# The finite product of $L^p$ spaces is reflexive ($1<p<\infty$)

I 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 $1 \leq p < \infty$.

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 $$L(x,y)=L(x,0)+L(0,y)=L_1x+L_2y.$$
• Okay, so your proof will give us the biyective isometry: $\begin{matrix} I: \hspace{0.6cm} X^*\times Y^* \longrightarrow (X\times Y)^*\\ (L_1,L_2) \longrightarrow L \end{matrix}$ Right? – D1X Dec 17 '15 at 22:05
• Then we have $( L^{p} \times L^{p})^{**} = L^{p ** } \times L^{p **}=L^p \times L^p$ for $1 < p < \infty$. – D1X Dec 17 '15 at 22:27
• @D1X Right.${}{}{}{}{}$ – David C. Ullrich Dec 18 '15 at 0:00