let 1 ≤ p, q < ∞ and let F = ( f k ) ∈ H(D, q ). Then F ∈ p Hweak (D, q ) if and only if ( f k ) ∈ qw (H p (D)). Moreover F H p
weak (D,q )
= q (( f k )).
Proof Let (βk ) ∈ (q )∗ . Then we have (βk ), F (z) =
∞
βk f k (z), z ∈ D.
(2.1)
k=1
Hence F H p
weak (D,q )
∞ = sup βk f k (βk )( )∗ ≤1 q
k=1
= q (( f k )). H p (D)
Since H p (D, p ) = p (H p (D)) we can apply Lemma 2.1 to get the following observation. p
Corollary 2.2 Let 1 ≤ p < ∞. Then CφX : Hweak (D, p ) → H p (D, p ) is bounded if and only if Cφ ∈ p (H p (D), H p (D)). Also we would like to mention (see [10, page 225]) that if T ∈ p (X , Y ) and ( , , μ) is a measure space then p
T F(w)Y dμ(w)
1/ p ≤ π p (T ) sup
x ∗ ≤1
| x ∗ , F(w) | p dμ(w)
1/ p
(2.2) for any Bochner integrable function F : → X . Let us point out a way to produce functions in vector-valued Hardy spaces by means of summing operators.
Composition Operators from Weak to Strong Vector-Valued…
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Proposition 2.3 Let 1 < p < ∞ and T ∈ L(H p (D), X ). Denote u n (z) = z n , set xn = T (u n ) ∈ X for n ≥ 0 and define FT (z) =
∞
xn z n , |z| < 1.
n=0
If T ∈ p (H p (D), X ) then F
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