Reflexive space which is not uniformly convex

I found this beautiful theorem due to Milman and Pettis:

Every uniformly convex Banach space is reflexive.

I think it's a remarkable statement, since uniformly convexity is a geometric property of the norm and therefore need not to be true for an equivalent norm. While reflexivity is a topological statement and therefore a reflexive space remains reflexive for an equivalent norm.

My first question is:

Could someone give an example of a space with two equivalent norms such that the space is uniformly convex for just one of the two norms.

And my second question is:

Could someone give me an example of a reflexive space that admits no uniformly convex equivalent norm.

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t.b. answered both your questions. Much more interesting (and difficult) is: give an example of a reflexive space which is uniformly convex for no equivalent norm. – GEdgar Oct 9 '11 at 16:45
Oh boy! I see it's Sunday eve! Actually this should be my second question! Sorry for the mistake! I edited my question! – math Oct 9 '11 at 16:50
Okay, I updated my answer – t.b. Oct 9 '11 at 16:56

The answer to the first question is probably disappointingly simple: Take $\mathbb{R}^2$, once with the Euclidean norm $\|(x,y)\|_2 = (|x|^2+|y|^2)^{1/2}$ and once with the $\ell^1$-norm $\|(x,y)\|_1 = |x| + |y|$ for example. The norms are equivalent, the Euclidean norm is uniformly convex, while the $\ell^1$-norm isn't. Of course, $\mathbb{R}^2$ is reflexive with respect to both norms.

The answer your updated second question is a classical result of M.M. Day:

Theorem. There exist Banach spaces which are separable, reflexive, and strictly convex, but are not isomorphic to any uniformly convex space.

This is the main result of his paper bearing the statement of the theorem in its title:

M.M. Day, Reflexive Banach spaces not isomorphic to uniformly convex spaces, Bull. Amer. Math. Soc. Volume 47, Number 4 (1941), 313-317, MR0003446.

Since the paper is freely available via the above link, there is little sense in my elaborating on it.

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Thx again! I will try to work through the paper tomorrow and I'm sorry for having posted the wrong question first. – math Oct 9 '11 at 17:03
eponymous = "named after a particular person" ... Is this called "Day's space"? – GEdgar Oct 9 '11 at 17:08
@GEdgar: No, not as far as I know. It's my (virtually non-existent) Greek and English failing on me at the same time... I corrected it. Thanks for pointing it out. – t.b. Oct 9 '11 at 17:14

Note that every separable space can be given an equivalent strictly convex norm. Indeed, if $X$ is separable then you can easily construct an injective operator $T\colon X\to \ell_2$. Then define a new norm on $X$ by $\|x\|^\prime = \|x\|+\|Tx\|_{\ell_2}$, which is strictly convex. (I guess this trick is attributed to Victor Klee.)

Now, to answer your second question we need a reflexive space which is separable but not super-reflexive because then we can endow it with a strictly convex norm, yet it cannot be uniformly convex. (Actually, every reflexive space can be given an equivalent strictly convex norm but this is more involved.) For this take

$$X = \big( \bigoplus_{n=1}^\infty \ell_1^n \big)_{\ell_2},$$

where $\ell_1^n$ stands for $\mathbb{R}^n$ endowed with the $\ell_1$-norm. By Enflo's thoerem, it cannot be given an equivalent uniformly convex norm because it is not super-reflexive. Of course, Enflo's theorem characterising super-reflexive spaces as precisely those which can be renormed in a uniformly convex manner is very deep.

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every finite dimensional Banach space is a reflexive bot it need not be uniformly convex .$X=\mathbb{R}^n , n\geq2$with the norm \$\Vat x\Vat_{1}=\sigma {i=1}^\n\vert x{i}\vert

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