I came across the following statements in a math book without proof.
Denote $M_k$ as the set of functions from $C[a,b]$ that is K-Lipschitz continous. i.e $\forall x,y,|f(x)-f(y)|\le K|x-y|$
1) The closure of $M_k$ is also $M_k$. (This I already have a proof.)
2) $M_k$ is also the closure of the set of differentiable functions that $\forall x,|f'(x)|\le K$. (This I don't know why)
It's clear that $\{ f: |f'(x)|\le K \}$ is a subset of $M_k$.
What is not clear is that why $\forall f \in M_k$, $f$ would be a contact point of $\{ f: |f'(x)|\le K \}$. i.e. how to prove that every $f \in M_k$ counld be arbitary approximate by $f \in \{ f: |f'(x)|\le K \}$?
Extend $f$ to the whole real real line, so that $f(x)=f(a)$, for $x<a$ and $f(x)=f(b)$ for $x>b$. Next set $$ f_\varepsilon(x)=\frac{1}{2\varepsilon}\int_{-\varepsilon}^\varepsilon f(x+t)\,dt. $$ Then $f_\varepsilon\to f$ uniformly in $[a,b]$, as $\varepsilon\to 0$, as $$ \lvert\, f_\varepsilon(x)-f(x)\rvert\le\frac{1}{2\varepsilon}\int_{-\varepsilon}^\varepsilon \lvert\,f(x+t)-f(x)\rvert\,dt\le\sup_{t\in [-\varepsilon,\varepsilon]}\lvert\,f(x+t)-f(x)\rvert, $$ due to the uniform continuity of $f$ in $[a,b]$.
Next $f_\varepsilon$ is $k-$Lipschitz as $$ \lvert\,f_\varepsilon(x)-f_\varepsilon(y)\rvert\le\frac{1}{2\varepsilon}\int_{-\varepsilon}^\varepsilon \lvert\,f(x+t)-f(y+t)\rvert\,dt\le k\lvert x-y\rvert, $$ and finally $f_\varepsilon$ is differentiable since $$ \frac{f_\varepsilon(x+h)-f_\varepsilon(x)}{h}=\frac{1}{2h\varepsilon}\left( \int_{x+h-\varepsilon}^{x+h+\varepsilon} f(t)\,dt-\int_{x-\varepsilon}^{x+\varepsilon}f(t)\,dt\right)=\frac{1}{2h\varepsilon} \left(\int_{x+\varepsilon}^{x+\varepsilon+h} f(t)\,dt-\int_{x-\varepsilon}^{x-\varepsilon+h}f(t)\,dt\right)\to \frac{1}{2\varepsilon}\big(f(x+\varepsilon)-f(x-\varepsilon)\big), $$ as $h\to 0$.