The space $ \left( \sum \ell_p^n \right)_2$ is reflexive.

Question

Let $\ell_p^n:= (\mathbb R^n, \|\cdot\|_p)$. I want to show that the space $$ \left(\sum_{n=1}^\infty\ell_p^n \right)_2 := \left(\left\{ (x_n)_{n \in \mathbb N} : x_n \in \ell_p^n \right\},\|\cdot\|_2 \right)$$ is reflexive. Here the $2$-norm of some $(x_n) \in \left(\sum_{n=1}^\infty\ell_p^n \right)_2 $ is $$ \left\|(x_n)\right\|_2:= \left\|\left(\|x_n\|_p\right)_{n \in \mathbb N}\right\|_2. $$

Obviously all the $\ell_p^n$ are reflexive as there are finite-dimensional.

Answer 1

It's a general fact that the dual of $\left(\oplus_{n=1}^\infty X_n\right)_2$ is $\left(\oplus_{n=1}^\infty X_n^*\right)_2$, with the natural pairing: $(x_n^*)$ acts on $(x_n)$ by $\sum_{n=1}^\infty x_n^*(x_n)$. The proof is basically the same as the proof that the dual of $\ell_2$ is $\ell^2$:

1. The pairing is well-defined, using Cauchy-Schwarz.
2. Given a continuous functional $\phi$ on $\left(\oplus_{n=1}^\infty X_n\right)_2$, let $x_n^*$ denote its restriction to $X_n$. Observe that $\sum\|x_n^*\|^2<\infty$.
3. Note that $\phi$ agrees with $(x_n^*)$ on the dense subspaced formed by the sequences where all but finitely many elements are zero.
4. By continuity, $\phi$ agrees with $(x_n^*)$ everywhere.

From the general fact it follows that the $\ell_2$ sum of any family of reflexive spaces (and in particular, of any finite dimensional spaces) is reflexive.

Naturally, in your example you need $p=1$ or $p=\infty$ to obtain a non-superreflexive example.