Are the Hahn sequence spaces really just isometrically isomorphic to l1?

Question

I believe I have proved this, but something isn't sitting right with me about the result. There's a whole body of literature out there about the Hahn sequence spaces. If they really are just isometrically isomorphic to $\ell_1$, that would seem to make the literature on it superfluous. So, I am wondering if my "proof" contains an error.

Definition. Let $\textbf{w}=(w_n)_{n=1}^\infty$ be a sequence of positive real numbers. For a scalar (real or complex) sequence $(a_n)_{n=1}^\infty$, we set $$ \|(a_n)_{n=1}^\infty\|_h:=\sum_{n=1}^\infty w_n|a_n-a_{n+1}| $$ A Hahn sequence space for $\textbf{w}$, denoted $h_{\textbf{w}}$, is defined as the completion of the normed space of all such sequences satisfying $\|(a_n)_{n=1}^\infty\|_h<\infty$ and $\lim_{n\to\infty}a_n=0$. (Usually, $h_\textbf{w}$ is defined as just the underlying normed space, which supposedly can be proved to be complete already. But that is beyond our needs, so we have simply added completeness to the usual definition.)

Theorem (maybe). The spaces $h_\textbf{w}$ and $\ell_1$ are isometrically isomorphic (although not in the natural way).

Proof (maybe). Let $(e_n)_{n=1}^\infty$ be the canonical basis for $\ell_1$. Setting $w_0=0$ for convenience, we can now define the vectors $h_1=e_1$, and $$ h_n=\left(\frac{w_n}{w_n+w_{n-1}}\right)e_n-\left(\frac{w_{n-1}}{w_n+w_{n-1}}\right)e_{n-1},\;\;\;n\geq 2. $$ It is clear that $[h_n]_{n=1}^\infty=\ell_1$.

For $(a_n)_{n=1}^\infty\in c_0$ with $\|(a_n)\|_h<\infty$, let us set $$ U(a_n)_{n=1}^\infty =\sum_{n=1}^\infty w_n(a_n-a_{n+1})e_n, $$ so that $U$ is a well-defined linear isometry. We claim that its image contains $\text{span}(h_n)_{n=1}^\infty$, in which case it can be extended to an isometric isomorphism $U:h_\textbf{w}\to\ell_1$. Indeed, any $x\in\text{span}(h_n)_{n=1}^\infty$ can be written in the form $$ x =\sum_{n=1}^\infty(w_n+w_{n-1})a_nh_n =\sum_{n=1}^\infty w_n(a_n-a_{n+1})e_n $$ so that $\|(a_n)_{n=1}^\infty\|_h=\|x\|<\infty$. Since also $a_n\to 0$, the proof is complete. $\;\;\;\square$

So, what do you guys think? Is it valid? It looks valid to me, but it's important that I be very, very sure. So, I would very much appreciate a second opinion.

Thanks!

Answer 1

Your proof is flawed. The structure of your argument is as follows:

1. Suppose $a\in h_w$, i.e. $a\in c_0$ with $\|a\|_{h_w}<\infty$.
2. Any $x\in\mathrm{span}_n{(h_n)}$ can be written in the form $$x=\sum_n{(w_n+w_{n-1})a_nh_n}\tag{*}$$ for some $a$.
3. Find an isometric isomorphism $U$ sending the $a$ from (*) to the corresponding $x$.
4. Profit!

The problem is part (2) already determines $a$ in its entirety; an $a$ satisfying (*) need not satisfy the conditions you impose on $a$ in part (1).

For example, take $w_n=\frac{1}{n^3}$, and consider $$x=\sum_{n=1}^{\infty}{\frac{h_n}{n^2}}$$ (the sum converges because $\|h_n\|_1\leq2$ and $\sum_n{n^{-2}}<\infty$). From (*), we must have $$a_n=\frac{1}{n^2(n^{-3}+(n-1)^{-3})}=2n(1+o(1))\not\to0$$