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A function $f$ is Log-Lipschitz if there exists a constant $C >0$ such that \begin{equation} |f(x) - f(y)| \le C|x-y| |\log|x-y|| \end{equation}

I am trying to construct two functions with the following properties.

First function is $f\in \mathcal{C}^1((0,a])$ ,$a>0$ and log-Lipschitz continuous on $[0,a]$ such that \begin{equation} \limsup_{x \to0^+} x^q|f'(x)|=+\infty, \forall q \geq 1 \end{equation} Second function is $g\in \mathcal{C}^1((0,a])$ continuous on $[0,a]$ but Holder continuous on $[0,a]$ for no $\alpha<1$ such that \begin{equation} \limsup_{x \to0^+} x|g'(x)|<+\infty. \end{equation} I tried constructing but not getting through much(I came across these functions in context of log-lipschitz regularity of certain hyperbolic pdes). Thanks in advance.

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  • $\begingroup$ Just make them a bit wild instead of trying something too nice. For instance, you can put $g(x)=\frac{\sin(\log x)}{\log x}$ extended to $0$ by $0$, etc. $\endgroup$
    – fedja
    Apr 18, 2018 at 8:20
  • $\begingroup$ The title of this question is misleading, since $g$ is precisely not Log-Lipschitz...(being Hölder for no $\alpha<1$). Also, do you really want $f$ to satisfy the limsup condition for every $q\ge1$ ? $\endgroup$ Apr 18, 2018 at 10:18
  • $\begingroup$ @Jean Dunchon I have corrected the title and $f$ needs to satisfy the limsup condition for every $q\ge1$ $\endgroup$ Apr 18, 2018 at 10:30

1 Answer 1

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More generally: if $\omega$ is a modulus of continuity with $\omega'(0)=\infty$ there is an $\omega$-continuous, smooth function $f$ on $\mathbb{R}_+$, with prescribed derivative $p_k\in \mathbb{R}$ at points of a prescribed discrete subset $(x_k)_{k\ge1}$ of $\mathbb{R}_+$: $f'(x_k)=p_k$ for all $k\ge1$.

Construction. We may assume w.l.o.g. $\omega$ is a concave modulus of continuity, so that $\omega(t)/t$ is decreasing with $\sup_{t>0}\omega(t)/t=+\infty$ . Fix a smooth function $\phi$ with support in $[-1/2,1/2]$ and with $\phi'(0)=1= \|\phi'\|_\infty$. Define positive numbers $\delta_k$ such that $2^k|p_k|\le \omega(\delta_k)/\delta_k,$ and such that the intervals $I_k$ of length $\delta_k$ centered at $x_k$ are pairwise disjoint.

For any $k\ge1$ consider $\phi_k(x):=\delta_k p_k \phi\big({x-x_k\over\delta_k}\big)$, a smooth function supported in $I_k$ such that $\phi_k'(x_k)=p_k$ and $\|\phi_k'\|_\infty=|p_k|$. It satisfies $|\phi_k(x)-\phi_k(y)|\le2^{-k}\omega(|x-y|)$ for all $x$ and $y\in\mathbb{R}$: indeed, to check the latter, it is sufficient to look at points $x$ and $y$ both in $I_k$, and for these points $|x-y|\le\delta_k$, so $$|\phi_k(x)-\phi_k(y)|\le |p_k||x-y|\le 2^{-k} {\omega(\delta_k)\over\delta_k}|x-y|\le 2^{-k} \omega( |x-y|)$$

Consider $\sum_{k=1}^\infty \phi_k$, a locally finite sum in $\mathbb{R}_+$, thus defining an $\omega$-continuous function $f\in C^\infty(\mathbb{R_+})$, with $f'(x_k)=p_k$. $$*$$ The other question: the function $g(x):=\exp(-|\log x|^{1/2})$ is smooth on $(0,1)$ and continuous on $[0,1]$ with $g(0)=0$. It is not Hölder at $0$ because for all $\alpha>0$, $x^\alpha=o(g(x))$ as $x\to0$. But, $2xg'(x)={g(x)|\log x|^{-1/2} }=o(1)$ as $x\to0$.

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