Mollification is Lipschitz Assume that $f$ is $L$-Lipschitz on $\mathbb{R}^n$ And consider a mollifier $$ \eta \geq 0,\ {\rm supp}\ \eta \subset B_1(O),\ \int_{\mathbb{R}^n} \eta(y)\ dy=1 $$ If $$f_1(z)=\int_{\mathbb{R}^n} \eta(z-y)f(y)\ dy$$ then note that $$ |f_1(z)-f(z)|\leq \bigg|\int_{\mathbb{R}^n} \eta(z-y)[f(y) -f(z)] dy \bigg|\leq L $$ **Question** : $f_1$ is Lipschtz Proof : Hence for $|z-x|\geq 1$, $$ |f_1(z)-f_1(x)| \leq |f_1(z)-f(z)| + |f(z)-f(x)| + |f(x)-f_1(z)| $$ $$\leq 2L +L|z-x| \leq 3L |x-z| $$ For $|z-x| <1$, how do we finish the proof ? Thank you in advance

I don't understand your reason for splitting the space. Here's my solution. Observe \begin{align} |f_1(x)-f_1(y)| \leq \int_{\mathbb{R}^n}\eta(z)|f(x-z)-f(y-z)|\ dz \leq L|x-y|\int_{\mathbb{R}^n}\eta(z)\ dz. =CL|x-y|. \end{align}