My question is :
As shown below, to prove that the $\overline{\mathcal{F}}$ is equal to the set on the right side of the equation we should prove that they contain each other. I've only proved that the right side contains the left side and I don't know how to prove that the left side contains the right side. That is, I don't know how to prove that any function that is bounded and Lipschitz continuous can be approximated by a function that is bounded and continuously differentiable.
The original question is here:
Let $C^1[0,1]$ denote the space of all continuous functions $f$ on $[0,1]$ with continuous derivative $f^{\prime}$. For constants $M>0$ and $N>0$, we define the subset $\mathcal{F}$ of $C[0,1]$ by $$ \mathcal{F}=\{f\in C^1[0,1]\ |\ \|f\|\leq M,\ \|f^{\prime}\|\leq N\}, $$ where $\|\cdot\|$ denotes the sup-norm. Then $\mathcal{F}$ is precompact in $C(K)$. It is not closed, however, because the unifrom limit of continuously differentialbe functions need not be differentiable. Thus, $\mathcal{F}$ is not compact. Its closure in $C[0,1]$ is the compact set $$ \overline{\mathcal{F}}=\{f\in C[0,1]\ |\ \|f\|\leq M,\ \mathrm{Lip}(f)\leq N\}. $$