Here is a curious (already submitted) homework problem I had in analysis some time ago:
Let $\Omega$ be a convex domain in $\mathbb{R}^n$ with $C^1$ boundary. Let $C^{0,\alpha}(\overline{\Omega})$ denote those functions that are Holder continuous on $\overline{\Omega}$ with coefficient $\alpha$ and with finite Holder norm: $$ \|f\|_{\alpha} = \|f\|_\infty + \sup_{x \neq y}\frac{|f(x) - f(y)|}{|x-y|^\alpha} $$ Let $C^{1}(\overline{\Omega})$ denote the continuously differentiable functions on $\overline{\Omega}$. Show that the continuously differentiable functions are not dense in $C^{0,\alpha}$ under the Holder norm, and moreover that a function $f \in C^{0,\alpha}$ lies in the closure of $C^{1}$ (w.r.t. Holder Norm) if and only if for any $x \in \overline{\Omega}$:
$$ \lim_{(y,z) \to (x,x)}\frac{|f(y) - f(z)|}{|y-z|^\alpha} = 0 $$ A hint is provided, stating that because $\overline{\Omega}$ is closed + convex, we have a projection $\eta(x)$ that allows us to extend $f$ to all of $\mathbb{R}^n$, and moreover, the projection is Lipschitz ($|\eta(x) - \eta(y)| \leq|x-y|$) thanks to projections being distance non-increasing.
It would appear that $C^1$ is not dense in $C^{0,\alpha}$ for any $\Omega.$ To start, take $n=1$ and $\Omega = (-1,1).$ Fix $\alpha \in (0,1).$ Let $f(x)=|x|^\alpha.$ Then $f \in C^{0,\alpha}(-1,1).$ Suppose $g\in C^1(-1,1).$ We have
$$\|f-g\|_\alpha \ge \frac{|(f-g)(x) - (f-g)(0)|}{|x-0|^\alpha} $$ $$= \frac{||x|^\alpha - (g(x)-g(0))|}{|x|^\alpha} \ge 1- \frac{|g(x)-g(0)|}{|x|^\alpha}$$
for $x$ close to $0.$ As $x\to 0,$ the last fraction $\to 0$ since $g'(0)$ exists. It follows that $\|f-g\|_\alpha \ge 1.$
This idea should work for any $n$ by taking $f(x_1,\dots ,x_n)=|x_1|^\alpha.$