# How to use the Mean Value Inequality to prove the following statement?

I am trying to do the following question. Let $E$ be an open subset of $\mathbb{R}^{n}$ containing the origin. Use the fact that if $f \in C^{1}(E)$ then for all $x,y \in N_{\delta}(0) \subset E$ (where $N_{\delta}(0)$ is a $\delta$-neighbourhood of the origin) there exists a $\xi \in N_{\delta}(0)$ such that

$$|f(x) - f(y)| \leq|| DF(\xi)|| |x-y|$$

to prove that if $f(0) = Df(0)=0$ then given any $\epsilon>0$ there exists a $\delta>0$ such that for all $x,y \in N_{\delta}(0)$ we have

$$|f(x) - f(y)| < \epsilon |x-y|.$$

Now the first inequality is obviously the so-called Mean Value Inequality (or Theorem). I don't know how to use it to prove the required statement. I could obviously take $\epsilon = \mbox{max} || DF(\xi)||$ but that doesn't help to establish a relation between $\delta$ and $\epsilon$.

Any help would be appreciated!

Hint: Let $g(t) = f(x + t(y-x)), t\in [0,1].$ Apply the mean value theorem to $g(1) - g(0).$
• You have $\nabla f (0) = 0.$ Choose $\delta > 0$ such that $|x|<\delta$ implies $|\nabla f (x)| < \epsilon.$ – zhw. Nov 20 '16 at 20:12
• I don't see how $|x|< \delta$ implies $|Df(x)|< \epsilon$ - what do we take $\epsilon$ to be? Heuristically, the given condition implies that $f$ and its first derivative cannot be very large near 0. How to justify this rigorously? What is the bound on $\delta$ for a given $\epsilon$? @zhw – Alex Nov 20 '16 at 20:19
• Hmm, I can't really tell what's bothering you. There's nothing tricky going on here. We have $f\in C^1$ which implies all partial derivatives of $f$ are continuous. Thus $|\nabla f(x)|=\|Df(x)\|$ is a continuous function. And we have $|\nabla f(0)|=0.$ Thus given $\epsilon>0$ there exists $\delta > 0$ … – zhw. Nov 20 '16 at 20:26