Let $f\in (C_c(\mathbb{R}))^*$ be a distribution. Show that $f\in C^{0,1}(\mathbb{R})$ if and only if $f\in L^\infty(\mathbb{R})$, and the distributional derivative $f'$ of $f$ also lies in $L^\infty(\mathbb{R})$. Here $C^{0,1}(\mathbb{R})$ is the space of continuous functions with bounded $C^{0,1}(\mathbb{R})$-norm defined by $$\|f\|_{C^{0,1}(\mathbb{R})}=\sup_{x\in\mathbb{R}}|f(x)|+\sup_{x,y\in\mathbb{R}:x\neq y}\frac{|f(x)-f(y)|}{|x-y|}.$$
I can prove that if $f\in C^{0,1}(\mathbb{R})$, then $f$ and $f'$ lie in $L^\infty(\mathbb{R})$, but I don't know how to prove the reverse. Now suppose that $f$ and $f'$ (in the distributional sense) lies in $L^\infty(\mathbb{R})$, How to show that $f$ is Lipschitz continuous almost everywhere?
If $f\in W^{1,\infty}(\mathbb R)$, then $f'\in L^\infty(\mathbb R)$, now observe that $f$ is continuous by Morrey's inequality. Then $$ |f(x)-f(y)|\leqslant\int_y^x |f'(t)|\,\mathrm dt $$ Does that help you establishing the Lipschitz condition?