Köthe–Toeplitz dual

Remarks

Both duals are sequence spaces, and $X^\alpha \subseteq X^\beta$ because absolutely convergent series converge. If $X \subseteq Y$, then $Y^\alpha \subseteq X^\alpha$ and $Y^\beta \subseteq X^\beta$.

Example: $(\ell^1)^\alpha = (\ell^1)^\beta = \ell^\infty$, with $\ell^1$ and $\ell^\infty$ as in the spaces ℓ^p. If $a$ is bounded, then $\sum \abs{a_n x_n} \le \sup_n \abs{a_n} \sum \abs{x_n}$ for every $x \in \ell^1$. If $a$ is unbounded, choose indices $n_1 < n_2 < \cdots$ with $\abs{a_{n_k}} \ge k^2$, and let $x_{n_k} = \overline{a_{n_k}} / \abs{a_{n_k}} \cdot k^{-2}$ and $x_n = 0$ otherwise. Then $x \in \ell^1$, but the terms $a_{n_k} x_{n_k} = \abs{a_{n_k}} k^{-2} \ge 1$ do not tend to $0$, so $a \notin (\ell^1)^\beta$.

Non-example: $a = (n)_{n \ge 1}$ is not in $(\ell^1)^\beta$; for $x_n = 1/n^2$ the series $\sum a_n x_n = \sum 1/n$ diverges.

Other standard values: $\omega^\beta$ is the space $\varphi$ of sequences with finitely many nonzero terms, $\varphi^\beta = \omega$, and $(c_0)^\beta = (\ell^\infty)^\beta = \ell^1$.