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 $L^p(I)?$

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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.

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The same method works for any sum of Banach spaces: $\bigoplus_p X_i$ is the space $\left\{(x_i)_{i \in I}\,:\,\sum_{i} \|x_i\|^p \lt \infty\right\}$. Its dual is $\bigoplus_q X_{i}^\ast$, etc. –  t.b. May 24 '12 at 14:31
Dear Theo Buehler, thanks a lot for your nice answer. Retrospectively, I missed the isometric identification $(f_1,f_2)\in X_1^\ast\times_q X_2^\ast\mapsto f\in (X_1\times_p X_2)^\ast$ induced by the pairing $\langle(f_1,f_2),(x_1,x_2)\rangle=\langle f_1,x_1\rangle+\langle f_2,x_2\rangle.$ From that I recknow the coincidence of the canonical injection $\iota_{X_1\times X_2}$ with $(\iota_{x_1},\iota_{x_2})$. Bye. –  Giuseppe Tortorella May 24 '12 at 14:56
Dear Giuseppe, You're welcome and that's exactly the point! By the way: feel free to call me Theo. Best wishes, –  t.b. May 24 '12 at 14:59

Sorry but i don't understand well.

0) What does mean the symbol $\oplus$ ? (i never saw it. (neither $\oplus_p$))

1) First when you write $(X \times Y)'=X' \times Y'$ you mean that you have a linear isometric homeomorphism ? If you change of norms on the product you can loose the isometric property i think, so is the isometric property really necessary ?

2) To prove that an set $E$ is reflexive i believe that exhibing a linear homéomorphism with its bidual isn't enough. It should be the particular "canonical inclusion". So we have to show that $J:(x,y) \in X \times Y \to \psi_{x,y} \in (X \times Y)''$ defined by $\psi_{x,y} (F)=<F,(x,y)>;\forall F \in (X \times Y)'$ is surjective. But i don't understand why it "conincide" with:$(x,y) \in X \times Y \to (J_x,J_y) \in X' \times Y'$. What do you mean by conincide

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