Let $g \in C^{\alpha} (B_1)$ be given. Can we find a sequence $(f_n) \subset C^{\infty} (B_1)$ such that $f_n \rightarrow g$ in $C^{\alpha}(\overline{B_1})$? If so, how can it be done?

I have tried using the method to show that given $g \in C_0(B_1)$, there exists $(f_n) \subset C^{\infty} (B_1)$ such that $f_n \rightarrow g$ in $C(\overline{B_1})$, that is, I make use of the modllifier $\varphi_\varepsilon$ and consider $g \star \varphi_\varepsilon$, then

$$|g \star \varphi_\varepsilon (x) - g(x)| \leq \int_{B_1} \varphi(z)|g(x - \varepsilon z) - g(x)|dz \rightarrow 0$$ uniformly since $g$ is uniformly continuous.

Using the approach, the only fact I need to show is $[g \star \varphi_\varepsilon - g]_\alpha \rightarrow 0$ as well. However, this is where I got stuck. If I try to calculate the semi-norm directly, then we have $$[g \star \varphi_\varepsilon - g]_\alpha = \sup_{x \neq y} |\int_{B_1} \varphi(z) \frac{(g(x - \varepsilon z) - g(x)) - (g(y - \varepsilon z) - g(y))}{|x - y|^{\alpha}}dz|$$ and I have no idea how can I proceed.

I believe this is a standard and elementary approximation problem, but I have yet to find any text that give light to this problem and spent quite some time on it. So, could anyone give me some kind of directions or answers? It would then be a great help to my studies and thanks in advance.

  • $\begingroup$ If $x-\epsilon z$ is outside $\overline{B_1}$, how do you define $g(x-\epsilon z)$? $\endgroup$ – Tomás Nov 12 '13 at 21:07
  • $\begingroup$ That's why I consider $g \in C_0^{\alpha}(\overline{B_1})$ instead of $g \in C^{\alpha}(\overline{B_1})$. If $g \in C_0^{\alpha}(\overline{B_1})$, then $g(x) = 0$ for $|x|$ close to 1, say $|x| > 1 - \delta$ for some small $\delta$. So, for $\varepsilon < \delta/ 2$, $g_\varepsilon (x) = g \star \varphi_\varepsilon (x)$ is defined on $B_{1 - \delta + \varepsilon}$ and we can extend the function to be zero outside. $\endgroup$ – user43378 Nov 13 '13 at 0:36
  • $\begingroup$ Of course, one should also take care of the case where one point is inside $B_{1 - \delta + \varepsilon}$ and one point is outside $B_{1 - \delta + \varepsilon}$ in the semi-norm. But I think it can be done so I haven't pay to much attention to it. $\endgroup$ – user43378 Nov 13 '13 at 0:46
  • 1
    $\begingroup$ I see. Usually the notation $C_0^\alpha (\overline{B_1})$ means that the function is zero on the boundary. The standard notation for functions with compact support is $C_0^\alpha (B_1)$. $\endgroup$ – Tomás Nov 13 '13 at 10:31
  • $\begingroup$ I could prove the convergence for every $0<\beta<\alpha$, but until now, I could not prove the convergence for $\alpha$... $\endgroup$ – Tomás Nov 14 '13 at 11:36

You can find a counterexample in this link, however I would like to note that for every $0<\beta<\alpha$ we do have convergence. Indeed, define for $x\neq y$ $$T(\epsilon,x,y)=\int_{B_1}\left|\varphi(z)\frac{g(x-\epsilon z)-g(x)-(g(y-\epsilon z)-g(y))}{|x-y|^\beta}\right|.$$

Case 1: $|x-y|\ge \delta_1$

In this case we have that for any $\eta>0$, there is $\epsilon_1>0$ such that if $\epsilon<\epsilon_1$ then

$$T(\epsilon,x,y)\leq 2\frac{\|\varphi\|_\infty}{\delta_1^{\alpha-\beta}}\left|B_1\right|\eta.\tag{1}$$

To prove $(1)$, use the uniform continuity of $g$ and the fact that $\frac{1}{|x-y|}\leq \frac{1}{\delta_1}$.

Case 2: $|x-y|<\delta_1$

In this case we use the $\alpha$-Hölder continuity of $g$ to conclude that for any $\epsilon>0$, $$T(\epsilon,x,y)\leq 2\|\varphi\|_\infty \left|B_1\right|\left|x-y\right|^{\alpha-\beta}\leq 2\|\varphi\|_\infty \left|B_1\right| \delta_1^{\alpha-\beta}.\tag{2}$$

We choose a suitable $\delta_1$ and combine $(1)$ and $(2)$ to conclude that $$g\star\varphi_\epsilon\to g\ \mbox{in}\ C^{\beta}(\overline{B_1}).$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.