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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 $1 \leq p < \infty$.

Dean Miller
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D1X
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  • 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 $L^p([0,1])\times L^p([0,1])\approx L^p(]0,1])\times L^p(]1,2])\approx L^p([0,2])$. – David C. Ullrich Jan 18 '21 at 19:28

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

David C. Ullrich
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  • 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 00:00
  • Very simple argument! Nice! +1 –  Feb 04 '17 at 16:32
  • Sorry but why having a bijective isometry from $X \times Y$ to the double dual implies reflectivity? The map must be James embedding, not any isometry. There exist spaces that are isometrically isomorphic to their double dual but not reflexive. – Son Gohan Jan 16 '21 at 00:17
  • @EroSennin I didn't say anything (explicitly) about a bijective isometry to the double dual. – David C. Ullrich Jan 16 '21 at 11:15
  • Look at the comment above of the OP. I cannot see why from this bijective isometry we conclude the reflexivity of the cartesian product. – Son Gohan Jan 16 '21 at 11:23
  • We need some Eberlein-Smulian type of argument, like as bounded sequences admit convergent subsequences in the weak topology. Otherwise, I do not see the argument. – Son Gohan Jan 17 '21 at 00:40
  • Perhaps. It still seems to me that it must be just as trivial as above: all the isomorphisms implicit in the equals signs are "natural", so the final bijection must be the natural one. Haven't quite written that down... – David C. Ullrich Jan 17 '21 at 12:36
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    @EroSennin Fixed - as I suspected, nothing but definitions. – David C. Ullrich Jan 18 '21 at 12:58
  • Perfect, now I definitely see why the isomorphisms were all natural. Thanks for the clarification! – Son Gohan Jan 18 '21 at 15:04
  • @David C. Ullrich Hi,to conclude the isometric isomorphism,we use $\ell_1$ norm on each summand correct?That is $|(x,y)|_1 = |x|_X +|y|_Y$ correct? – yi li May 18 '21 at 09:02
  • @yili In fact if $||\cdot||*$ is any norm on $\Bbb R^2$ we can say $||(x,y)||{X\times Y}=||(||x||X,||y||_Y)*||$, because any two norms on $\Bbb R^2$ are equivalent. – David C. Ullrich May 18 '21 at 11:49
  • @David C. Ullrich ,thanks I have notice this , isometric is stronger than equivalent norm, but this does not influce the properties of the space on both side,since metric are equivalent. – yi li May 18 '21 at 11:55
  • See this answer for a more detailed explanation that $X$ and $Y$ reflexive implies $X \times Y$ reflexive. – Dean Miller Aug 05 '25 at 06:12