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.

Consider a Holder function $f \in C^\alpha(\mathbb{T}^2, \mathbb{R})$, $\alpha \in (0,1)$. I would like to approximate $f$ with $f_\epsilon \in C^k(\mathbb{T}^2, \mathbb{R})$, $k \in \mathbb{N}$, in the $L^1(\mathbb{T}^2)$ norm.

I was wondering how one does such approximation? In particular, how one get estimates on $\|f-f_\epsilon\|_{L^1}$ and $\|f_\epsilon\|_{C^k}$ in terms of $\epsilon$?

Thank you in advance.

share|improve this question
1  
The usual trick would be to use a mollifier (compactly-supported $C^{\infty}$ "bump" function) and a convolution-type integral. I can't see why that couldn't be adapted here, but maybe I'm missing something or am misinterpreting your notation. en.wikipedia.org/wiki/Mollifier –  T.A.E. Jun 8 at 16:00
    
Could you show me the calculation? –  Cantor Jun 9 at 21:42
    
Have you been exposed to compactly-supported $C^{\infty}$ functions used to smooth other functions through convolution? –  T.A.E. Jun 9 at 23:51
    
Yes. Here I want a $C^k$, $k<\infty$ approximation, also I want the approximation in $L^1$. Given these milder restrictions (as opposed to $C^\infty$ and approximation in $L^\infty$) I wonder what is the best bound one can get for $\|f_\epsilon\|_{C^k}$. –  Cantor Jun 10 at 14:17
    
"also I want the approximation in $L^{1}$" does not make sense to me. An approximation has to be respect to something, and you haven't specified anything else other than with respect to the $L^{1}$ norm. Are you expecting the approximation to be 'close' in some other way that you didn't specify? As for wanting the approximation $f_{\epsilon}$ to be in $C^{k}$ as opposed to $C^{\infty}$, are you trying to suggest that $f_{\epsilon}$ cannot be allowed to have too many derivatives? I can't see how to guarantee such a thing, especially not knowing anything more about $f$. –  T.A.E. Jun 10 at 17:49

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.