Let $f: \Bbb R^n \to \Bbb R^m$ be a function with the property that for all $v \in \Bbb R^n$ an $L=L(v) \gt 0$ exists so that for all $x \in \Bbb R^n$ the function $t \longmapsto f(x+tv)$ is $L$-Lipschitz continuous.
Now i have to show that $f$ is Lipschitz-continuous.
Any ideas or tips on how to do this? Thanks in advance
Choose a basis of vectors $e_i$ of $\mathbb{R}^n$ and, for given $x\neq y $ let $$ x- y = \sum_i a_i e_i$$ Then $$|f(x)-f(y)| = |f(x)- \sum_i (f(z_{i})-f(z_i)) - f(y)|$$ where $z_1= x - a_1 e_1$, and $z_2 = z_1 -a_2 e_2$ and so on, so $y =z_{n-1}- a_n e_n$ so by rearranging
(i.e. by writing
$$ \begin{eqnarray} |f(x)-f(y)|&\le & |(f(x) - f(z_1)) +(f(z_1) -f (z_2)) + \ldots +(f(z_n)-f(y)) | \\ &\le & |f(x) - f(z_1)| +|f(z_1) -f (z_2)| + \ldots +|f(z_n)-f(y) | \\ & = & |f(x) - f(x-a_1 e_1)| + \ldots +|f(z_i) - f(z_i - a_{i+1}e_{i+i})| + \ldots \end{eqnarray} $$ and now applying the Lipshitz contidition
$$|f(z_i) - f(z_i - a_{i+1}e_{i+i})| \le L(e_{i+1})| a_{i+1}| $$ for each $i$) you get
$$|f(x)-f(y)|\le \sum L(e_i) |a_i| \le \max_i{L(e_i)}\sum |a_i|$$ But $a_i = x_i - y_i$ and it is a well known fact that for $x, y$ $$|x-y|_* :=\sum |x_i-y_i|$$ is a norm on $\mathbb{R}^n$. Since, on $\mathbb{R}^n$, any two norms are equivalent (if you don't know this search for this here or google it), the result follows.