2
$\begingroup$

I came across the following problem:

Let $f: \mathbb{R}^n \to \mathbb{R}$ be partial differentiable to every variable and let $\nabla f(x)=0, \forall x \in \mathbb{R}^n $

Proof that $f$ is constant.

Intuitively I understand that this is true and I sort of understand how to proof it. For a function from $\mathbb{R}$ to $\mathbb{R}$ I know this is true, but now I need to go to a N-dimensional function.

My idea is as follows. Let $x,y \in \mathbb{R}^n$, now we need to show that $f(x)=f(y)$. For the one dimensional case this is already proven. So I would start by proving that $f(x_1,...,x_n) =f(y_1,x_2,...,x_n)$ and then the next step would be $f(y_1,x_2,...,x_n)=f(y_1,y_2,x_3...,x_n)$ etc.

How would I do this formally and is this the correct approach?

My second question is what would happen if we have a function $g : \mathbb{R}^n \to \mathbb{R}^p$ and the total derivative $Dg(x)=0, \forall x\in \mathbb{R}^n$. This function $g(x)$ should also be constant. How can I use this question to proof that?

My thoughts on this case are that I should use a family of functions defined by:

$\phi _i: \mathbb{R}^n \to \mathbb{R}, x \mapsto g_i(x)$

Then by using the first question all $g_i(x)$ are constant, but can I than say that the composition of all $g_i(x)$ is also constant?

$\endgroup$
  • 1
    $\begingroup$ I think you could use the mean value theorem: en.wikipedia.org/wiki/… (and the next paragraph for your second question) $\endgroup$ – Augustin Sep 16 '15 at 8:44
  • 1
    $\begingroup$ Yes I've been told that the mean value theorem can also be used, but the method I describe should also be possible (perhaps a little harder). I would really like to know how to do it my way $\endgroup$ – DeanTheMachine Sep 16 '15 at 8:53
2
$\begingroup$

Your approach is correct and "formal" enough as it stands. But there is a more elegant solution: Since all partial derivatives are $\equiv0$ they are in particular continuous, which implies that $f$ is differentiable in the "proper" sense, so that we may apply the chain rule. Given any two points $x$, $y\in{\mathbb R}^n$ consider the auxiliary function $$\phi(t):=f\bigl((1-t)x+ty\bigr)\qquad(0\leq t\leq1)\ .$$ Then $$\phi'(t)=\nabla f\bigl((1-t)x+ty\bigr)\cdot(y-x)\equiv0\ ,$$ and this implies $f(y)=\phi(1)=\phi(0)=f(x)$.

Concerning your second question: A vector valued function $g$ whose component functions $g_i$ are constant, is of course constant.

$\endgroup$
2
$\begingroup$

Let $x_{0} \in \mathbb{R}^{n}$ and let $A := \{ x \in \mathbb{R}^{n} \mid f(x) = f(x_{0}) \}$. Since $f(x_{0}) = f(x_{0})$, so $A \neq \varnothing$. Since $f$ is differentiable by assumption, so $f$ is continuous, and hence $A$ is closed in $\mathbb{R}^{n}$.

Let $x \in A$; let $U \subset \mathbb{R}^{n}$ be an open ball centered at $x$; and let $L(x,y)$ be the line segment joining $x$ and $y$ for all $y \in U$. Then $L(x,y) \subset U$ for all $y \in U$. Since $\nabla f = 0$ on $\mathbb{R}^{n}$ by assumption, so $ | f(y) - f(x) | = 0 $ for all $y \in U$ by the mean-value theorem, whence $f(y) = f(x) = f(x_{0})$ for all $y \in U$. This says that $A$ is open in $\mathbb{R}^{n}$. Since $\mathbb{R}^{n}$ is connected, and since $A \neq \varnothing$ and is clopen in $\mathbb{R}^{n}$, it follows that $A = \mathbb{R}^{n}$, so $f$ is constant on $\mathbb{R}^{n}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.