Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $L^p(I)?$

share|cite|improve this question
up vote 5 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.