I'm having difficulty getting started on the following question:

Prove that as $\mathbb{R}$ - vector spaces $(\mathbb{R}^{< \infty})'\cong \mathbb{R}^{\infty}$.

$\mathbb{R}^{\infty}$ is the vector space of all sequences $(a_i)_{i=1}^\infty$ with each $a_i$ in $\mathbb{R}$, and $\mathbb{R}^{<\infty}$ is a subspace of $\mathbb{R}^{\infty}$ consisting of those $(a_i)_{i=1}^\infty$ for which only finitely many $a_i$ are nonzero.
Also here $(\mathbb{R}^{< \infty})'$ is the dual space of $\mathbb{R}^{< \infty}$.

What I have so far:

Since only finitely many $a_i$'s in $\mathbb{R}^{<\infty}$ are nonzero we can find a finite basis for $\mathbb{R}^{<\infty}$. Then the basis of its dual space is $\delta_{ij}$ (the kronecker delta). So we can find a function $\varphi$ that sends each element of the dual basis to an different element in $\mathbb{R}^\infty$, so we see that $\varphi$ is injective.

From that point on I'm not sure how to proceed for surjectivity. I usually have a hard time with finding isomorphisms, and in this case its worst since I'm having a hard time seeing why these two spaces should be isomorphic at all.

Note that what I have so far is meant to be more of a sketch than a proof. Any hints would be incredibly helpful!

  • 1
    $\begingroup$ Are you certain there is a finite basis for $\mathbb R^{<\infty}?$ This space is clearly isomorphic to the space of real polynomials. Is that finite dimensional? $\endgroup$ – saulspatz Feb 21 '18 at 6:16
  • $\begingroup$ No I guess I was wrong to assert that. But the canonical basis is a basis for this space? $\endgroup$ – one over pie Feb 21 '18 at 6:28

First, the criticism to your arguments. It is not true that $\mathbb R^{<\infty}$ is finite-dimensional. Other than that, your argument basically says nothing, since you don't mention at all who the supposed isomorphism will be ("we can find a function $\varphi$" is really not a proof).

Now for a proof.

What is true is that the canonical basis $\{e_n\}$, where $e_n$ is the sequence with $1$ in the $n^{\rm th}$ coordinate an zero elsewhere, is a basis for it.

Given any linear $f:\mathbb R^{<\infty}\to\mathbb R$, for any $x=\sum_nx_ne_n$ we have (since only finitely many coefficients are nonzero) $$ f(x)=\sum_n x_n f(e_n). $$ So $f$ is determined by the numbers $\{f(e_n)\}$. To get formal, define $\gamma:(\mathbb R^{<\infty})'\to\mathbb R^{\infty}$ by $\gamma(f)=\{f(e_n)\}$. This is clearly linear. It is one-to-one: if $\gamma(f)=\gamma(g)$, then $f(e_n)=g(e_n)$ for all $n$, and so by linearity $f=g$. And it is onto: given $a\in \mathbb R^{\infty}$, let $f_a\in (\mathbb R^{<\infty})'$ be given by $$ f_a(x)=\sum x_n a_n. $$ Then $\gamma(f_a)=\{a_n\}$.

  • $\begingroup$ Thank you this is very helpul! Ya I guess what I had was a bit silly... $\endgroup$ – one over pie Feb 21 '18 at 6:30

I'm guessing that linear functionals on $\mathbb{R}^{<\infty}$ turn out to be the following: a linear functional is a function which multiplies every term in an infinite sequence by some position-dependent constants:

$$f: \langle a_1,a_2,\ldots,\rangle \mapsto\langle c_1a_1,\, c_2a_2,\, c_3a_3,\,\ldots\rangle$$

then adds up the results: $\sum_{i=1}^\infty c_ia_i$. As long as the original sequence $\langle a_i\rangle$ had only finitely many nonzero terms, this sum will always result in a number. You can also prove that this sum is linear with respect to the input sequence; hence this sum is genuinely a linear functional.

If my guess is correct and these are the linear functionals, then the dual of $\mathbb{R}^{<\infty}$ is isomorphic to $\mathbb{R}^\infty$ just because each linear functional on $\mathbb{R}^{<\infty}$ corresponds uniquely to some infinite list of constants $\langle c_i\rangle \in \mathbb{R}^\infty$, and because that correspondence is linear.

  • $\begingroup$ That's a pretty illuminating answer, thank you! $\endgroup$ – one over pie Feb 21 '18 at 6:32

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