Extremum of Continuously Partially Differentiable Function

Question

If $f$ is continuously partially differentiable with open non-empty domain $D\subset \mathbb{R}^N$, s.t. the determinant of its jacobian is non-zero for all $x\in D$, show the mapping: $D\rightarrow \mathbb{R}, \ \ \ x\mapsto \|f(x)\|$ has no local maximum.

Likewise, if $D$ is now compact, and $f$ has a continuous extension $g:\overline{D}\rightarrow \mathbb{R}^N$. Show the continuous map: $\overline D\rightarrow \mathbb{R}, \ \ \ x\mapsto \|g(x)\|$ has a maximum on $\partial D$. (Boundary).

I know if $D\subset \mathbb{R}^N$ is non empty, and $x_0$ is an interior point of $D$, then the function $f:D\rightarrow \mathbb{R}$ has a local maximum at $x_0$, if there is $\epsilon >0$, s.t. $B_{\epsilon}(x_0)\subset D$ and $f(x)\leq f(x_0)$ for all $x\in B_{\epsilon}(x_0)$.

I tried making use of the fact that because $f$ is continuously partially differentiable with open domain $D$, such that the determinant of its jacobian is non-zero for all $x\in D$, implies there is a neighborhood $V\subset D$ of $x_0\in D$, and $C>0$, s.t. $\|f(x)-f(x_0)\|\geq C\|x-x_0\|$. Yet I can't show $\|f(x)\|\geq \|f(x_0)\|$, which would imply there doesn't exist a local maximum.

Another thought for the first question, may be to note that because the function is locally injective for all $x\in D$, then $f$ is injective. Thus I have to prove somehow, that injective multivariable functions don't attain a local maximum.

I am not sure if this is the right way to approach the question, and any help would be much appreciated. Thanks!

Answer 1

because $f$ is continuously partially differentiable with open domain $D$, such that the function of its jacobian is non-zero for all $x\in > D$, implies there is a neighborhood $V\subset D$ of $x_0\in D$, and $C>0$, s.t. $\|f(x)-f(x_0)\|\geq C\|x-x_0\|$. Yet I can't show $\|f(x)\|\geq \|f(x_0)\|$, which would imply there doesn't exist a local maximum.

The argument would go like this: Consider $\delta$ such that the ball $B_{\delta} \subset V$. Now consider $x_1=x_0+\delta e_1$ and $x_2=x_0-\delta e_1$ Then $f(x_1)- f(x_o)$ and $f(x_2)- f(x_o)$ are both non-zero as you have noted since the determinant is non-zero. And the two have opposite signs. That is, one of $f(x_1)$ or $f(x_2)$ must be greater than $f(x_0)$.

Answer 2

Assume that $x_{0}\in D$ is a local maximum of $\varphi:x\rightarrow\|f(x)\|$.

By Inverse Function Theorem, there is a small $\delta>0$ such that $f:B_{\delta}(x_{0})\rightarrow U$ is a $C^{1}$ one-to-one and onto map.

Now for small $r>0$, one has $f(x_{0})+\dfrac{f(x_{0})}{\|f(x_{0})\|}r\in U$, so $f(\eta)=f(x_{0})+\dfrac{f(x_{0})}{\|f(x_{0})\|}r$ for some $\eta\in B_{\delta}(x_{0})$.

Now $\varphi(\eta)=\|f(\eta)\|\leq\varphi(x_{0})=\|f(x_{0})\|$, but $\|f(\eta)\|=\|f(x_{0})\|\cdot\left|1+\dfrac{r}{\|f(x_{0})\|}\right|>\|f(x_{0})\|$, a contradiction.

Note that $f(x_{0})\ne 0$, if not, $f$ is not one-to-one on $B_{\delta}(x_{0})$.

For the second question, $g$ must attain its maximum on $\overline{D}$, but by the first result, the point cannot be attained in $D$.