Ratio Mathematica Volume 41, 2021, pp. 173-180 Tikhonov type regularization for unbounded operators E Shine Lal* P Ramya† Abstract In this paper, we introduce a Tikhonov type regularization method for an ill-posed operator equation Tx = y, where T is a closed densely defined unbounded operator on a Hilbert space H. Keywords: densely defined operator, closed operator, Tikhonov type regularization. 2020 AMS subject classifications: 47A10, 47A52 1 *E Shine Lal, (Department of Mathematics, University College, Thiruvananthapuram, Kerala, India -695034); shinelal.e@gmail.com †P Ramya, (Department of Mathematics, N.S.S College, Nemmara, Kerala, India-678508); ramyagcc@gmail.com 1Received on September 20, 2021. Accepted on December 10, 2021. Published on December 31, 2021. doi: 10.23755/rm.v41i0.663. ©The Authors. This paper is published under the CC-BY licence agreement. 173 E Shine Lal, P Ramya 1 Introduction Most of the problems arise in the field of science and engineering can be mod- elled as an operator equation Tx = y (1) where T : X → Y is a bounded linear map from a normed linear space X to a normed linear space Y . In most of the cases (1) is Ill-posed. Certain regularization procedures are known for solving ill-posed operator equation (1). For example Tikhonov regularization, Mollifier method, Ritz method [5, 3]. In this paper we introduce a Tikhonov type regularization method for solving an ill-posed operator equation (1), where T is a closed densely defined operator on a Hilbert space H and we study the order of convergence. 2 Preliminaries Let L(H),C(H) and B(H) denote the space of all linear, closed linear and bounded linear operators on a Hilbert space H respectively. For T ∈ L(H), the domain, range of T are denoted by D(T),N(T) respectively. An operator T ∈ L(H) is said to be densely defined if D(T) = H. For example let T : l2(N) → l2(N) defined by T(x1,x2,x3, ....,xn, ....) = (x1, 2x2, 3x3, .....,nxn, ....) with domain D(T) = {(x1,x2,x3, ....,xn, ....) ∈ H : Σ∞j=1|jxj| 2 < ∞}. Then T is closed and unbounded. Since c00 ⊆ D(T) and c00 is dense in l2(N), D(T) is dense in l2(N). Proposition 2.1. Let T ∈ C(H) be a densely defined operator. Then there exist a unique operator T∗ ∈ C(H) such that 〈Tx,y〉 = 〈x,T∗y〉 ∀x ∈ D(T), ∀y ∈ D(T∗). Proof. Let D(T∗) = {y ∈ H : 〈Tx,y〉 is continuous for every x ∈ D(T) }. For y ∈ D(T), define f : D(T) → C by f(x) = 〈Tx,y〉 ∀x ∈ D(T). Extend f to f0 : H → C by f0(x) = lim n→∞ 〈Txn,y〉 where (xn) is a sequence in D(T) such that xn → x. 174 Tikhonov type regularization for unbounded operators Next we prove that f0 is well defined. For, let (xn) and (yn) be two sequences in D(T) converges to x. Since T is closed, T(xn−yn) → 0. If 〈Txn,y〉→ 〈x,y〉, then |〈Tyn,y〉−〈x,y〉| = |〈Tyn −Txn + Txn −x,y〉| ≤ ‖T(yn −xn)‖‖y‖ + |〈Txn −x,y〉| → 0 as n →∞ Hence f0 is well defined. Since f0 is a bounded linear functional on the Hilbert space H, by Riesz rep- resentation theorem there exist a unique y∗ ∈ H such that f0(x) = 〈x,y∗〉. Thus 〈Tx,y〉 = 〈x,y∗〉 ∀x ∈ D(T). Define T∗ : D(T∗) → H by T∗y = y∗. Then T∗ is well-defined. Also 〈Tx,y〉 = 〈x,T∗y〉 ∀x ∈ D(T), ∀y ∈ D(T∗). Consider an ill-posed operator equation Tx = y (2) where T is a closed densely defined operator on H. Definition 2.1. [7] Let T ∈ C(H) be densely defined. Then there exist a unique densely defined operator T† ∈ C(H) with domain D(T†) = R(T)⊕R(T)⊥ satisfies the following properties (i) TT†y = P R(T) y for all y ∈ D(T†), (ii) T†Tx = QN(T)⊥x for all x ∈ D(T). (iii) N(T†) = R(T)⊥. where P and Q are the orthogonal projection on to R(T) and N(T⊥) respec- tively. The operator T† is called the Moore-Penrose inverse of T . For y ∈ D(T†), let Sy = {x ∈ D(T) : ‖Tx−y‖ ≤ ‖Tu−y‖ ∀u ∈ D(T)}. Then u ∈ Sy is called least square solution of the operator equation (2). Note that ‖T†y‖ ≤ ‖x‖ ∀x ∈ Sy, is called least square solution of minimal norm and is denoted by x̂ [7]. If R(T) is not closed, then T† is not continuous. Now we introduce a Tikhonov type regularization procedure for finding an approximate solution for T†y. 3 Tikhonov type regularization In this section we introduce a Tikhonov type regularization procedure for solv- ing (2). 175 E Shine Lal, P Ramya Lemma 3.1. Let T ∈ C(H) be densely defined and α > 0. Then T∗T + αI and TT∗+αI are bijective closed densely defined operators on H. Also (TT∗+αI)−1 and (T∗T + αI)−1 are bounded, self adjoint operators on H. Proof. Let T ∈ C(H) and α > 0. By proposition 2.1, we have T∗ ∈ C(H). Hence, (TT∗ + αI) and (T∗T + αI) are closed densely defined operators on H. Since 〈(TT∗ + αI)x,x〉 = 〈T∗x,T∗x〉 + α〈x,x〉 ≥ 0,∀x ∈ D(T∗), we have (TT∗+αI) is a positive operator. Similarly (T∗T +αI) is also a positive operator. Since T∗T + αI is positive, ‖(T∗T + αI)x‖‖x‖≥ 〈(T∗T + αI)x,x〉 = 〈T∗Tx,x〉 + α‖x‖2 ≥ α‖x‖2 ∀x ∈ H Thus ‖(T∗T + αI)x‖≥ α‖x‖∀x ∈ H (3) Since T∗T + αI is bounded below, it is one-one and its inverse from the range is continuous. Also R(T∗T + αI) is closed. Since T∗T + αI is also self adjoint, R(T∗T + αI) = N(T∗T + αI)⊥ = H. Hence T∗T + αI is onto. Therefore (T∗T + αI)−1 ∈ B(H). Similary (TT∗ + αI)−1 ∈ B(H). From (3), ‖(T∗T + αI)−1‖≤ 1 α . Theorem 3.1. Let T ∈ C(H) be densely defined. Then T∗(TT∗ + αI)−1 and T(T∗T + αI)−1 are bounded operators on H. Also ‖ T∗(TT∗ + αI)−1 ‖≤ 1√ α and ‖ T(T∗T + αI)−1 ‖≤ 1√ α . Proof. We have (T∗T + αI)−1T∗T = I −α(T∗T + αI)−1 Since 〈(T∗T + αI)−1x,x〉≥ 0 ∀x ∈ H, 〈(T∗T + αI)−1T∗Tx,x〉 = 〈I −α(T∗T + αI)−1x,x〉 = 〈x,x〉−α〈(T∗T + αI)−1x,x〉≤ 〈x,x〉. Since (T∗T + αI)−1T∗T self adjoint, ‖(T∗T + αI)−1T∗T‖≤ 1. Let x ∈ H. ‖T∗(TT∗ + αI)−1x‖2 = 〈T∗(TT∗ + αI)−1x,T∗(TT∗ + αI)−1x〉 = 〈TT∗(TT∗ + αI)−1x, (TT∗ + αI)−1x〉 = 〈(TT∗ + αI)−1TT∗x, (TT∗ + αI)−1x〉 ≤ ‖(TT∗ + αI)−1TT∗x‖‖(TT∗ + αI)−1x‖ ≤ 1 α ‖x‖2 176 Tikhonov type regularization for unbounded operators we have ‖T∗(TT∗ + αI)−1x‖2 ≤ 1 α ‖x‖2 ∀x ∈ H. Thus ‖T∗(TT∗ + αI)−1‖≤ 1 √ α . Hence T∗(TT∗ + αI)−1 is bounded. Similarly T(T∗T + αI)−1 is bounded. Lemma 3.2. [7] Let T ∈ C(H) be densely defined. Then (i) (TT∗ + I)−1T ⊆ T(T∗T + I)−1 (ii) (T∗T + I)−1T∗ ⊆ T∗(TT∗ + I)−1 Remark 3.1. From Theorem 3.1 , we have T∗(TT∗ + αI)−1 and T(T∗T + αI)−1 are bounded. Therefore from Lemma 3.2, we have (TT∗ + αI)−1T and (T∗T + αI)−1T∗ are