Approximate Holder continuous functions by smooth functions

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