Consider $f(x)=x $ if $x \ge0$ and $f=0$ if $x \le 0$
For each $r \gt 0$ we can find a function $f_r $ such that $f_r = f$ when $|x| \ge 2r$ , $f_r \in C^{\infty} $ and $||f_r-f||_{H^1} \to 0$
We can choose $\phi \in C_c^{\infty}$ , $\phi=1$ when $|x|\le 1$ and $\phi=0$ when $|x|\ge 2$ , then $f_r := (1-\phi(\frac{x}{r})f(x)$ is what we want . And it can be shown that $||f-f_r||_{H^1} \to 0$ .
Since $f'(0-)=0,f'(0+）=1$ it seems that we can construct $f_r$ with $||f'_r||_{C^0}$ uniformly bounded , although $||f''_r||_{C^0}$ must tend to $\infty$ .
For each $r \gt 0$ can we find a function $f_r $ such that $f_r = f$ when $|x| \ge 2r$ , $f_r \in C^{\infty} $ and $||f'_r||_{C^0}\le 2$ ?