Smooth functions with vanishing normal derivatives Let $B_1 := \{z ∈ C : |z| \le 1\}$, and let $C_0(B_1,\mathbb C)$ be the space of continuous complex-valued functions on $B_1$ equipped with the uniform convergence topology.
How to show that, the set of smooth functions $f: B_1 \rightarrow \mathbb C$ whose normal derivative vanishes along the boundary ∂$B_1$ are dense in $C_0(B_1,\mathbb C)$?
I tried to use the Stone-Weierstrass theorem. Couldn't make much progress. Kindly share your views. Thank you.
 A: To make it a bit quantitative, let $\omega$ be a modulus of continuity of $u$. Extend $u\in C^0(B_1)$ to $u_0(x):=u\big(\frac x{ \|x\|\vee1 }\big)$, which incidentally has modulus of continuity  $ \omega $, since it is a composition with a $1$-Lip map. For $ \epsilon>0$ approximate uniformly $u_0$ by smooth $u_\epsilon\in C^\infty(\mathbb{R}^n)$, mollifying by a smooth kernel $\phi_\epsilon\in C_c^\infty(B_\epsilon)$, which  gives $u_\epsilon:=u_0*\phi_\epsilon\in C^\infty(\mathbb{R}^n)$ with $\|u -u_\epsilon\|_{\infty,B_1}\le\omega(\epsilon)$. Again, all $u_\epsilon$ have modulus of continuity  $ \omega $, since they are convex mixtures of translates of $u_0$. Let $\sigma_\epsilon\in C^\infty(\mathbb{R})$ an approximation of $\text{id}_\mathbb{R}$ with $\sigma_\epsilon(t)=t$ for $t<1/2$, $\|\sigma_\epsilon-\text{id}_\mathbb{R}\|_{\infty,_\mathbb{R}}\le\epsilon$ and $\sigma_\epsilon'(1)=0$.
Finally define
$$v_\epsilon(x):=u_\epsilon\big(x\frac {\sigma_\epsilon(\|x\|) }{\|x\|}\big),$$
so that $v_\epsilon\in C^\infty(\mathbb{R}^n)$, $\|v_\epsilon-u_\epsilon\|_{\infty,B_1}\le\omega(\epsilon)$ and $\partial_r v _{\epsilon}{ |_{r=1}}=0$.
