Let $V$ be a convex open set $\mathbb{R}^2$ and let $f:V \to \mathbb{R}$ be continuously differentiable function in $V$. Show that if there is a positive number $M$ such that $|\nabla f(x)| \leq M$ for all $x \in V$, then there is a positive number $L$ such that $$ |f(x)-f(y)| \leq L|x-y|.$$
Answer:
Since $f:V \to \mathbb{R}$ be continuously differentiable function in $V$, we have
$$|\nabla f(x)-\nabla f(y)| \leq L|x-y|$$ for some $L$.
But how to conclude the answer?
Help me
Let $x,y\in V$. Since $V$ is convex, we have $x + t (y-x) \in V$ for any $t \in [0,1]$. Therefore, the function $$ g\colon [0,1]\to\mathbb R,\quad t\mapsto f(x+ t (y-x)),$$ is well-defined and since $f\in C^1(V)$, we can calculate $$|g'(t)|= |\nabla f(x+ t (y-x)) \cdot (y-x)|\le M|y-x| \quad \text{for all $t\in[0,1]$}.$$ By the (one-dimensional) mean-value-theorem, we can find some $\xi\in[0,1]$ such that $$ g(1)-g(0)=g'(\xi)(1-0)$$ and hence $$ |f(y)-f(x)|=|g(1)-g(0)|=|g'(\xi)|\le M|y-x|.$$
Multi-dimensional variants of the mean-value-theorem (as suggested in the comments) are in fact nothing else than suitable applications of the one-dimensional mean-value-theorem - like the one we just used.