I'm stuck trying to solve the following exercise:
Let $f:\mathbb R^n \to \mathbb R^m$ a function with the property that, for all $v \in \mathbb R^n$, there is $L=L(v) > 0$ such that the function $t \mapsto f(x+tv)$ is $L$-Lipschitz continuous for every $x\in\mathbb R^n$. Show that f is Lipschitz continuous.
First note that for $v_1,v_2\in S^{n-1}$ we have that $$ |f(x+v_1t_1+v_2t_2)- f(x)| \\= |f(x+v_1t_1+v_2t_2)-f(x+v_1 t_1)+f(x+v_1 t_1) -f(x)| \\\leq |t_2|L(v_2) + |t_2|L(v_1) $$
Now take an orthonormal basis $(e_i)$ of $\mathbb{R}^n$ and define $$ L_i = L(e_i) $$ to get $$ |f(x+ \sum_i t_i e_i)-f(x)| \leq \sum_i |t_i| L_i \leq \sqrt{\sum_i |t_i|^2}\sqrt{\sum_i L_i^2}= |\sum_i t_i e_i|\sqrt{\sum_i L_i^2}. $$ hence $f$ is Lipschitz continuous with Lipschitz constant $\sqrt{\sum_i L_i^2}$.