Lipschitz in $\mathbb R^1$ implies Lipschitz along any line in $\mathbb R^k$ (for convex functions) I'm trying to solve a homework problem (baby rudin course) and realized that Lipschitz continuity really makes the problem easier for me. However, since it is not defined in Rudin I have to prove the properties if I want to use them.
I have just argued that convex functions are (locally) lipschitz in $\mathbb R^1$. I'm trying to avoid arguing that it holds in $\mathbb R^k$ directly since the proof is messy and too difficult. However, I really need it to apply to any line in $\mathbb R^k$. I wrote:
"ii) Convex functions are locally Lipschitz along any line in $\mathbb R^n$.
Let $f$ be defined on some set $C \in R^n$ let $I=\{u| a< u<b\} \subset K$ be a line with $a<s<t<u<b$. We have assumed without loss of generality that the line is a linear combination of $e_1$. This is further justified in c) but really it is just a matter of notation in this case. Then by convexity (c.f. excercise R4.23)
$$\frac{f(t)-f(s)}{t-s}\leq \frac{f(u)-f(s)}{u-s}\leq \frac{f(u)-f(t)}{u-t}$$
Now choose
$$K=\max\left[\left|\frac{f(t)-f(s)}{t-s}\right|,\left|\frac{f(u)-f(t)}{u-t}\right|\right]$$
It follows that
$$  \frac{\left|f(u)-f(s)\right|}{\left|u-s\right|}\leq K \rightarrow |f(u)-f(s)|\leq K|u-s|$$
And thus $f$ is locally Lipschitz on $I$ and we have the result desired."
I feel that my argument that it applies to $\mathbb R^k$ is not very strong at the moment. I was considering induction, but I'm unsure what to induct on really.
Thanks in advance.
EDIT:
I think I solved it:
For $\mathbb R^n$ consider the function $f$ restricted to any line $I \in R^n$, this can be considered a convex function on a subset of $\mathbb R^1$, and then the result follows.
 A: 
For $\mathbb R^n$ consider the function $f$ restricted to any line $I\in \mathbb R^n$, this can be considered a convex function on a subset of $\mathbb R^1$, and then the result follows.

You should be more careful. It is true that convexity is determined by line restrictions, i.e., a function $f:\mathbb R^n\to \mathbb R$ is convex if and only if its restriction to every line is. But for the Lipschitz condition this is false unless we have uniform control on Lipschitz constant. The problem is that a function may be Lipschitz on every line, but with a constant depending on the line; since there are infinitely many lines to consider, we might not be able to pick a Lipschitz constant that works for all of them, even locally. This is not just an abstract concern: the function 
$$f(x,y) = \frac{xy^2}{x^2+y^4},\qquad f(0,0)=0$$ 
is locally Lipschitz on every line, but is not continuous at $(0,0)$. 
You can still fix the argument by making the one-dimensional version quantitative. That is, show that if $f:\mathbb R\to \mathbb R$ is convex, then it is Lipschitz on $[-a,a]$ with a constant that depends only on $a$ and $\sup_{[-2a,2a]} |f|$. (It is  important to have the larger interval in the supremum.) The proof goes like this: if some secant of $f$ on $[-a,a]$ has a large slope, then it grows large either at $2a$ or at $-2a$, and $f$ will be even bigger there.
With this quantitative version, the argument by line restriction works. Let $M$ be the supremum of $|f|$ on a ball of radius $2R$ cantered at the origin. Consider the restriction of $f$ to the ball of radius $R$ centered at the origin: this restriction will be Lipschitz on every line, with a uniformly bounded constant, independent of line.
