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A Volterra-Type Integration Operator on Fock Spaces

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A Volterra-Type Integration Operator on Fock Spaces PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 140, Number 12, December 2012, Pages 4247–4257 S 0002-9939(2012)11541-2 Article electronically published on April 18, 2012 A VOLTERRA-TYPE INTEGRATION OPERATOR ON FOCK SPACES OLIVIA CONSTANTIN (Communicated by Richard Rochberg) Abstract. We study certain spectral properties and the invariant subspaces for some classes of integration operators of Volterra type on the Fock space. 1. Introduction For analytic functions f,g we consider the Volterra-type integration operator given by z Tgf(z)= f(ζ)g (ζ) dζ. 0 The boundedness and compactness, as well as some spectral properties (such as Schatten class membership) of Tg acting on various spaces of analytic functions of the unit disc D in C have been extensively investigated (see [2, 6] for Hardy spaces, [3, 5, 8, 17, 19] for weighted Bergman spaces, and [9, 10] for Dirichlet spaces, or the surveys [1, 20] and the references therein). Furthermore, the spectrum of Tg on weighted Bergman spaces was recently characterized in [3]. p In this paper we consider the operator Tg acting on the Fock spaces F ,p>0 (here, the symbol g is an entire function with g(0) = 0). Recall that, for p>0, the Fock space F p consists of entire functions f for which 1 | |2 p p − z p fp = f(z)e 2 dA(z) < ∞. 2π C The boundedness and compactness of Tg follow by classical methods, and we include a sketch of their proofs in Section 3 for the sake of completeness. It turns out that, p q if p ≤ q, Tg : F →F is bounded if and only if the symbol g is a polynomial of degree ≤ 2, while the necessary and sufficient condition for compactness is: g is a polynomial of degree ≤ 1. For p>q, the operator Tg is bounded if and only if it is compact, and these are equivalent to the condition 2p q> and g(z)=az for some a ∈ C. p +2 In particular, this shows that the primitive of a function in F p is in F q for q> 2p/(p + 2), and this result is sharp. In Section 4 we turn to the spectral properties of Tg. Regarding the Schatten class membership of Tg|F 2 , we show that, provided p it is compact, Tg fails to be Hilbert-Schmidt, but it belongs to all classes S for Received by the editors May 30, 2011. 2010 Mathematics Subject Classification. Primary 30H20, 47B38. Key words and phrases. Fock spaces, integration operator, spectrum, invariant subspaces. c 2012 American Mathematical Society Reverts to public domain 28 years from publication 4247 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4248 OLIVIA CONSTANTIN p>2. Moreover, we prove that the spectrum of Tg|F 2 is a closed disc centred at the origin whose radius depends on the symbol g. In the last section of this paper we characterize the invariant subspaces of the Volterra operator z Vf(z)= f(ζ) dζ, 0 which is a particular case of the operator Tg, corresponding to g(z)=z.Acomplete description of the invariant subspaces of V , when acting on various classical spaces of analytic functions on the unit disc (Hardy spaces, weighted Bergman spaces, Dirichlet spaces) was obtained in [4]. We prove the Fock space analogue of these results, showing that the proper invariant subspaces of V : F p →Fp, p>0, are given by F p p { k ≥ } AN = Span z : k N +1 , where N is an arbitrary nonnegative integer. 