bounded. Lemma 3.3. Let T ∈ C(H) be densely defined. For every x ∈ D(T) ∩ N(T)⊥ ‖α(T∗T + αI)−1x‖−→ 0,as α → 0. Proof. Let Tα = α(T∗T + αI)−1, α > 0. From (3.1) we have ‖(T∗T + αI)−1‖≤ 1 α . Hence ‖Tα‖≤ 1 for every α > 0. Let u ∈ R(T∗T) then there exist v ∈ D(T∗T) such that T∗Tv = u. ‖Tαu‖ = ‖TαT∗Tv‖ = α‖(T∗T + αI)−1T∗Tv‖ ≤ α‖(T∗T + αI)−1T∗T‖‖v‖ ≤ α‖v‖ Hence ‖Tαu‖≤ α‖v‖∀u ∈ R(T∗T). Thus for every u ∈ R(T∗T), ‖α(T∗T + αI)−1u‖ −→ 0 as α −→ 0. Since R(T∗T) = N(T)⊥, ‖α(T∗T + αI)−1x‖−→ 0, ∀x ∈ D(T) ∩N(T)⊥. Theorem 3.2. Let T ∈ C(H) be densely defined and Rα = (T∗T + αI)−1T∗. Then {Rα}α>0 is a regularization family for (2). Proof. Let y ∈ D(T∗). Then (T∗T + αI)x̂ = T∗y + αx̂. Hence x̂ = (T∗T + αI)−1(T∗y + αx̂). Thus T†y −Rαy = x̂− (T∗T + αI)−1T∗y = (T∗T + αI)−1(T∗y + αx̂) − (T∗T + αI)−1T∗y = (T∗T + αI)−1αx̂ 177 E Shine Lal, P Ramya Hence ‖T†y −Rαy‖ = α‖(T∗T + αI)−1x̂‖. Since x̂ ∈ D(T) ∩N(T)⊥, by Lemma 3.3, ‖T†y −Rαy‖−→ 0 as α −→ 0. Thus {Rα}α>0 is a regularization family for (2). 4 Order estimate In this section we find an error estimate for the regularization family Rα = (T ∗T + αI)−1T∗, where T is a closed densely defined operator. We use the following lemmas. Lemma 4.1. [7] For T ∈ C(H) we have the following (i) If µ ∈ C and λ ∈ σ(T) then λ + µ ∈ σ(T + µI) (ii) If α ∈ C and λ ∈ σ(T) then αλ ∈ σ(αT ) (iii) σ(T2) = {λ2 : λ ∈ σ(T)} Lemma 4.2. [7] Let T ∈ L(H) be a positive operator. Then the following results bold. (i) T† is positive. (ii) σ(T) = σa(T) (iii) 0 /∈ σ(I + T) that is (I + T)−1 ∈ B(H) (iv) If 0 /∈ σ(T) then 0 6= λ ∈ σ(T) if and only if 1 λ ∈ σ(T−1) Theorem 4.1. Suppose T ∈ C(H) is densely defined positive operator. Then for every α > 0 σ ( (T + αI)−2T ) = { λ (λ + α)2 : λ ∈ σ(T) } Proof. Since T is positive, T + αI is bijective. Also (T + αI)−2T = (T + αI)−1 −α(T + αI)−2. From Lemmas 4.1, 4.2 for α,λ > 0, we have λ ∈ σ(T) if and only if (λ + α)−1 ∈ σ ( (T + αI)−1 ) . Hence σ ( (T + αI)−2T ) = { µ−αµ2 : µ ∈ σ ( (T + αI)−1 )} = { 1 λ + α − α (λ + α)2 : λ ∈ σ(T) } = { λ (λ + α)2 : λ ∈ σ(T) } . 178 Tikhonov type regularization for unbounded operators Corolary 4.1. Let T ∈ C(H) be densely defined and α > 0. Then ‖(T∗T + αI)−1T∗‖ = sup { √λ λ + α : λ ∈ σ(T∗T) } 6 1 2 √ α Proof. We have Rα = (T∗T + αI)−1T∗. Hence R∗αRα = T(T ∗T + αI)−2T∗. From Lemma 2.2 in [2], we have R∗αRα = (TT ∗ + αI)−2TT∗. Since R∗αRα is self adjoint and bounded, ‖Rα‖2 = ‖R∗αRα‖ = Sup { |k| : k ∈ σ(R∗αRα) } = Sup { λ (λ + α)2 : λ ∈ σ(TT∗) } ‖Rα‖ = Sup { √λ λ + α : λ ∈ σ(TT∗) } . Since 2 √ αλ(λ + α)−1 6 1 for λ,α > 0, we have ‖Rα‖ 6 1 2 √ α . Now we find an order estimate for Rα. Corolary 4.2. Let T ∈ C(H) is densely defined and Rα = (T∗T + αI)−1T∗. For every α > 0 and δ > 0, let yδ ∈ H be such that ‖y − yδ‖ 6 δ. Then ‖Rαy −Rαyδ‖ 6 δ 2 √ α . Proof. For ‖y −yδ‖ 6 δ, ‖Rαy −Rαyδ‖ 6 ‖Rα‖‖y −yδ‖ 6 1 2 √ α ‖y −yδ‖ 6 δ 2 √ α Theorem 4.2. Let T ∈ C(H) is densely defined and Rα = (T∗T +αI)−1T∗. Then ‖x̂−Rαyδ‖ 6 ‖x̂−Rαy‖ + δ 2 √ α . If α = α(δ) is chosen such that α(δ) −→ 0 and δ √ α(δ) −→ 0 as δ −→ 0, then ‖x̂−Rδα(δ)‖−→ 0 as δ −→ 0. 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