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).
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. kreyszig functional analysis solutions chapter 3
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 - \frac^2y - \frac\langle x, y \rangle^2 + \frac\langle x, y \rangle^2 ] Wait – compute carefully: So (y_n = 0) for all odd (n)