Lower bound the best $\alpha$-Hölder constant of a convolution

Question

Let $\mathcal D_1$ be the set of bounded probability density functions on $\mathbb R^d$. This means $f \in \mathcal D_1$ if and only if $f$ is non-negative measurable such that $\int_{\mathbb R^d} f (x) \, \mathrm d x=1$ and $\|f\|_\infty < \infty$. We fix $\alpha \in (0, 1)$. For $f \in \mathcal D_1$, let $[f]_\alpha$ be the best $\alpha$-Hölder constant of $f$, i.e., $[f]_\alpha := \sup_{x \neq y} \frac{|f(x) - f(y)|}{|x-y|^\alpha}$.

Now we fix $\rho \in \mathcal D_1$ such that $[\rho]_\alpha < \infty$. We define a function $\sigma : \mathbb R^d \times \mathcal D_1 \to \mathbb R$ by $$ \sigma (x, f) := (f*\rho) (x) := \int_{\mathbb R^d} f(x-y) \rho (y) \, \mathrm d y. $$

Are there constants $C, \varepsilon>0$ such that $[\sigma (\cdot, f)]_{\alpha} \ge C [f]_{\alpha} - \varepsilon$ for all $f \in \mathcal D_1$ with $[f]_{\alpha} < \infty$?

Thank you so much for your elaboration!

Answer 1

No. Take $f_1$ an $\alpha$-Hölder compactly supported convolution kernel (non-negative, with $\int_{\mathbb R}^nf_1dx=1$), and consider the usual approximation of identity by convolution with $f_\epsilon(x):=\frac1{\epsilon^n}f(\frac x\epsilon)$, for all $\epsilon>0$. Then $[f_\epsilon]_\alpha=\frac1{\epsilon^{n+\alpha}}[f_1]_\alpha$ is unbounded (or also, without computing it: it couldn’t be bounded, by Ascoli-Arzelà, since $f_\epsilon$ is not compact in the uniform convergence). On the other hand, since $f_\epsilon\in\mathcal D_1$, for every $\alpha$-Hölder function $\rho$ one has $[\rho*f_\epsilon]_\alpha\le[\rho]_\alpha$.