In the exercises sheet of a student to whom I teach, there is the following statement:

Let $\|\cdot\|$ be a norm on $\mathbb{R}^n$, $U$ be a connected open subset of $\mathbf{R}^n$ and $f\colon U\to\mathbb{R}$ be a differentiable map such that there exists $k\in\mathbf{R}$ such that for all $x\in U$ and $h\in\mathbf{R}^n$, it holds $|D_xf(h)|\leqslant k|f(x)|\|h\|$. Prove that if $f$ vanishes at one point of $U$, then it vanishes everywhere on $U$.

I am able to prove the result when the bound is $|D_xf(h)|\leqslant kf(x)\|h\|$, without the absolute value on $f$, but my proof does not work for the original statement. I even suspect that the original statement is false, but I did not succeed to cook up a counterexample in dimension $1$.

The proof for the modified statement goes as follows:

Let $Z=\{f=0\}$, then $Z$ is nonempty and closed in $U$, then it suffices to prove that $Z$ is open to conclude. Let $x\in Z$, there exists $\varepsilon>0$ such that $B(x,\varepsilon)\subset U$ and I claim that $B(x,\varepsilon)\subset Z$.

Indeed let $y\in B(x,\varepsilon)$ and introduce $g\colon [0,1]\to\mathbf{R}$ defined by: $$g(t)=f((1-t)x+ty).$$ then, $g$ is differentiable and for all $t\in[0,1]$, $|g'(t)|\leqslant\varepsilon kg(t)$.

From there it is straightforward to show, computing derivatives, that $t\in[0,1]\mapsto e^{-\varepsilon kt}g(t)$ is decreasing while $t\in[0,1]\mapsto e^{\varepsilon kt}g(t)$ is increasing, so that $g(1)=f(y)=0$ and $y\in Z$.


1 Answer 1


Consider the function $q(x)=|f(x)|$. At points $x$ where $f(x)\ne 0$ you know that $q$ is differentiable and $|D_xq(h)|\le |D_xf(h)|$ and at points where $f$ is zero, your inequality tells you that the differential of $q$ is zero. So $q$ satisfies your inequality and you can apply your proof to $q$.

  • $\begingroup$ I am probably missing something obvious, but $q$ is non differentiable which is a required assumption for my proof to apply. $\endgroup$
    – C. Falcon
    Commented Dec 28, 2019 at 14:09
  • $\begingroup$ It is differentiable. At points where $f\ne0$ it is just the chain rule. At the other points you should apply the definition of differentiability. Since the gradient is zero at those points, having the absolute value does not change anything. Let me know if I should write more details $\endgroup$
    – Gio67
    Commented Dec 28, 2019 at 16:51
  • 1
    $\begingroup$ In general it is not true that the absolute value of a differentiable function is differentiable, but if you know that at the points where the function is zero also the gradient is zero, then it is true. $\endgroup$
    – Gio67
    Commented Dec 28, 2019 at 16:53

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