Proof $\left\lVert g(x) \right\rVert_2 \leq \ln(1+\left\lVert x \right\rVert_2^2) \text{ for all x}\in \mathbb{R}^n $ differentiable

Question

Let $g:\mathbb{R}^n \to \mathbb{R}^m$ with

$$\left\lVert g(x) \right\rVert_2 \leq \ln(1+\left\lVert x \right\rVert_2^2) \text{ for all x}\in \mathbb{R}^n $$

How can I prove that $g$ is totally differentiable in $0$ and calculate it's total derivate in this point?

I don't even know what I should use for $g(x)$ since it is not given.

I know that regarding the total derivative it is calculated by

Answer 1

You must have $g(0) = 0$ since $\ln 1 = 0$. Thus $$ 0 \le \frac{\|g(x) - g(0)\|}{\|x-0\|} = \frac{\|g(x)\|}{\|x\|} \le \frac{\ln(1 + \|x\|^2)}{\|x\|}.$$ It is readily checked that $$\lim_{ \|x\| \to 0} \frac{\ln(1 + \|x\|^2)}{\|x\|} = 0,$$ so that $$\lim_{\|x\| \to 0} \frac{\|g(x) - g(0)\|}{\|x-0\|} = 0.$$ It follows that $g$ is differentiable at $0$ with $dg(0) = 0$.