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.

I have a question which is quite "candid" and for which I can hardly formalize properly every concepts but nonetheless, I was wondering if there were some topoligical and/or algebraic structure properties over spaces (to be defined) that make the set of continuous functions and the set of differentiable functions the same.

Maybe this is simply impossible or only for trivial examples.

Best regards

PS : Motivation is simple curiosity

share|improve this question
add comment

2 Answers 2

up vote 2 down vote accepted

Lie groups almost provide another example.

A continuous homomorphism between Lie groups is automatically smooth. Of course, one needs to assume it's a homomorphism which is very strong condition (hence, the "almost" above).

The idea of the proof is in two big steps. First one proves that any closed subgroup of a Lie group is automatically smooth (known as Cartan's theorem). The idea of this proof is that one can identify what the tangent space to the identity should be using the group exponential map and show this tangent space is actually closed under addition using closedness.

The second step is a bit easier: given $f:G\mapsto H$, consider the map $g:G\mapsto G\times H$ given by $g(x) = (x,f(x))$. Using hypothesis on $f$, one proves that the image of $g$ is a closed subgroup, hence smooth by step 1. This implies the two projection maps, when restricted to the image of $g$, are smooth. With a bit more work, one shows $f$ can be written as a composition of projection maps and their inverses, so is smooth.

share|improve this answer
    
De Vito : Hi thank's for this usefull example which makes apparent that the property of homomorphism is "alike" linerarity reducing the st of admissible function to this set entails the result I beleive. The question is how to put sufficient structure for the set of all functions being automatically "linear" at some level. Best Regards. –  TheBridge Oct 19 '11 at 12:09
    
If "a continuous homomorphism between Lie groups is automatically smooth" that's because "homomorphism" is quite a strong property. –  Michael Hardy Oct 20 '11 at 19:58
    
@Michael: Agreed, hence my "it's a homomorphism which is very strong condition" (wow, my English was terrible!). Related, I've heard, but failed to dig up a reference, that a homomorphism between compact Lie groups is automatically continuous (hence smooth). Compactness is certainly necessary since $\mathbb{R}^2$ and $\mathbb{R}$ are isomorphic as groups. –  Jason DeVito Oct 20 '11 at 20:50
1  
Apparently, it was proven by van der Waerden at least when the domain has finite center. That is, if $f:G\rightarrow H$ is a homomorphism where $G$ is a compact Lie group with finite center and $H$ is a compact Lie group, then $f$ is automatically continuous (hence smooth). I'm not sure about general compact Lie groups. The original article is in German and costs money, so I won't be reading it soon. –  Jason DeVito Oct 20 '11 at 21:06
add comment

I seem to recall that this works for p-adic-valued functions of a p-adic variable: they're differentiable if they're continuous.

share|improve this answer
    
@Micheal Hardy : and do you happen to know what make them so ? –  TheBridge Oct 18 '11 at 21:57
    
@TheBridge: Without knowing the exact details, I'd guess it is the same reason why every point in an open ball is its center (i.e. $p$-adic metrics are ultrametrics), and you could probably rewrite the limits for the derivative as a sum $\sum a_n$, and the continuity is enough to show $a_n\to 0$, which is enough to conclude it converges, so the derivative would exist. But then again, this is just a gut instinct. –  Asaf Karagila Oct 18 '11 at 22:50
add comment

Your Answer

 
discard

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.