Suppose I have a $L^2(\mathbb{R}^n)$ fucntion $f(x)$ such that $|f(x)-f(y)|<C|x-y|^a$ for almost every $x,y \in \mathbb{R}^n$ and some $a \in (0,1]$, can we find a continuous representative $f^*$ of $f$ such that $f=f^*$ a.e.?
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1How do you define $|f(x)-f(y)|$ ? – Stratos supports the strike Apr 22 '23 at 19:35
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Yes, this is true. To see this, let $(\eta_\epsilon)_{\epsilon}$ be a family of smooth mollifiers, in particular $\int_{\mathbb{R}^n} \eta_{\epsilon}\,dx=1$, and set $f_{\epsilon} = \eta_{\epsilon} * f$ for $\epsilon >0$. Then, we compute for any $x,y \in \mathbb{R}^n$ that $$|f_{\epsilon}(x)-f_{\epsilon}(y)| \leq \int_{\mathbb{R}^n} \eta_{\epsilon}(z)|f(x-z)-f(y-z)|\,dz \leq C\int_{\mathbb{R}^n} \eta_{\epsilon}(z)|x-y|^{a}\,dz=C|x-y|^{a}.$$ Therefore, the sequence $(f_{\epsilon})_{\epsilon}$ is uniformly equicontinuous and locally uniformly bounded, so that the Arzela-Ascoli theorem shows that (up to a non-relabeled subsequence) $f_{\epsilon} \to g$ locally uniformly as $\epsilon \to 0$, where $g \in C(\mathbb{R}^n)$ is a continuous function. At the same time, by standard properties of mollification, we have that $f_{\epsilon} \to f$ in $L^2(\mathbb{R}^n)$ as $\epsilon \to 0$. We deduce that $g=f$ almost everywhere, or in other words, $g$ is a continuous representative of $f$.
Hidde
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4I believe we can avoid appealing to Arzela-Ascoli. By a similar calculation, $|f_{\epsilon}-f|{\infty}\leq C\epsilon^a$, and so by the triangle inequality, $|f{\epsilon_1}-f_{\epsilon_2}|{C(\Bbb{R}^n)}\leq C\epsilon_1^a+C\epsilon_2^a$ vanishes as $\epsilon_1,\epsilon_2\to 0$, which shows the family is uniformly Cauchy, so has a limit $g$, which is necessarily continuous, and $f{\epsilon}\to g$ uniformly. – peek-a-boo Apr 22 '23 at 20:28
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2@peek-a-boo ah yes, good point, that would definitely be a more elementary way of proving it – Hidde Apr 22 '23 at 20:48