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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!

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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.

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  • $\begingroup$ 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. $\endgroup$
    – CPJ
    Commented May 1, 2011 at 13:05
  • $\begingroup$ 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 $$ \partial_xf_\epsilon(0)=\int_{-1}^1\rho_\epsilon(y)(\log|y|+1)dy= \int_{-\infty}^\infty\rho(y)(\log|\epsilon y|+1)dy= $$ $$ \int_{-\infty}^\infty\rho(y)\log|y|dy+\log\epsilon+1. $$ $\endgroup$
    – Andrew
    Commented May 1, 2011 at 14:23

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