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13.1 Symmetric linear compact operator Recall that eigenvalue problems were written in variational form in Chapter 10. In this chapter we consider a generalization. Let H be a real Hilbert space with inner product (,) and norm || ||. Let V be a linear subspace of H and b a symmetric bilinear form on V. Eigenvalue problem EV Variational form Find A ER and r EV, such that x = 0 and b(x, y) = X(x, y) for each y € V. The eigenvalue problem must be written in operator form to apply the Hilbert- Schmidt theory. (The differential operator in the original problem cannot be used since it is unbounded.) In the weak variational problems considered in previous lectures the linear sub- space V is a Hilbert space with inner product b. This serves as motivation for the assumptions for the abstract eigenvalue problem. Assumptions A1 The linear subspace V is a Hilbert space with inner product b. A2 There exists a c> 0 such that b(u, u) ≥ clu||2 for each u E V. A3 V is dense in H. Notation Norm for V: ||u||v = √b(u, u). Theorem 1 For each fE H there exists a unique element u € V such that b(u, v) = (f, v) for each v EV (13.1) and ||u|| v ≤ c¹||f||. 1

13.1 Symmetric linear compact operator Recall that eigenvalue problems were written in variational form in Chapter 10. In this chapter we consider a generalization. Let H be a real Hilbert space with inner product (,) and norm || ||. Let V be a linear subspace of H and b a symmetric bilinear form on V. Eigenvalue problem EV Variational form Find A ER and r EV, such that x = 0 and b(x, y) = X(x, y) for each y € V. The eigenvalue problem must be written in operator form to apply the Hilbert- Schmidt theory. (The differential operator in the original problem cannot be used since it is unbounded.) In the weak variational problems considered in previous lectures the linear sub- space V is a Hilbert space with inner product b. This serves as motivation for the assumptions for the abstract eigenvalue problem. Assumptions A1 The linear subspace V is a Hilbert space with inner product b. A2 There exists a c> 0 such that b(u, u) ≥ clu||2 for each u E V. A3 V is dense in H. Notation Norm for V: ||u||v = √b(u, u). Theorem 1 For each fE H there exists a unique element u € V such that b(u, v) = (f, v) for each v EV (13.1) and ||u|| v ≤ c¹||f||. 1

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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prove theorem 4 and theorem 5 Pls do it fast! Within 20 min and i will rate instantly for sure Solution must be in typed form
Proof: Exercise Definition The operator T For each fE H let Tf be the unique solution of (13.1). Theorem 2 The operator T is linear. Proof: Exercise Theorem 3 The operator T is a bounded linear operator from (H, ||- ||) to (V, ||- ||v), i.e. ||Tx|v ≤e¹|||| for each EH. (13.2) Proof: Exercise Remark There exists a constant K> 0 such that ||Tx|| K|||| for each ΤΕΗ. Theorem 4 The null space N(T) of the operator T is trivial. Proof: If x EN(T), then Tx = 0. Consequently, (x, y) = b(Tx, y) = 0 for each y E V. Since V is dense in H, (x, y) = 0 for each y EV and hence x = 0. Definition A linear operator L with domain D(L) = H, is called symmetric if (Lx, y) = (x, Ly) for all x, y in H. Theorem 5 The operator T is symmetric. Proof: Exercise Proposition 1 (Tx, x) > 0 for each x H, x 0. Proof: For each x H, 0≤ b(Tx, Tx) = (Tx,x). The result follows from Theorem 4. The definition of compactness of a linear operator is Definition 8.1-1 in Kreyszig. We need only the equivalent Compactness criterion, Theorem 8.1-3, and in a simpler form. Theorem A linear operator L with domain D(L) = H, is compact if and only if it maps every bounded sequence (n) in H onto the sequence (Lan) which has a convergent subsequence. Assumption A4 Every bounded set in V contains a sequence that converges in H.
Transcribed Image Text:Proof: Exercise Definition The operator T For each fE H let Tf be the unique solution of (13.1). Theorem 2 The operator T is linear. Proof: Exercise Theorem 3 The operator T is a bounded linear operator from (H, ||- ||) to (V, ||- ||v), i.e. ||Tx|v ≤e¹|||| for each EH. (13.2) Proof: Exercise Remark There exists a constant K> 0 such that ||Tx|| K|||| for each ΤΕΗ. Theorem 4 The null space N(T) of the operator T is trivial. Proof: If x EN(T), then Tx = 0. Consequently, (x, y) = b(Tx, y) = 0 for each y E V. Since V is dense in H, (x, y) = 0 for each y EV and hence x = 0. Definition A linear operator L with domain D(L) = H, is called symmetric if (Lx, y) = (x, Ly) for all x, y in H. Theorem 5 The operator T is symmetric. Proof: Exercise Proposition 1 (Tx, x) > 0 for each x H, x 0. Proof: For each x H, 0≤ b(Tx, Tx) = (Tx,x). The result follows from Theorem 4. The definition of compactness of a linear operator is Definition 8.1-1 in Kreyszig. We need only the equivalent Compactness criterion, Theorem 8.1-3, and in a simpler form. Theorem A linear operator L with domain D(L) = H, is compact if and only if it maps every bounded sequence (n) in H onto the sequence (Lan) which has a convergent subsequence. Assumption A4 Every bounded set in V contains a sequence that converges in H.
13.1 Symmetric linear compact operator Recall that eigenvalue problems were written in variational form in Chapter 10. In this chapter we consider a generalization. Let H be a real Hilbert space with inner product (,) and norm || ||. Let V be a linear subspace of H and b a symmetric bilinear form on V. Eigenvalue problem EV Variational form Find A ER and rE V, such that x = 0 and b(x, y) = A(x, y) for each y € V. The eigenvalue problem must be written in operator form to apply the Hilbert- Schmidt theory. (The differential operator in the original problem cannot be used since it is unbounded.) In the weak variational problems considered in previous lectures the linear sub- space V is a Hilbert space with inner product b. This serves as motivation for the assumptions for the abstract eigenvalue problem. Assumptions A1 The linear subspace V is a Hilbert space with inner product b. A2 There exists a c> 0 such that b(u, u) ≥ clu2 for each u € V. A3 V is dense in H. Notation Norm for V: |||u|| v = Theorem 1 For each fE H there exists a unique element u € V such that b(u, v) = (f, v) for each ve V (13.1) = √b(u, u). and ||u|| v ≤ c ¹||f||. 1
Transcribed Image Text:13.1 Symmetric linear compact operator Recall that eigenvalue problems were written in variational form in Chapter 10. In this chapter we consider a generalization. Let H be a real Hilbert space with inner product (,) and norm || ||. Let V be a linear subspace of H and b a symmetric bilinear form on V. Eigenvalue problem EV Variational form Find A ER and rE V, such that x = 0 and b(x, y) = A(x, y) for each y € V. The eigenvalue problem must be written in operator form to apply the Hilbert- Schmidt theory. (The differential operator in the original problem cannot be used since it is unbounded.) In the weak variational problems considered in previous lectures the linear sub- space V is a Hilbert space with inner product b. This serves as motivation for the assumptions for the abstract eigenvalue problem. Assumptions A1 The linear subspace V is a Hilbert space with inner product b. A2 There exists a c> 0 such that b(u, u) ≥ clu2 for each u € V. A3 V is dense in H. Notation Norm for V: |||u|| v = Theorem 1 For each fE H there exists a unique element u € V such that b(u, v) = (f, v) for each ve V (13.1) = √b(u, u). and ||u|| v ≤ c ¹||f||. 1
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