2. Preliminaries Recall that the point evaluation functionals are bounded on F p, p>0. More precisely, for every f ∈Fp,wehave |z|2/2 (1) |f(z)|≤e fp,z∈ C. √ Moreover, the monomials zn/ n!,n≥ 0, form an orthonormal basis for F 2,and 2 λz¯ therefore the reproducing kernel of F is given by Kλ(z)=e , i.e. 1 2 f(λ)= f(z) eλz¯ e−|z| dA(z),λ∈ C,f∈F2. π C The following straightforward fact will be used in our further considerations: for an entire function f, the maximum principle and the subharmonicity of |f|p ensure that p p |f(z)| 2 |f(z)| 2 (2) e−p|z| /2dA(z) ∼ e−p|z| /2dA(z) | | p | | p C (1 + z ) |z|>1 (1 + z ) f(z)p 2 ∼ e−p|z| /2dA(z). |z|>1 z Moreover, if f(0) = 0, the above quantities are comparable to p f(z) −p|z|2/2 e dA(z). C z The next proposition can be easily deduced from a result in [7]. We include a sketch of its proof for the sake of completeness. Proposition 1. For an entire function f : C → C and p>0, the following holds: | |p | |p −p|z|2/2 ∼| |p f (z) −p|z|2/2 (3) f(z) e dA(z) f(0) + p e dA(z). C C (1 + |z|) Proof. It is enough to prove the claim for entire functions f with f(0) = 0. Indeed, the result for a general entire function f follows by applying (3) to z → f(z)−f(0). License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use A VOLTERRA-TYPE INTEGRATION OPERATOR ON FOCK SPACES 4249 Suppose now that f(z)=zg(z), where g is entire. Then by Lemmas 5-6 in [7] we have 2 2 |zg(z)|pe−p|z| /2dA(z) ∼|g(0)|p + |g(z)|pe−p|z| /2dA(z). C C This is equivalent to − p | |p −p|z|2/2 ∼| |p zf (z) f(z) −p|z|2/2 (4) f(z) e dA(z) f (0) + 2 e dA(z). C C z The integral on the right side is comparable to − p zf (z) f(z) −p|z|2/2 2 e dA(z). |z|>1 z Using this fact, relation (2) and the obvious inequality (a + b)p ≤ 2p(|a|p + |b|p), we now obtain | |p f (z) −p|z|2/2 | |p −p|z|2/2 p e dA(z) f(z) e dA(z). C (1 + |z|) C Hence one inequality in (3) is proven. We prove the reverse inequality by contra- diction. Assume there exists a sequence of entire functions fn with fn(0) = 0 such | |p −p|z|2/2 that C fn(z) e dA(z)=1and | |p fn(z) −p|z|2/2 ≤ 1 ≥ (5) p e dA(z) ,n 1. C (1 + |z|) n The expression (1+|z|) is bounded above and below on any compact set in C.Then | |p by the subharmonicity of f (z) and relation (5) we deduce that the sequence fn(z) converges to zero uniformly on the compact sets in C and, since fn(0) = 0, the same holds for fn. We now write relation (4) for each fn and we use (2) to get − p p zfn(z) fn(z) −p|z|2/2 1 ∼|f (0)| + e dA(z) n z2 C p p | |p fn(z) −p|z|2/2 fn(z) −p|z|2/2 fn(0) + e dA(z)+ 2 e dA(z) | | z | | z z >1 z >1 p p 1 fn(z) −p|z|2/2 |f (0)| + + e dA(z) n n z2 R>|z|>1 1 | |p −p|z|2/2 + 2 fn(z) e dA(z) R | | z >R p ≤| |p 1 fn(z) −p|z|2/2 1 fn(0) + + 2 e dA(z)+ 2 , n R>|z|>1 z R for any R>1. Letting first n →∞and then R →∞above we obtain a contradic- tion. Thus (3) holds. 3. Boundedness and compactness The next theorem characterizes the boundedness and the compactness of Tg. p q Theorem 1. (i) For 0 <p≤ q the operator Tg is bounded from F to F if and 2 p q only if g(z)=az + bz for some a, b ∈ C. Moreover, Tg : F →F is compact if and only if g(z)=az for some a ∈ C. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 4250 OLIVIA CONSTANTIN p (ii) Assume 0 <q<pand g ≡ 0. Then the operator Tg is bounded from F F q 2p ∈ C \ to if and only if q>p+2 and g(z)=az,forsomea 0. Furthermore, if p q Tg : F →F is bounded, then it is compact. Proof. (i) We start by proving the sufficiency part of the statements in (i). Assume g(z)=az2 + bz.ThenProposition1gives q |fg | 2 T fq ∼ e−q|z| /2 dA(z) g q (1 + |z|)q C | |q | |q 2az + b −q|z|2/2 = f q e dA(z) C (1 + |z|) q ≤ q f q f p, p q and the boundedness of Tg follows from the inclusion F ⊆F . Letusnowshowthatifg is constant, then Tg is compact. To this end, let fn ∞ → be a sequence of entire functions such that supn fn p < and fn 0 uniformly on compact sets in C. Also, let R>0. An application of Proposition 1 yields | |q q ∼ | |q g (z) −q|z|2/2 lim sup Tgfn q lim sup fn(z) e dA(z) →∞ →∞ (1 + |z|)q n n C q |f (z)| 2 (6) ∼ lim sup n e−q|z| /2dA(z) →∞ (1 + |z|)q n |z|≤R q |f (z)| 2 + n e−q|z| /2dA(z) | | q |z|>R (1 + z ) ≤ 1 q q lim sup fn p.
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