4
$\begingroup$

Consider $\displaystyle(\ell_2, \lVert\cdot\rVert_\star), \lVert x\rVert_\star = \sum\limits_{k=1}^{\infty}\frac{|x(k)|}{k}$. What is its dual space? Is this space reflexive?

My idea is to consider $\displaystyle\varphi: (\ell_2, \lVert\cdot\rVert_\star) \rightarrow (\ell_1, \lVert\cdot\rVert_1), \varphi(\{x(k)\}_{k=1}^\infty) = \left\{ \frac{x(k)}{k} \right\}_{k=1}^\infty$. Then $\lVert\varphi(x)\rVert_1 = \lVert x\rVert_\star$ and $\varphi$ is isometric isomorphism. But $\text{Im}\varphi \subsetneq \ell_1$ becasue, for example, $\displaystyle\left\{\frac{1}{k^{3/2}}\right\} \in \ell_1, \left\{\frac{1}{k^{1/2}}\right\} \notin \ell_2$. So how to find dual space in this case?

$\endgroup$
4
  • 1
    $\begingroup$ Your $\varphi$ is isometric, but not an isomorphism (as you realized yourself). Note that $\mathrm{Im}\varphi$ is dense in $\ell^1$. Now you can use (or prove first) that the closure has isomorphic dual spaces to conclude. $\endgroup$ Commented Mar 31 at 5:23
  • $\begingroup$ Let $X=\mathrm{Im}(\varphi)$ and consider the map $$F:(\ell^1)^*\rightarrow X^*$$ where $f\in (\ell^1)^*$ gets mapped to its restriction on $X$, i.e. $$F(f): X\rightarrow \mathbb{C}, F(f)(x)=f(x).$$ We can prove that $F$ is an isomorphism of normed spaces. $\endgroup$ Commented Mar 31 at 5:29
  • $\begingroup$ The dual space consist of sequences $(a(k))$ with $\sum (|a(k))|/k)^{2}<\infty$. The norm is the square root of $\sum (|a(k))|/k)^{2}$. $\endgroup$ Commented Mar 31 at 5:34
  • $\begingroup$ In general for a positive sequence $\lambda_n$ let $\|x\|_*=\sum \lambda_k|x_k|,$ where $x$ has finitely many nonzero terms. Then the completion of the space with respect to this norm is isometrically isomorphic to $\ell^1.$ $\endgroup$ Commented Mar 31 at 9:08

1 Answer 1

4
$\begingroup$

As you have noticed, we can identify our space with $X:=\mathrm{Im}(\varphi)$ which is a dense subspace of $\ell^1$ (as $X$ contains the subspace of sequences with finitely many nonzero entries). The identification part gives us that $(\ell^2,\Vert \cdot \Vert_*)^*$ is isomorphic as normed space to $(X, \Vert \cdot \Vert_1)^*$ via the map $$G: (\ell^2,\Vert \cdot \Vert_*)^*\rightarrow (X, \Vert \cdot \Vert_1)^*, f \mapsto f\circ \varphi^{-1}.$$ Using the density we get that $$F: (\ell^1, \Vert \cdot \Vert_1)^*\rightarrow (X, \Vert \cdot \Vert_1)^*$$ yields an isomorphism of normed spaces, where $F$ is the restriction to $X$, i.e. for $f\in (\ell^1)^*$ $$F(f): X \rightarrow \mathbb{C}, F(f)(x)=f(x).$$ The linear map $F$ is clearly bounded. It is injective (if a continuous function vanishes on a dense subset, then it vanishes everywhere, i.e. $F$ has trivial kernel). Furthermore, $F$ is surjective too as every continuous linear $g: X\rightarrow \mathbb{C}$ can be extended to $\tilde{g}:\ell^1\rightarrow \mathbb{C}$ with $\tilde{g}=g$ on $X$ (this is the fact that a lipschitz function can be extended to a lipschitz function on the closure of the domain). Finally, by the inverse mapping theorem $F$ is an isomorphism of normed spaces (using the fact that all dual spaces of normed spaces are Banach spaces).

Thus, the dual space in question is isomorphic to $(\ell^1,\Vert \cdot \Vert_1)^*$. In particular the space in question is not reflexiv.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .