Dual spaces of $\ell^2$ of an arbitrary normed space Let $(V,\|\cdot\|_V$ be a normed space, and consider $\ell^2(V)$ to be the standard $l^2$ space using sequences in $V$ (instead of $\mathbb{R})$. We can consider the dual space pair $(V,V')$. We know that $(\ell^p,\ell^q)$ are dual space pairs, when $(p,q)$ are conjugate pairs. What can we say about $\ell^2(V')$, or $(\ell^2(V))'$? Using the standard results of regular dual spaces, we know that these should be isometrically isomorphic. I would like to know why they are not. 
 A: Define $T:l^2(V') \to l^2(V)'$ by
$$
(Tv)(x) =\sum_k v_k(x_k).
$$
This $T$ is linear, bounded, injective. To argue surjectivity, let $f\in l^2(V)'$ be given. Define the functionals $v_k:V\to \mathbb R$ by $v_k(x):= f( e_kx)$, where $e_k$ is the unit vector in $l^2(\mathbb R)$. Clearly, $v_k\in V'$.
For $\epsilon>0$ there is $x_{k,\epsilon}\in V$ with $\|x_{k,\epsilon}\|_V=1$ 
such that $v_k( x_{k,\epsilon} )\ge \|v_k\|_{V'}  - 2^{-k} \epsilon$.
For $n\in \mathbb N$ define $x = \sum_{k=1}^n e_kx_{k,\epsilon}$.
Then
$$
v(x) = \sum_{k=1}^n \|v_k\|_{V'}v_k(x_{k,\epsilon}) \ge \sum_{k=1}^n\|v_k\|_{V'}^2 - 2^{-k}\epsilon \|v_k\|_{V'} \ge 
\sum_{k=1}^n\left( \|v_k\|_{V'}^2 -\epsilon\|v_k\|_{V'}^2 - 2^{-2k-2}\epsilon\right)\\
\ge - \epsilon +(1-\epsilon) \sum_{k=1}^n\|v_k\|_{V'}^2.
$$
By boundedness of $v$ and the construction of $x$, we get
$$
v(x) \le \|v\|_{L^2(V)'} \|x\|_{l^2(V)} 
$$
and
$$
\|x\|_{l^2(V)} ^2 = \sum_{k=1}^n  \|v_k\|_{V'}^2.
$$
Putting everything together, we obtain
$$
(1-\epsilon)\sum_{k=1}^n\|v_k\|_{V'}^2 \le \epsilon + v(x) \le\epsilon + \|v\|_{L^2(V)'}\left(\sum_{k=1}^n  \|v_k\|_{V'}^2\right)^{1/2}
\le \epsilon +\frac12 \|v\|_{L^2(V)'}^2 +\frac12\left(\sum_{k=1}^n  \|v_k\|_{V'}^2\right)^{1/2},
$$
which implies 
$$
\frac12(1-2\epsilon)\sum_{k=1}^n\|v_k\|_{V'}^2 \le \epsilon + \frac12 
\|v\|_{L^2(V)'}^2.
$$
This holds for all $\epsilon>0$ and all $n$, hence, $(v_k)\in l^2(V')$, and $\|(v_k)_k\|_{l^2(V')}\le \|v\|_{L^2(V)'}
$.
The same proof should work with some modifications for the case $p\ne2$, $p\in(1,+\infty)$.
