I was wondering if the following is true and if this result can be strengthened to more general "growth rates" than logarithms. Let $\phi \in C_{0}^{\infty}(\mathbb{R}) = \{ f \in C^{\infty}(\mathbb{R}) | $ $supp(f)$ is compact$\}$. Then $ \lim_{\epsilon \downarrow 0} \left( \phi(\epsilon) - \phi(-\epsilon) \right)ln(\epsilon) = 0$. Below is what I believe to be a proof.

Since $\phi$ is smooth, let $K > 0$ be a Lipschitz constant for $\phi$ on $[-1,1]$. Then for $\epsilon < 1$, we have \begin{align*} |\phi(\epsilon) - \phi(-\epsilon)||ln(\epsilon)| &\leq 2 K \epsilon ln(\epsilon) \to 0 \end{align*} as $\epsilon \downarrow 0$.

In general, this should work for any function that goes as $\epsilon ^ p$ where $p > -1$ correct? Then $p = 1$ has the counterexample $\phi \equiv \epsilon$ in a neighborhood of $0$. Also, would this rate also be generalizable?

  • $\begingroup$ $\phi(\epsilon)-\phi(-\epsilon) = \int_{-\epsilon}^\epsilon \phi'(x)dx$ so $|(\phi(\epsilon)-\phi(-\epsilon))\ln(\epsilon)| \le 2 |\epsilon\ln(\epsilon)| \sup_{x \in [-1,1]} |\phi'(x)|$ $\endgroup$ – reuns Dec 8 '18 at 21:28
  • $\begingroup$ Yeah that would be essentially the same proof right? The Lipschitz constant for a differentiable function is in general $sup_{x \in [-1,1]} |\phi'(x)|$ right? By the mean-value theorem $\endgroup$ – TylerMasthay Dec 9 '18 at 21:35

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