Problem with differential of a function.

Question

Let $L: \mathbb{R^n} \rightarrow \mathbb{R^m}$ a linear transformation and $h: \mathbb{R^n} \rightarrow \mathbb{R^m}$ a transformation | for some $M$ $\ge 0$, $||h(x)||$ $\le$ $M$ $||x||^2$ $\forall$ $x \in \mathbb{R^n}$. Let $f(x) = L(x) + h(x).$

Prove $f$ is differentiable in $0$. $¿Df(0)?$

Answer 1

For $x\neq 0$, let $\epsilon(x)=\frac{h(x)}{||x||}$, then we have

$0\leq ||\epsilon(x)||\leq M||x||$

thus $\lim_{x\to 0}\epsilon(x)=0$

and

$f(x)=L(x)+||x||\epsilon(x)$ with $\lim_{x\to 0}\epsilon(x)=0$ thus $f$ is differentiable at $0$ and

$Df(0)=L$.

Answer 2

We have $\|h(0)\| \le M\|0\|^2 \therefore h(0)=0$. Now:

$$\frac{\|h(0+t) - h(0) - 0\|}{\|t\|} = \frac{\|h(t)\|}{\|t\|} \le M\|t\| \to 0$$

So $h$ is differentiable at $0$ and $Dh(0) = 0$. Also $L$ is differentiable at $0$ and $DL(0) =L$. Thus $f$ is differentiable at $0$ and $Df(0) = L$.