Since (e_k) is complete, (x = \sum_k=1^\infty \langle x, e_k \rangle e_k) in norm. Then [ |x|^2 = \langle x, x \rangle = \sum_k=1^\infty \sum_j=1^\infty \langle x, e_k \rangle \overline\langle x, e_j \rangle \langle e_k, e_j \rangle = \sum_k=1^\infty |\langle x, e_k \rangle|^2. ] 8. Problem (Riesz Representation Theorem): Let (f) be a bounded linear functional on a Hilbert space (H). Show there exists a unique (y \in H) such that (f(x) = \langle x, y \rangle) for all (x \in H), and (|f| = |y|). Solution (Sketch): If (f=0), take (y=0). Otherwise, let (M = \ker f), a closed subspace. Since (f \neq 0), (M^\perp \neq 0). Pick (z \in M^\perp) with (f(z)=1). For any (x \in H), write (x = m + \alpha z) with (m \in M), (\alpha = f(x)). Then (f(x) = \alpha f(z) = \alpha). But (\langle x, z \rangle = \langle m, z \rangle + \alpha |z|^2 = \alpha |z|^2). So (\alpha = \frac\langle x, z \rangle^2). Hence (f(x) = \langle x, \fraczz \rangle). Set (y = \fraczz). Uniqueness: If (\langle x, y_1 \rangle = \langle x, y_2 \rangle) for all (x), then (\langle x, y_1-y_2 \rangle =0) for all (x), so (y_1-y_2=0). Also (|f| = |y|) by Schwarz. Summary of Key Results (Chapter 3, Kreyszig) | Concept | Formula / Statement | |---------|----------------------| | Schwarz inequality | (|\langle x,y\rangle| \le |x||y|) | | Triangle inequality | (|x+y| \le |x|+|y|) | | Parallelogram law | (|x+y|^2+|x-y|^2=2(|x|^2+|y|^2)) | | Orthogonal complement | (M^\perp = x \in H : \langle x,m\rangle=0\ \forall m\in M) | | Projection theorem | (H = M \oplus M^\perp) for closed (M) | | Bessel’s inequality | (\sum |\langle x,e_k\rangle|^2 \le |x|^2) | | Parseval’s identity | (|x|^2 = \sum |\langle x,e_k\rangle|^2) if orthonormal basis | | Riesz representation | (f(x)=\langle x,y\rangle), (|f|=|y|) |
For any (n), [ 0 \le | x - \sum_k=1^n \langle x, e_k \rangle e_k |^2 = |x|^2 - \sum_k=1^n |\langle x, e_k \rangle|^2. ] Thus (\sum_k=1^n |\langle x, e_k \rangle|^2 \le |x|^2). Let (n \to \infty) gives the inequality. 7. Problem: Parseval’s identity. In a Hilbert space with complete orthonormal set (e_k), prove [ |x|^2 = \sum_k=1^\infty |\langle x, e_k \rangle|^2 \quad \forall x. ]
Expand: [ |x+y|^2 = |x|^2 + \langle x, y \rangle + \langle y, x \rangle + |y|^2 = |x|^2 + 2\Re\langle x, y \rangle + |y|^2. ] [ |x-y|^2 = |x|^2 - 2\Re\langle x, y \rangle + |y|^2. ] Add: (|x+y|^2 + |x-y|^2 = 2|x|^2 + 2|y|^2). 4. Problem: In (\ell^2), find the orthogonal complement of the subspace (M = (x_n) : x_2k=0 \ \forall k ) (sequences with zeros at even indices). kreyszig functional analysis solutions chapter 3
Thus (M^\perp =) sequences with zeros at odd indices. Solution (Outline): Let (d = \inf_y \in M |x - y|). Choose sequence (y_n \in M) s.t. (|x - y_n| \to d). By parallelogram law, show ((y_n)) is Cauchy, so converges to some (m \in M) (since (M) closed). Define (n = x - m). Show (n \perp M). Uniqueness: If (x = m_1 + n_1 = m_2 + n_2), then (m_1 - m_2 = n_2 - n_1 \in M \cap M^\perp = 0). So (m_1=m_2), (n_1=n_2). 6. Problem: Bessel’s inequality. Let (e_k) be orthonormal in inner product space (X). Prove [ \sum_k=1^\infty |\langle x, e_k \rangle|^2 \le |x|^2. ]
If (y = 0), both sides are 0. Assume (y \neq 0). For any scalar (\lambda), [ 0 \le |x - \lambda y|^2 = \langle x - \lambda y, x - \lambda y \rangle = |x|^2 - \lambda \langle y, x \rangle - \overline\lambda \langle x, y \rangle + |\lambda|^2 |y|^2. ] Choose (\lambda = \frac\langle x, y \rangley). Then [ 0 \le |x|^2 - \frac\langle x, y \rangley - \frac\langle x, y \rangle^2 + \fracy ] Wait – compute carefully: Since (e_k) is complete, (x = \sum_k=1^\infty \langle
(M = (x_1, 0, x_3, 0, x_5, \dots) ). For (y = (y_n) \in M^\perp), we need (\langle x, y \rangle = \sum_n=1^\infty x_n \overliney_n = 0) for all (x \in M).
If you need solutions to (e.g., 3.1, 3.2, ..., 3.10) from the book, just provide the problem statement, and I will solve them step by step. Problem (Riesz Representation Theorem): Let (f) be a
Take (x = e_2k-1) (1 at odd index (2k-1), zero elsewhere). Then (\langle e_2k-1, y \rangle = y_2k-1 = 0) for all (k).
So (y_n = 0) for all odd (n). Therefore (M^\perp = (y_n) : y_2k-1=0 \ \forall k ) (sequences nonzero only at even indices).