The finite product of $L^p$ spaces is reflexive ($1I am trying to understand the proof that the Sobolev Space $W^{1,p}$ is reflexive given in Functional Analysis, Sobolev Spaces and Partial Differential Equations, by Haim Brezis.
There it is used that the space $L^p(I) \times L^p(I)$ is reflexive, where I is an open interval (possibly not bounded). I understand this comes from the fact that $L^p(I)$  is indeed reflexive for $1<p<\infty$.
I need help showing that result. I haven't really worked with products of $L^p$ spaces and can't seem to find any basic information on it on the Internet.
Note: The same thing is done for proving the separability of $W^{1,p}$ for $1 \leq p < \infty$.
 A: If $X$ and $Y$ are reflexive Banach spaces then $X\times Y$ is reflexive. This follows from the fact that $(X\times Y)^*=X^*\times Y^*$.
Which is very easy. First, if $L_1$ and $L_2$ are bounded linear functionals on $X$ and $Y$ respectively and we define $$L(x,y)=L_1x+L_2y$$then it's clear that $L$ is a bounded linear functional on $X\times Y$. Conversely, if $L$ is a bounded linear functional on $X\times Y$, define $L_1:X\to\Bbb C$ and $L_2:Y\to\Bbb C$ by $$L_1x=L(x,0)$$and$$L_2y=L(0,y),$$ and it's clear that $L_1$ and $L_2$ are bounded, and that $$L(x,y)=L(x,0)+L(0,y)=L_1x+L_2y.$$
Edit: So $(X\times Y)^*=X^*\times Y^*$, in the sense that an element $(x^*,y^*)\in X^*\times Y^*$ defines a functional on $X\times Y$ via the pairing $\newcommand\ip[2]{\langle #1,#2\rangle}$ $$\ip{(x^*,y^*)}{(x,y)}=\ip{x^*}x+\ip{y^*}y.$$
So if $X$ and $Y$ are reflexive then $(X\times Y)^{**}=X^{**}\times Y^{**}=X\times Y$, so $X\times Y$ is reflexive.
It's been objected that that's not complete, since saying $X^{**}$ is isomorphic to $X$ does not imply $X$ is reflexive; for that we need that the canonical injection from $X$ into $X^{**}$ is surjective. It seems clear to me that since all of the mappings constructed above are "natural", the final one we constructed from $X$ to $X^{**}$ must also be "natural", so this really can't be a problem.
Luckily we don't have to try to prove that, since it's easy to verify directly that $X\times Y$ is reflexive. Indeed, suppose $L=(x^{**},y^{**})\in X^{**}\times Y^{**}$: $$L(x^*,y^*)=\ip{x^{**}}{x^*}+\ip{y^{**}}{y^*}.$$
Since $X$ and $Y$ are reflexive there exist $\alpha\in X$ and $\beta\in Y$ with $$\ip{x^{**}}{x^*}=\ip{x^*}\alpha,
\ip{y^{**}}{y^*}=\ip{y^*}\beta.$$Hence $$L(x^*,y^*)=\ip{x^*}\alpha+\ip{y^*}\beta=\ip{(x^*,y^*)}{(\alpha,\beta)}.$$
Which says "$L=(\alpha,\beta)$" (with the equals sign in the same spirit as in "$X^{**}=X$").
