Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear community.

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.

$|f(x-\tau)+f(x+\tau)-2f(x)| \leq C |\tau|$ for all $x \in dom(f)$.

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 $f_\epsilon(t) := (f \ast_{s}\rho)(t)$.

If one uses the implication Zygmund $\Rightarrow$ LogLipschitz, the one easily obtains the estimate $|\frac{d}{dt}f_\epsilon(t)| \leq C \log\left( 1+\frac{1}{\epsilon} \right)$.

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?

It would be nice to get something like $|\frac{d}{dt}f_\epsilon(t)| \leq C \left(\log\left( 1+\frac{1}{\epsilon} \right)\right)^{1/2}$

Thanks!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
But this function is also log-Lipschitz. I know that the estimate on the regularization for such functions is optimal. I would like to know if there is some kernel with which we might get better estimates for Zygmund-functions, just using the Zygmund property. –  CPJ May 1 '11 at 13:05
    
In terms of the first difference Zygmund functions are generally no better than log-Lipschitz as the example for $f$ above shows. Ok, lets do the math. For this $f$ and small $\epsilon>0$ we have $$ \partial_xf_\epsilon(0)=\int_{-1}^1\rho_\epsilon(y)(\log|y|+1)dy= \int_{-\infty}^\infty\rho(y)(\log|\epsilon y|+1)dy= $$ $$ \int_{-\infty}^\infty\rho(y)\log|y|dy+\log\epsilon+1. $$ –  Andrew May 1 '11 at 14:23

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.