It is not possible: your condition $|f_n|<|f|$ implies that the zero set of $f_n$ is contained in the zero set of $f$. So $f(x)=|x|^{\alpha}$ cannot be approximated by a smooth function $f_n$, since $f_n(x)=(c+o(1))x$, and $\sup|f(x)-f_n(x)|\geq (1+o(x))|x|^\alpha$ for small $x$.

It is not possible: your condition $|f_n|<|f|$ implies that the zero set of $f_n$ is contained in the zero set of $f$. So $f(x)=|x|^{\alpha}$ cannot be approximated by a smooth function $f_n$, since $f_n(x)=(c+o(1))x$, and $\sup|f(x)-f_n(x)|\geq (1+o(x))|x|^\alpha$ for some small $|x|$.

It is not possible: your condition $|f_n|<|f|$ implies that $f$ and $f_n$ have the same zero set. So $f(x)=|x|^{\alpha}$ cannot be approximated by a smooth function $f_n$, since $f_n(x)=(c+o(1))x$, and $\sup|f(x)-f_n(x)|\geq (1+o(x))|x|^\alpha$ for small $x$.

It is not possible: your condition $|f_n|<|f|$ implies that the zero set of $f_n$ is contained in the zero set of $f$. So $f(x)=|x|^{\alpha}$ cannot be approximated by a smooth function $f_n$, since $f_n(x)=(c+o(1))x$, and $\sup|f(x)-f_n(x)|\geq (1+o(x))|x|^\alpha$ for small $x$.