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.
Best Answer
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 $$g\star\varphi_\epsilon\to g\ \mbox{in}\ C^{\beta}(\overline{B_1}).$$