Differential Maping in Elementary Analysis

Question

The problem I stuck was :

Let $ f = R^{n} \rightarrow R^{m} $ and suppose there is a constant $M$ such that for $ x \in R^{n} $, $ || f(x) || \leq M || x ||^{2} $. Prove that $ f $ is differentable at $ x_{0} = 0 $ and that $ Df(x_{0}) = 0 $

As it is begining of the semester, and I was never been through any exercises or examples yet, it is kind of hard for me to prove.

The source of this queistion is by Marsden, Page 330 Exercise 6.1.

Answer 1

Using Marsden's definitions, you need to show that for all $\epsilon>0$, there is some $\delta>0$ such that if $\|x\|< \delta$, then $\|f(x)-f(0)\| \le \epsilon \|x\|$ (since we want to show that $Df(0) = 0$).

What is $f(0)$? (You can work this out from the bound $\|f(x)\| \le M \|x\|^2$.)

Hint:

Setting $x=0$ gives $\|f(0)\| \le M \|0\|^2 = 0$, so $f(0) = 0$.

Given that, how can you choose $\delta>0$ (in terms of M) so that if $\|x\| < \delta$, then the above bound will be true?

Hint:

In order to get $\|f(x)-f(0)\| = \|f(x)\| \le \epsilon \|x\|$, we can just choose $\delta = \frac{\epsilon}{M}$. Then if $\|x\| < \delta$, we have $\|f(x)-f(0)\| \le M \|x\|^2 \le M \delta \|x\| = \epsilon \|x\|$.