Int. J. Contemp. Math. Sciences, Vol. 6, 211, no. 7, 345-351 Volterra Composition Operators Anupama Gupta G.C.W. Parade, Jammu, India anu ju8@yahoo.com B. S. Komal Department of Mathematics University of Jammu, Jammu, India bskomal2@yahoo.co.in Abstract In this paper we characterized the Volterra composition operator. It is shown that the spectrum of Volterra composition operator is consisting of zero only. Mathematics subject classification: Primary 47B2, Secondary: 46B38 Keywords: Composition operator, Volterra operator, Expectation operator, Idempotent oprator, Radon-Nikodym derivative 1 Preliminaries Let (X, S, μ) be a s-finite measure space and let φ : X X be a non-singular measurable transformation (μ(e) μφ 1 (E) ). Then a composition transformation, C φ : L p (μ) L p (μ), 1 p<, is defined by C φ f foφ for every f L p (μ). In case C φ is continuous, we call it a composition operator induced by φ. It is easy to see that C φ is a bounded operator if and only if dμφ 1 f dμ o, the Radon-Nikodym derivative of the measure μφ 1 with respect to the measure μ, is essentially bounded. For more detail about composition operators we refer to Nordgen [4],Shapiro[7],Cowen[3],Singhand Manhas[11],Singh and Komal [1]. For each f L p (μ), 1 p<, there exists a unique φ 1 (S) measurable function E(f) such that gfdμ ge(f)dμ for every φ 1 (S) measurable function g for which left integral exists. The function E(f) is called conditional expectation of f with respect to the sub algebra φ 1 (S). In particular E(f) goφ if and only if g (E(f)oφ 1 ). For more
346 A. Gupta and B. S. Komal properties of the expectation operator, see Parthasarthy [8 ]and Lambert [2]. The Volterra operator V is an integral operator induced by the kernel K(x,y) defined as {, if x y K(x, y) 1, if x>y i.e. (Vf)(x) f(y)dy y Let φ :[, 1] [, 1] be a measurable transformation. The Volterra composition operator V is defined by (V φ f)(x) (C φ Vf)(x) φx f(t)dtfor every f L p [, 1] Set K φ (x, y) f o (x)e(φ 1 (y)) 2 dμ(y)dμ(x). A great deal of work on voltera operators was done by mathematicians Sinnamen [5], Erdos [6],Stepanov ([13],[14]) Sunder and Halmos [9], Lybic s [15] conjecture was introduced by Whitley [12] and generalized it to Volterra composition operators on L p [, 1].Gupta and Komal [3] also studied composite integral operator on L p spaces. Our main purpose in this paper is to study Volterra composition operator. It is shown that spectrum of Volterra composition operator is equal to {}. Theorem 1.1 Suppose K φ L 2 (μ μ). Then V φ is bounded Volterra composition operator. Proof: For any f L 2 (μ), we have V φ f 2 (V φ f)(x) 2 dμ(x) f(φ(x))dμ(y) 2 dμ(x) ( f o (x)e(φ 1 (y)) 2 dμ(y) f(y) 2 dμ(y))dμ(x). f o (x)e(φ 1 (y)) 2 dμ(y)dμ(x). f 2. K φ 2. f 2 <. Hence V φ is bounded Volterra composition operator.
Volterra composition operators 347 Theorem 1.2 Let V φ B(L 2 (μ)). Then Vφ f f o E(V f)oφ 1 Proof: Let f,g L 2 (μ). Consider Thus for μ-almost all x. Hence V φ f,g f,v φ g 1 1 f(x)v φ g(x)dμ(x) f(x) goφ(y)( (goφ)(y)dμ(y)dμ(x) 1 y f(x)dμ(x)dμ(y) goφ(y)(v f)(y)dμ(y) V f,c φ g Cφ V f,g Vφ f(x) (CφV f)(x) f o E((V f)oφ 1 (x)) V φ f f o E(V f)oφ 1. Theorem 1.3 Let V φ B(L 2 [, 1]). Then V φ is idempotent if and only if V φ is the zero operator. Proof : Suppose V φ is idempotent. Then Vφ 2f V φf. Taking f χ [,1], implies that φ(x) φ(t)dt φ(x), for all x. Differentiating, we get φ(φ(x))φ (x) φ (x) Suppose φ (x) for μ-almost all x E,μ(E) >. Then φ(φ(x)) 1 for μ-almost all x E. Next f(x) x for every x [, 1], we have (V φ f)v φ (V φ f)(x) φ(x) (φ(t)) 2 2 dt (φ(x))2. 2
348 A. Gupta and B. S. Komal On differentiating with respect to x, This yields that φ(x) 1 2, so that φ (x), which is contradiction. Further, suppose φ (x) a.e. Then φ(x) c for μ-almost all x [, 1]. we have (V φ f)(x) c and V φ (V φ f)(x) c 2. This implies that c or c 1. Now taking c 1 and f (x) x, again from (1) and (2) we get c 2 2 c3. i.e. 1 2, which is absurd. Hence c must be zero. Thus φ(x) for μ-almost every x [.1]. φ(x) Hence (V φ f)(x) f(x) for every x X and f L 2 ([, 1]). This proves that V φ. The converse is trivial. Theorem 1.4 The Volterra composition operator is never an identity operator. Proof: If possible, suppose a Volterra composition operator V φ for some φ is the identity operator, I. Then V φ f f for every f L 2 ([, 1]). taking f χ [,1], we have φ(x) f(t)dt f(x) for μ-almost all x [, 1] Hence from (1) Hence V φ I. 1 f(t)dt f(x) for all x [, 1], which is absurd. Theorem 1.5 Suppose φ(). Then σ(v φ ){}. Proof : If possible, suppose σ(v φ ) {}. Let λ is an eigen value of V φ. For, f,f L 2 (μ). (V φ f)(x) λf(x) Differentiating with respect to x, we get f(φ(y))dμ(y)λf(x) d x dx ( f(φ(y))dμ(y)) λf (x)
Volterra composition operators 349 f(φ(x) λf (x) f (x) 1 λ (foφ)(x) 1 λ f(φ(x)) From (1) f(). From (2) f () Also f (),... and so on f n (). This implies that f is zero function, which is a contradiction Hence σ(v φ ) {}. Example 1.6 Suppose (i) φ(x) ax, 1 >a>: x [o, 1]. (ii) φ(x) x n,n>1. Then σ(v φ ){}. Solution : (i) Suppose o λ is an eigen value of V φ Then f(ay)dy λf(x) (1) Put ay t, a dy dt x 1 i.e. f(t)dt λf(x) α Differentiating with respect to x, we get f(φ(x)) λf (x) From (1), f(), From (2), f ()... and so on f n () Hence σ(v φ ){}. (ii) If φ(x) x n we have f(φ(y)dy λf(x) Put y n t ny n 1 dy dt Put this value in (3), we have f (x) 1 λ f (φ(x)) (2) f(y n )dy λf(x) (3) dy 1 n t1/n 1 dt 1 f(t)t 1/n 1 dt λf(x) n Differentiating both sides with respect to x. We have 1 n f(x)x1/n 1 λf (x) (4)
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