I want to prove this fact:
I consider a function $g:\mathbb{R}^m \rightarrow \mathbb{R}$ separatelly convex, i.e. convex in each variable. Then $g\in {\rm Lip}_{\rm loc}(\mathbb{R}^m)$.
My attempt:
I make this preliminar observation. I consider $k\in\{0,...,m \}$ and let $\xi^0:=z$ and $\xi^k:=(w_1,...,w_k,z_{k+1},...,z_m)$, with $z,w\in \mathbb{R}^m$.
I have that $$|g(z)-g(w)|=|g(\xi^0)-g(\xi^m)|\leq \sum_{i=0}^{m-1}|g(\xi^{i+1})-g(\xi^i)|$$
I observe that $\xi^{i+1}-\xi^i=(0,\dots,o,w_{i+1}-z_{i+1}, \dots, 0)$.
Now let be $R>0$. I consider $w,z\in \mathbb{R}^m$ such that $|\xi^{i+1}-\xi^i|=|w_{i+1}-z_{i+1}|\leq R$ for every $i\in \{0,\dots, m-1\}$.
Let be $g_{i+1} :(0,2R) \rightarrow \mathbb{R}$ defined as follows:
$$g_{i+1}(t)=g(w_1,\dots, w_i,t,z_{i+2},\dots, z_m).$$
I observe that $g_{i+1}(w_{i+1})=g(\xi^{i+1})$ and $g_{i+1}(z_{i+1})=g(\xi^i)$.
Now, using the convexity of $g_{i+1}$ in $(0,2R)$, I obtain:
$$\dfrac{g_{i+1}(w_{i+1})-g_{i+1}(z_{i+1})}{w_{i+1}-z_{i+1}}\leq \dfrac{g_{i+1}(2R)-g_{i+1}(z_{i+1})}{2R-z_{i+1}}\leq \dfrac{{\rm osc}(g_{i+1}, (0,2R))}{R}:=c_i(R),$$
where ${\rm osc}(f,S)={\rm sup}\{|f(t)-f(s)|\ {\rm with}\: s,t\in S\}$.
From this follows that $|g(z)-g(w)|\leq(\sum_{i=0}^{m-1}c_i(R) )|z-w|$.
I call $c(R)=\sum_{i=0}^{m-1}c_i(R)$ and I have finished.
My problem: Is my proof correct? It seems to me that in this way I have proved something more than the locally lipschitz condition, since $R$ is generic. Is this the Lipschitz condition on every compact set?
Edit I have proved that $\forall R$ if $z,w\in\mathbb{R}^m$ are such that $|z_i-w_i|\leq R$ $\forall i\in\{1,...,m\}$ then $|g(z)-g(w)|\leq c(R)|z-w|$. So I have proved the Lipschitz condition on every m-cube and so in every compact set. Is it right?
Thanks a lot for the help!