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As the title says: how can we prove that $L^p(\mathbb R^n)$ is not uniformly convex for $p=1$ and $p=\infty$.

Does anyone knows a counter-example for the cases $ p=1$ and $ p = \infty$ for the space $L^p( \mathbb R^n)$ ?

I am studying N.L. Carothers book "A short course to Banach space theory".

Any help?

passenger
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1 Answers1

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For $p=\infty$ and $n=2$, consider $\mathbf{x}=(1,1)$ and $\mathbf{y}=(0,1)$.

You might also make use of the fact that uniformly convex Banach spaces are reflexive.

Edit: As I was typing up a more detailed answer, I found the first comment to this answer, which puts things more succinctly than I was going to do (and saves me a bit of typing).

Community
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JohnD
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  • Thank you for your reply, but I think there is some misunderesttod about the notation. $L^p$ is the space of lebesgue measurable functions raied in $p-\text{th}$ power, not the space of seequences. – passenger Nov 26 '13 at 18:36
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    Even so, the $L^p$ spaces you mention have two-dimensional subspaces isometric to this $2$-dimensional $l^p$ space. And if the subspace is not uniformly convex, certainly the whole space is not either. – GEdgar Nov 26 '13 at 19:04