Invariant subspaces of bilateral shift on L2(S1)

In this Paper, we have proved that all the invariant subspaces of the bilateral shift on L2(S1) is of the form φHb2 where φ is a function in L∞ which is equals to 1 almost everywhere.

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Introduction

Notation S1 = {z ∈ C: |z| = 1} i.e S1 is the circle with center origin and radius 1. 1.1 Definition (L2(S1)) It is defined as the space of all the equivalence classes of functions [3] that are Lebesgue measurable on S1 and square integrable on S1 with respect to Lebesgue measure normalized such that measure of S1 is 1. Inner product on L2(S1) is given by – Note L2(S1) is an Hilbert-space with the orthonormal basis given by {en : n ∈ Z} where en(eιθ) = eιnθ. Therefore . 1.2 Definition (Hc2 space) Hc2 = {f ∈ L2(S1) :< f, en >= 0 for negative value of n} (∞) Hc2 = f ∈ L2(S1) : f = X < f,en > en. n=0 is a closed subspace of L2(S1) whose negative Fourier coefficients are 0 ∴ {en : n = 0,1,...} are orthonormal basis of 1.3 Wandering Subspace A closed N of a Hilbert-Space H is said to be a Wandering subspace of a bounded linear operator T on H if N is orthogonal to Tk(N) for all k ≥ 0 Maths 2020; 5(4): 186-190 Dr. KL Kaushik Associate Professor, (Head) Department of Mathematics Aggarwal College, Ballabgarh, Faridabad, Haryana, India Dr.

KL Kaushik Associate Professor, (Head) Department of Mathematics Aggarwal College, Ballabgarh, Faridabad, Haryana, India http://www.mathsjournal.com 1.4 The Wolde’s Decomposition Theorem Let T be an isometric operator on a Hilbert-Space H then Where H = T(H)MN 1.4 Definition (L∞) It is defined as the Collection of all the essentially bounded measurable functions on the circle S1 . For a function φ ∈ L∞ the essential norm is defined by ||φ||∞ = inf{r : m{eιθ;|φ(eιθ)| > r} = 0} where m is normalized lesbesgue measure. Note |φ(eιθ| ≤ ||φ||∞ a.e 1.6 Multiplication Operator by a L∞ function For a function φ ∈ L∞ the multiplication operator on L2(S1) is defined by Sφ : L2(S1) → L(S1) f → φf Note (∵ T(N) ⊆ N) ⇒ φf ∈ L2(S1) ∴ Sφ is well defined 1.7 Bilateral Shift Bilateral shift on L2(S1) is an operator B : L2(S1) → L2(S1) defined by Bf(eιθ) = eιθf(eιθ) Note B is an isometric Unitary Operator [2] whose adjoint is the bounded linear operator B? : L2 → L2 given by B?f(eιθ) = e−ιθf(eιθ) i.e. B?B = I = BB? 1.8 Invariant Subspace A Closed subspace N of a Hilbert-Space H is said to be an Invariant Subspace of s Bounded linear operator T on H if T(N) ⊆ N Theorem 1.1.

N is an invariant subspace of T iff N⊥ is an invariant subspace of T? Proof. Let N is an invariant subspace of T ⇒ T(N) ⊆ N Let y ∈ T?(N⊥) be arbitrary then ∃ x ∈ N⊥ such that y = T?(x) ⇒ < y,n > = < T?(x),n > = < x,T(n) > = 0 ∀ n ∈ N ⇒ y ∈ N⊥ ⇒ T?(N⊥) ⊆ N⊥ ∴ N⊥ is an invariant subspace of T? Conversely Let N⊥ is an invariant subspace of T? ⇒ T?(N⊥) ⊆ N⊥ http://www.mathsjournal.com (∵ T?(N⊥) ⊆ N⊥) Let y ∈ T(N) be arbitrary then ∃ x ∈ N such that ⇒ y = T(x) ⇒ < y,n > = < T(x),n > = < x,T?(n) > = 0 ∀ n ∈ N⊥ y ∈ N ⇒ T(N) ⊆ N ∴ N is an invariant subspace of T 1.9 Reducing Subspace An invariant Subspace N of a bounded linear operator T on a Hilbert Space H if T(N) ⊆ N and T(N⊥) ⊆ N⊥ i.e.

T(N) ⊆ N and T?(N) ⊆ N Theorem 1.2. If N is a reducing subspace of a unitary operator T on a hilbert space H then T(N) = N Proof. Since N is a reducing subspace of T ⇒ T(N) ⊆ N and T?(N) ⊆ N [1] Since T is a Unitary Operator ⇒ TT? = I = T?T Then N = TT?(N) ⊆ T(N) (∵ T?(N) ⊆ N) ⇒ T(N) = N

Result

(?) If E is a non zero Reducing Subspace of bilateral shift B on L2(S1) then ∃ a subset E of S1 of positive measure such that E = XEL2(S1) 2. Invariant subspace of bilateral shift on L2(S1) Theorem 2.1. Let N be an invariant(not reducing) subspace of bilateral shift B on L2(S1) then there exist a function φ in L∞ such that N = φHc2 with essential norm 1. Proof.

Since N is an Invariant but not reducing subspace of bilateral shift B : L2(S1) → L2(S1) ⇒ B(N) ( N (∵ Theorem1.2) Since B is an Isometry and N is a closed subspace of L2(S1) . Then by Wolde’s Decomposition Theorem (1.3) we have ∞ N = \ Bk(N)MK MB(K)MB2(K)M......... n=0 Where N = K LB(N) (∵ since B is an isometry then B(N) is a closed ) Since K 6= {0} then we can find φ ∈ N such that ||φ||L2 = 1 1. Claim : φ is a function in L∞ with essential norm 1 Since φ ∈ K ⊂ B(N)⊥ (1) Taking conjugate of (1) we get (2) Since φ ∈ L2 ⇒ |φ|2 ∈ L1(S1) Then from(1) and (2) we have < |φ|2,ek >= 0 ∀ k = ±1,±2,......... http://www.mathsjournal.com And ∴ φ is a function in L∞ with essential norm 1. 2. Claim: K = span(φ)= < φ > Suppose not then ∃ a function ψ ∈ K such that ψ ⊥ φ and ||ψ||L2 = 1 Then same as claim 1 we can also show that ψ is a function in L∞ with essential norm 1 Since φ ⊥ ψ ⇒ φ ⊥ Bk(ψ) and Bk(φ) ⊥ ψ ∀ k ∈ N (∵ BK(N) ⊂ B(N)) which is a contradiction since |φ(eιθ)| = 1 = |ψ(eιθ| a.e. ∴ K = span(φ) =< φ > Then we have (∵ By Definition1.2) 3.

Claim: Let let f ∈ L.H.S then f = α1φ + α2φe1 + α3φe2 + ..... ∈ φHc2 Conversely , let f ∈ φHc2 ∞ ⇒ f = X < f,en > φen n=0 which clearly lies in the L.H.S. Hence the claim Then ∞ N = \ Bk(N)MφHc2 n=0 4. Claim: Clearly ∞ \ k B (N) n=0 is a reducing subspace of bilateral shift B. Then by the Result (?) ∃ an subset E of S1 such thatt http://www.mathsjournal.com ⇒ |φ| is zero on E But |φ| = 1 ⇒ m(E) = 0 a.e.

Hence where φ is a function in L∞ of essential norm 1.

Conclusion

We proved the results that help to identify all the invariant subspaces of Bilateral shift on L2 which is of the form φHb2 where φ is a function in L∞ which is equals to 1 almost everywhere which results to make the study of theseinvariant subspaces much easier.

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