Bourbaki

Exercise 2, § 1

Functions of a Real Variable, VI, 291

Let E be the vector space over C generated by the functions xn (n∈N)x^n \ (n \in \mathbf{N}), eλx (λ∈C,λ≠0)e^{\lambda x} \ (\lambda \in \mathbf{C}, \lambda \neq 0) and ∣x+μ∣ (μ∈R)|x+\mu| \ (\mu \in \mathbf{R}).

a) Show that the preceding functions form a basis for E.

b) Let U be the composition operator defined on E by the conditions: U0ξ(ξn)=(n!)2U_0^\xi(\xi^n) = (n!)^2, U(eλx)=eλxU(e^{\lambda x}) = e^{\lambda x} for λ≠0\lambda \neq 0, U(∣x+μ∣)=∣x+μ∣U(|x+\mu|) = |x+\mu| for μ∈R\mu \in \mathbf{R}. The indicatrix u(λ)u(\lambda) is the constant 1 but the series with general term n!λnn! \lambda^n does not converge for any value λ≠0\lambda \neq 0.

c) Let V be the composition operator defined on E by the conditions V(xn)=xnV(x^n) = x^n, V(∣x+μ∣)=∣x+μ∣V(|x+\mu|) = |x+\mu|, V(eλx)=0V(e^{\lambda x}) = 0 for λ≠0\lambda \neq 0; the indicatrix v(λ)v(\lambda) is equal to 1 for λ=0\lambda = 0, to 0 for λ≠0\lambda \neq 0, so is different from the sum of the series ∑n=0∞αnλn\sum_{n=0}^\infty \alpha_n \lambda^n where αn=V0ξ(ξn)/n!\alpha_n = V_0^\xi(\xi^n)/n!.

d) Let W be the composition operator defined on E by

W(xn)=xn,W(eλx)=eλx,W(∣x+μ∣)=eλ+μ; W(x^n) = x^n, \qquad W(e^{\lambda x}) = e^{\lambda x}, \qquad W(|x+\mu|) = e^{\lambda+\mu};

show that VW≠WVVW \neq WV.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.