The reflexivity of the product $L^p(I)\times L^p(I)$ I am starting to read about the Sobolev spaces $W^{1,p}(I),$ where $I$ is an open interval in $\mathbb{R}.$
In order to establish  the reflexivity of $W^{1,p}(I)$ for $p\in ]1,\infty[,$ I need the reflexivity of $L^p(I)\times L^p(I).$
My question is: how to derive the reflexivity of $L^p(I)\times L^p(I)$ starting from the reflexivity of $L^p(I)?$
 A: There are many norms that you can put on the product $X \times Y$ of two Banach spaces. The most common ones are the $\ell^p$-sum norms resulting in the space $X \mathbin{\oplus_p} Y$. For $1 \leq p \leq \infty$, they are given by
$$
\lVert (x,y) \rVert_p = 
\left(
  \lVert x\rVert^p + \lVert y\rVert^p
\right)^{1/p},
\quad \text{and} \quad
\lVert (x,y) \rVert_\infty = \max\{\lVert x \rVert, \lVert y\rVert\}
$$
From
$$
\lVert (x,y) \rVert_\infty \leq \lVert (x,y) \rVert_p \leq \lVert (x,y) \rVert_1 \leq 2\lVert (x,y) \rVert_\infty
$$
we see that all the $\ell^p$-sum norms are equivalent.
As in the duality between $\ell^p$ and $\ell^q$, using Hölder's inequality, one shows that
$(X \mathbin{\oplus_p} Y)^\ast = X^\ast \mathbin{\oplus_q} Y^\ast$
whenever $\frac1p+\frac1q = 1$. 
Given the identification $(X \mathbin{\oplus_p} Y)^\ast = X^\ast \mathbin{\oplus_q} Y^\ast$, you can verify that the canonical inclusions
$\iota_{X}\colon X \to X^{\ast\ast}$ and 
$\iota_Y\colon Y \to Y^{\ast\ast}$ give a map 
$$
X \mathbin{\oplus_p} Y \to X^{\ast\ast} \mathbin{\oplus_p} Y^{\ast\ast}, (x,y) \mapsto (\iota_X(x),\iota_Y(y))
$$
which coincides with the canonical inclusion 
$$
\iota_{X \mathbin{\oplus_p} Y}\colon X \mathbin{\oplus_p} Y \longrightarrow \left(X \mathbin{\oplus_p} Y\right)^{\ast\ast} = \left(X^\ast \mathbin{\oplus_{q}} Y^\ast\right)^{\ast} = X^{\ast\ast} \mathbin{\oplus_p} Y^{\ast\ast}.$$
From this it follows that $X \mathbin{\oplus_p} Y$ is reflexive  if and only if both $X$ and $Y$ are reflexive.
I'll leave it at that for the moment, but if you need more details, I can add them.
