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Exercise 7.A.21

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Proof. (a) Let $e_{j} = \frac{\cos jx}{\sqrt{π}}$ and $f_{j} = \frac{\sin jx}{\sqrt{π}}$ . By Exercise 4 in section 6B, $\frac{1}{\sqrt{2 π}}, e_{1}, \dots, e_{n}, f_{1}, \dots, f_{n}$ is an orthonormal basis of $V$ . Note that $D e_{j} = - j f_{j}$ and $D f_{j} = j e_{j}$ . Then, for any $v, w \in V$ , we have

\begin{aligned} ⟨ v, D^{∗} w ⟩ & = ⟨ Dv, w ⟩ \\ = ⟨ D (⟨ v, \frac{1}{\sqrt{2 π}} ⟩ \frac{1}{\sqrt{2 π}} + \sum_{j = 1}^{n} (⟨ v, e_{j} ⟩ e_{j} + ⟨ v, f_{j} ⟩ f_{j})), w ⟩ \\ = ⟨ \sum_{j = 1}^{n} (- j ⟨ v, e_{j} ⟩ f_{j} + j ⟨ v, f_{j} ⟩ e_{j}), ⟨ w, \frac{1}{\sqrt{2 π}} ⟩ \frac{1}{\sqrt{2 π}} + \sum_{j = 1}^{n} (⟨ w, e_{j} ⟩ e_{j} + ⟨ w, f_{j} ⟩ f_{j}) ⟩ \\ = \sum_{j = 1} (- j ⟨ v, e_{j} ⟩ ⟨ w, f_{j} ⟩ + j ⟨ v, f_{j} ⟩ ⟨ w, e_{j} ⟩) \\ = ⟨ ⟨ v, \frac{1}{\sqrt{2 π}} ⟩ \frac{1}{\sqrt{2 π}} + \sum_{j = 1}^{n} (⟨ v, e_{j} ⟩ e_{j} + ⟨ v, f_{j} ⟩ f_{j}), \sum_{j = 1}^{n} (- j ⟨ w, f_{j} ⟩ e_{j} + j ⟨ w, e_{j} ⟩ f_{j}) ⟩ \\ = ⟨ v, - Dw ⟩ . \end{aligned}

Hence $D^{∗} = - D$ . Obviously $D$ is normal but not self-adjoint.

(b) Note that $T = D^{2}$ . Thus $T^{∗} = {(DD)}^{∗} = D^{∗} D^{∗} = (- D) (- D) = D^{2} = T$ . □

celiopassos

2017-10-06 00:00

Exercise 7.A.21

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