Problem : If $f$ is continuous on metric space $(X,|\ |)$, then for $\varepsilon >0$ there is locally Lipschitz $f_\varepsilon$ s.t. $$|f(x)- f_\varepsilon (x)|<\varepsilon\ \ast $$
Proof : (1) If $f$ is continuous then it is difference of continuous functions bounded below by $1$ : $$f=({\rm max}\ \{f,0\} +1)-({\rm max}\ \{ -f,0\} +1) $$
(2) Assume that $f\geq 1$ Define $\rho$ on $X$ : $$ |x-y|\leq \rho (x)\Rightarrow |f(x)-f(y)|<\varepsilon $$
Set $$f_\varepsilon (y):=\sup_x \ \bigg\{ f(x)\bigg(1- \frac{|x-y|}{\rho(x)} \bigg) \bigg\} $$ Then $$ f(y)-f_\varepsilon (y) \leq f(y)- \bigg[f(x) - f(x) \frac{|x-y|}{\rho (x) } \bigg]_{x=y} =0 $$
If $$ f_\varepsilon (y) -f (y) =A,\ A:= \sup_x\ B,\ B:= f(x)-f(y) -f(x) |x-y|/\rho(x) $$ we will find an upper bound for $A$ : Then $X =X_1\cup X_2\cup X_3 $ where $$ X_1=\{ x\in X| |x-y|\leq \rho(y) \} $$
$$ X_2=\{x\in X| \rho(y)<|x-y| \leq \rho(x) \} $$
$$ X_3 =\{ x\in X| \rho(y),\ \rho(x) < |x-y| \} $$
Then $${\rm sup}_{x\in X_1 \cup X_2 } B \leq \varepsilon $$
$$ {\rm sup}_{x\in X_3} B \leq -f(y) \leq -1 $$
Here we have $\ast$
But to prove that $f_\varepsilon$ is locally Lipschitz is difficult How can we prove this ? Thank you in anticipation