# 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 $1 \leq p < \infty$.

• Btw to get what you want for $L^p$ you don't need anything. Note that for example $L^p([0,1])\approx L^p([1,2])$; hence $L^p([0,1])\times L^p([0,1])\approx L^p(]0,1])\times L^p(]1,2])\approx L^p([0,2])$. Commented Jan 18, 2021 at 19:28

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 "$$X^{**}=X$$").

• 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
Commented Dec 17, 2015 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
Commented Dec 17, 2015 at 22:27
• @D1X Right.${}{}{}{}{}$ Commented Dec 18, 2015 at 0:00
• Very simple argument! Nice! +1
– user194469
Commented Feb 4, 2017 at 16:32
• @EroSennin Fixed - as I suspected, nothing but definitions. Commented Jan 18, 2021 at 12:58