Regularization of Zygmund functions - MathOverflow most recent 30 from http://mathoverflow.net2013-05-24T04:26:34Zhttp://mathoverflow.net/feeds/question/60725http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/60725/regularization-of-zygmund-functionsRegularization of Zygmund functionsCJ2011-04-05T18:28:25Z2011-04-30T18:06:33Z
<p>Dear community.</p>
<p>I would like to derive a "good" estimate on $\frac{d}{dt}f_\epsilon(t)$, where $f_\epsilon$ is a regularization of a Zygmund-continuous function $f$, i.e.</p>
<p>$|f(x-\tau)+f(x+\tau)-2f(x)| \leq C |\tau|$ for all $x \in dom(f)$.</p>
<p>The regularization is defined as usual. We use an even function $\rho \in C_0^\infty(\mathbb R)$ with $\int_{\mathbb R} \rho(\tau)d\tau = 1$, set $\rho_\epsilon(t) := \frac{1}{\epsilon}\rho\left( \frac{t-s}{\epsilon} \right)$ and define $f_\epsilon$ by convolution as $f_\epsilon(t) := (f \ast_{s}\rho)(t)$.</p>
<p>If one uses the implication Zygmund $\Rightarrow$ LogLipschitz, the one easily obtains the estimate $|\frac{d}{dt}f_\epsilon(t)| \leq C \log\left( 1+\frac{1}{\epsilon} \right)$.</p>
<p>I would like to have a better estimate than this. This means I would like to use the fact, that $f$ is Zygmund, not just LogLipschitz. The obvious ideas don't work, if I'm not mistaken. Maybe one should use a special mollifier?! Does anyone have some experience in this direction? Or can one point to literature?</p>
<p>It would be nice to get something like $|\frac{d}{dt}f_\epsilon(t)| \leq C \left(\log\left( 1+\frac{1}{\epsilon} \right)\right)^{1/2}$</p>
<p>Thanks!</p>
http://mathoverflow.net/questions/60725/regularization-of-zygmund-functions/63549#63549Answer by Andrew for Regularization of Zygmund functionsAndrew2011-04-30T18:06:33Z2011-04-30T18:06:33Z<p>I don't think it is possible to make this estimate better. A proof can be done imho considering some simple function from Zygmund space, for example $f(x)=x\log|x|$, $x\in[-1,1]$. It has one point of non-smoothness and the derivative of the regularisation can be written out explicitly.</p>