Take the 2-minute tour ×
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.

This is to get some clarity, I know the two words have nothing to do with each other but they sound so similar that I caught myself saying "locally convex" when I really meant "uniformly convex" and the other way around. Perhaps it would help me if you could tell me whether my understanding of why they are called that is correct and correct me if it's wrong. So here goes:

A uniformly convex space is a space such that for any two points $x,y$ that I pick on the unit ball, if the line between them is $\varepsilon > 0$ long then the midpoint $\frac{x+y}{2}$ is $\delta > 0$ away from the boundary of the ball. (by that I mean that the the absolute value of the midpoint is less than $1$)

I am not sure how this is much different from saying that the unit ball is convex. It makes me think that every point in the space has a convex neighbourhood basis that is, that for every neighbourhood $N$ I can find a ball inside it (by scaling and shifting of the unit ball) that is contained in $N$. Not sure I am using the right words. This understanding must be wrong because if that was the case one should call the space "locally convex". Perhaps this is the cause of my confusion?

On the other hand, as far as I understand, a locally convex space is a vector space together with a collection of seminorms that is endowed with the weakest topology such that all the seminorms are continuous. I think this means that any neighbourhood of any point in the space contains an intersection of open balls $B_{\|\cdot\|_\alpha}$ where $\|\cdot\|_\alpha$ are seminorms. So here it seems more obvious that I should call the space locally convex, since if I pick a neighbourhood I can intersect convex open sets to get a convex open set inside it.

So my question is: which parts of my current understanding are right and which aren't? Thanks for your help.

share|improve this question
1  
Most of this is right. Uniform convexity is a geometric property (it depends on the exact norm of a normed space) while being locally convex is a topological/linear property (preserved under all linear homeomorphisms). Uniform convexity says that every segment between boundary points goes far inside the interior of the unit ball. Look at the $p$-norms on $\mathbb{R}^2$ for $1 \leq p \leq \infty$ and determine which ones are uniformly convex. The unit ball of a normed space always is convex while uniform convexity of a Banach space implies its reflexivity by Milman's theorem. –  t.b. Aug 15 '12 at 11:06
1  
I was mildly objecting to the formulation "nothing to do with each other". Normed spaces are the easiest family of examples of locally convex spaces as you argue in your middle paragraph. And yes, James A. Clarkson introduced the notion of uniformly convex spaces in ... surprise, surprise ... his paper Uniformly convex spaces, Trans. Amer. Math. Soc. 40, no. 3 (1936), pp. 415-420, where he proved some basic things about uniformly convex spaces and, most importantly, his famous inequalities. –  t.b. Aug 15 '12 at 11:29
    
@t.b. I misread you, I'm sorry. In $\mathbb R^2$, I get a uniformly convex space for $p=2$ but not for $p=1$. So I conjecture that if $d$ is the dimension, I get a uniformly convex space for $p \geq d$. I'd rather you didn't remove comments that contain information relevant to the question but up to you. I consider this question more or less answered. –  Matt N. Aug 15 '12 at 12:21
1  
No, your conjecture isn't really true: what about $1 \lt p \lt 2$? The point of Clarkson's inequalities is... The pictures here should help a bit. –  t.b. Aug 15 '12 at 12:23
1  
Yes, your suspicion is correct and holds independently of the dimension (as long as it is at least two, possibly infinite): you get uniform convexity for $1 \lt p \lt \infty$ (the reflexive range) and no uniform convexity for $p = 1,\infty$ (the unit sphere contains straight lines!). –  t.b. Aug 15 '12 at 12:56
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.