(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 (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 \rangle^2). Then [ 0 \le |x|^2 - \fracy - \frac\langle x, y \rangley + \frac ] Wait – compute carefully:
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).
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.
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. ]
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).
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).