5
$\begingroup$

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 $L^p(I)?$

1 Answers 1

5

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.

  • 0
    Dear Giuseppe, You're welcome and that's exactly the point! By the way: feel free to call me Theo. Best wishes,2012-05-24