Let act on with domain

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Linear algebra and vector spaces · Linear algebra and vector spaces

Conditions of Validity, Ch Inner products, orthogonality and the adjoint

Problem

Let Ty=y′Ty = y' act on [0,2][0,2] with domain D(T)={y∈C1[0,2]:y(0)=0}\mathcal{D}(T) = \{y \in C^1[0,2] : y(0) = 0\}, and use the inner product ⟨f,g⟩=∫02f∗g dx\langle f,g\rangle = \int_0^2 f^*g\,\mathrm{d}x. Someone argues: "For f,g∈D(T)f,g \in \mathcal{D}(T), ⟨f,Tg⟩=∫02f∗g′ dx=[f∗g]02−∫02f∗′g dx=f∗(2)g(2)−f∗(0)g(0)−⟨f′,g⟩.\begin{aligned} \langle f, Tg\rangle &= \int_0^2 f^*g'\,\mathrm{d}x = [f^*g]_0^2 - \int_0^2 f^{*\prime}g\,\mathrm{d}x \\ &= f^*(2)g(2) - f^*(0)g(0) - \langle f', g\rangle. \end{aligned} Since g(0)=0g(0) = 0 for every g∈D(T)g \in \mathcal{D}(T), this boundary term vanishes and ⟨f,Tg⟩=−⟨f′,g⟩\langle f, Tg\rangle = -\langle f', g\rangle. Hence T†=−d/dxT^\dagger = -\mathrm{d}/\mathrm{d}x with domain D(T†)=D(T)={f:f(0)=0}\mathcal{D}(T^\dagger) = \mathcal{D}(T) = \{f : f(0) = 0\}, and TT is therefore its own adjoint up to an overall sign." ((a)) Find the error in this working. ((b)) State the correct domain D(T†)\mathcal{D}(T^\dagger) and say whether (T,D(T))(T, \mathcal{D}(T)) is self-adjoint.

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