Is the countable direct sum of reflexive spaces reflexive? Let $(X_n)$ be a sequence of reflexive spaces. We define the $\ell_2$-direct sum $\bigoplus_n X_n$ as the normed space with elements $(x_n)\in \prod_n X_n$ such that 
$$
\|(x_n)\|=\left(\sum_{n}\|x_n\|^2\right)^{\frac{1}{2}}<\infty.
$$
Is $\bigoplus_n X_n$ reflexive? What if we define analogously a $\ell_p$-direct sum for a different $p$? 
Thank you. 
 A: Let $1<p<\infty$ with $\frac1p+\frac1q=1$, and let normed vector spaces $X_1, X_2, \dots$ be given. Define the $\ell_p$ direct sum  $\bigoplus_i^{\ell_p} X_i$. We claim there is a isometric isomorphism $\psi: \left(\bigoplus_i^{\ell_p} X_i\right)' \to \bigoplus_i^{\ell_q} X_i'.$ From this it follows immediately that if the $X_i$ are reflexive then so is $\bigoplus_i^{l^p} X_i$.
Observe that the inclusions $X_i \to \bigoplus_j^{\ell_p} X_j$ are isometric, so we may identify each $X_i$ with its image in $\bigoplus_j^{\ell_p} X_j$. Given a bounded linear functional $f$ on $\bigoplus_i^{\ell_p} X_i$, $f$ restricts to bounded linear functionals $f_i$ on $X_i$. Define $\psi$ by $f \mapsto \sum_{i=1}^\infty f_i$. But we must show that $\psi$ actually maps into $\bigoplus_i^{\ell_q} X_i'$, i.e. that $\sum_i \|f_i\|^q<\infty$. Fix a positive constant $C<1$. Choose $x_i\in X_i$ such that $\|x_i\|=1$ and $f_i(x_i) \geq C\|f_i\|$. For each $n\geq 1$, define an element $y_n \in \bigoplus_j^{\ell_p} X_j$ by $$y_n=\sum_{i=1}^{n} \frac{\|f_i\|^{q/p}}{\left(\sum_{j=1}^n \|f_j\|^q\right)^{1/p}}x_i,$$
so that 
$$\|y_n\|^p = \sum_i \left\|\frac{\|f_i\|^{q/p}}{\left(\sum_{j=1}^n \|f_j\|^q\right)^{1/p}}x_i\right\|^p
=\sum_i \frac{\|f_i\|^q}{\sum_{j=1}^n \|f_j\|^q}\|x_i\|^p = 1
$$
And,
\begin{align*}
\|f\| \geq f(y_n) &= \sum_{i=1}^n \frac{\|f_i\|^{q/p}}{\left(\sum_{j=1}^n \|f_j\|^q\right)^{1/p}}f_i(x_i)\\
&\geq \sum_{i=1}^n \frac{\|f_i\|^{q/p}}{\left(\sum_{j=1}^n \|f_j\|^q\right)^{1/p}}C\|f_i\|\\
&= C \sum_{i=1}^n \frac{\|f_i\|^{q/p+1}}{\left(\sum_{j=1}^n \|f_j\|^q\right)^{1/p}}\\
&= C \sum_{i=1}^n \frac{\|f_i\|^{q}}{\left(\sum_{j=1}^n \|f_j\|^q\right)^{1/p}}\\
&= C \left(\sum_{j=1}^n \|f_j\|^q\right)^{1/q}
%\to C \left(\sum_{j=1}^\infty \|f_j\|^q\right)^{1/q}
\end{align*}
Taking the limit as $n\to\infty$ shows that $(\sum_{j=1}^\infty \|f_j\|^q)^{1/q}\leq C^{-1}\|f\| < \infty$, as required. Taking the limit as $C\to 1$ shows that $\|\psi(f)\|=(\sum_{j=1}^\infty \|f_j\|^q)^{1/q}\leq \|f\|$, so that $\|\psi\| \leq 1$.
Now define a map in the opposite direction, $\phi : \bigoplus_i^{\ell_q} X_i' \to \left(\bigoplus_i^{\ell_p} X_i\right)'$, by $\phi(\sum_{i=1}^\infty f_i)(\sum_{j=1}^\infty x_j)_= \sum_{i=1}^\infty f_i(x_i)$, for $f_i\in X_i'$ and $x_i\in X_i$. We need to show that the linear functionals $\phi(\sum_{i=1}^\infty f_i)$ are bounded. This follows from Holder's inequality :
\begin{align*}
\left|\phi\left(\sum_{i=1}^\infty f_i\right)\left(\sum_{j=1}^\infty x_j\right)\right|
&= \left|\sum_{i=1}^{\infty} f_i(x_i)\right| \\
&\leq \sum_{i=1}^\infty \|f_i\|\|x_i\| \\
&\leq \left(\sum_{i=1}^{\infty}\|f_i\|^q \right)^{1/q}
\left(\sum_{i=1}^{\infty}\|x_i\|^p \right)^{1/p} \\
&= \left\|\sum_{i=1}^\infty f_i\right\|\left\|\sum_{j=1}^\infty x_j\right\|
\end{align*}
which shows that $\phi\left(\sum_{i=1}^\infty f_i\right)$ is bounded, with $\|\phi\| \leq 1$. Now, it is straightforward to check that $\phi$ and $\psi$ are inverses of one another. Then since $\|\phi\|\leq 1$ and $\|\psi\|\leq 1$ we have $1 = \|\phi \circ \psi\| \leq \|\phi\|\|\psi\| \leq \|\phi\| \leq 1$, so that $\|\phi\|=1$, and by a similar argument $\|\psi\|=1$. Thus $\phi$ and $\psi$ are isometric isomorphisms.
