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Exercise 3.F.35

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Proof. Define $τ \in L ({(P (ℝ))}^{'}, ℝ^{\infty})$ by

τ (σ) = (σ (1), σ (x), σ (x^{2}), \dots)

You can check that $τ$ is indeed linear. To prove $τ$ is injective, suppose there are $σ, η \in {(P (ℝ))}^{'}$ such that $τ (σ) = τ (η)$ . By definition of $τ$ we have that $σ (x^{j}) = η (x^{j})$ . Let $p \in P (ℝ)$ . Clearly, $p$ takes the form

p (x) = a_{0} + a_{1} x + \dots + a_{m} x^{m}

For some $a_{0}, \dots, a_{m} \in 𝔽$ where $\deg p = m$ . Then

\begin{aligned} σ (p) & = σ (a_{0} + a_{1} x + \dots + a_{m} x^{m}) \\ = σ (a_{0}) + σ (a_{1} x) + \dots + σ (a_{m} x^{m}) \\ = a_{0} σ (1) + a_{1} σ (x) + \dots + a_{m} σ (x^{m}) \\ = a_{0} η (1) + a_{1} η (x) + \dots + a_{m} η (x^{m}) \\ = η (a_{0}) + η (a_{1} x) + \dots + η (a_{m} x^{m}) \\ = η (a_{0} + a_{1} x + \dots + a_{m} x^{m}) \\ = η (p) \end{aligned}

Because $p$ was arbitrary, it follows that $σ = η$ and thus $τ$ is injective.

To prove surjectivity, let $a = (a_{0}, a_{1}, a_{2}, \dots) \in ℝ^{\infty}$ . Define $σ \in {(P (ℝ))}^{'}$ by

σ (x^{n}) = a_{n}

You can check that $σ$ is also linear. By definition of $τ$ we have that $τ (σ) = a$ and, because $a$ was arbitrary, it follows that $τ$ is surjective.

Hence $τ$ is an isomorphism between ${(P (ℝ))}^{'}$ and $ℝ^{\infty}$ . □

celiopassos

2017-10-06 00:00

Exercise 3.F.35

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