Mean Value Theorem inequality

Question

Suppose $\mathbb{R}^n$ space with norm $\|\cdot\|$ and $I$ open interval and $f: I \rightarrow \mathbb{R}^n$ derivable function.

let $k$ be a constant and $x_0$ element in $I$.

we have $\|f\,'(x)\|\leq k\|f(x)\|$ for every $x \in I$ and we have $ f(x_0)=0$

show that (by applying the mean value theorem on $f(x)$ in the interval $[x_0-h, x_0+h]$ for $h$ very small )there exist $h>0$ such that $f$ is equal zero on the interval $[x_0-h, x_0+h]$.

Am not sure how to make it, hope to get a hint.

Thanks alot

Sahar

Answer 1

Since all norms on $\mathbb{R}^n$ are equivalent to the infinity norm, we'll assume that $||\cdot ||$ is the infinity norm. We denote the components of the vector valued function $f$ by $f_1,\ldots,f_n$.

Set $h<1/k$ and small enough such that $f$ is defined on $[x_0-h,x_0+h]$. Pick $y\in [x_0-h,x_0+h]$ such that $|f_i(y)|=\max_{z\in[x_0-h,x_0+h]}||f(x)||$ for some $i$. By the mean value theorem there exists a $z\in(x_0,y)$ with $|f_i'(z)(y-x_0)|=|f_i(y)|$. So if $f_i(y)\neq 0$ then $||f'(z)||\geq |f_i'(z)|>k|f_i(y)|=k||f(y)||\geq k||f(z)||$, a contradiction. Hence $f_i(y)=0$ which implies that $f$ is constantly $0$ on the interval.