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Prove or disprove if $f:\mathbb R^m \to \mathbb R^n$ is a differentiable function and $Df=0$ then $f$ is constant function

From my understanding, $Df$ is just the linearly transformation of $f$, so $Df=0$ doesn't mean that $f$ doesn't change at all, but I can't find any counter example for this.

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    $\begingroup$ Try using mean value theorem. $\endgroup$
    – hrkrshnn
    Mar 22, 2014 at 14:37
  • $\begingroup$ are you saying it's true? $\endgroup$
    – Diane Vanderwaif
    Mar 22, 2014 at 14:39
  • $\begingroup$ "$Df$ is just the linearly transformation of $f$" ???? $Df(x_0)$ is the best lineat approximation of $f$ near $x_0$. What is "the linearly transformation"? $\endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 22, 2014 at 14:39
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    $\begingroup$ "linearly transformation is also know as vector multiplication" ??????????????????????????????????????????????????????? $\endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 22, 2014 at 14:42
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    $\begingroup$ Matrix product $\ne$ vector product. $\endgroup$
    – Martín-Blas Pérez Pinilla
    Mar 22, 2014 at 14:48

2 Answers 2

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$Df$ is the mape $a \mapsto Df_a$. Forall $a$ in $\Bbb R^m$, we know that $Df_a$ is linear map from $\Bbb R^m$ to $\Bbb R^n$

To answer your question you can consider $a,b \in \Bbb R^m$ and prove that $f(a)=f(b)$.

To do this Let : $$\begin{array}{lcrcl}g &:& [0,1]& \to & \Bbb R^n \\ &&t&\mapsto&f((1-t)a+tb)\end{array}$$ for all $t \in[0,1]$, then we have:

$$(\forall t \in [0,1])\quad g'(t)=Df_{((1-t)a+tb)}(b-a)=0$$

That gives $g$ is constant in $[0,1]$ and $g(0)=g(1)$ then $f(a)=f(b)$ (since $g(0)=f(a)$ and $g(1)=f(b)$.)

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  • $\begingroup$ may I have a question, you let $g:[0,1] \to \mathbb R^n$ but $[0,1] \in \mathbb R$ not $\mathbb R^m$. is it just one dimension but not all $m$ dimension? $\endgroup$
    – Diane Vanderwaif
    Mar 23, 2014 at 12:18
  • $\begingroup$ we have using one dimension property to prove that $f(a)=f(b)$ using the mape $g$ who depends on $a,b \in \Bbb R^m$. $\endgroup$
    – Mohamed
    Mar 23, 2014 at 21:49
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For one varaible, if $\frac {df}{dx}=0$ then $\frac {f(x)-f(y)}{x-y}=f^{'}(c)=0$ by mean value so $f(x)=f(y)$.

Now,Let $F:R^n\to R^m$ then $\frac{\partial F}{\partial u}|_a=0$ since $dF=0$ where it denotes directional derivatives of $F$. From first argument $F$ is constant on given direction.Since it is true for all $\vec u$, $F$ is constant.

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