Function whose gradient is of constant norm

Question

Let $f:\mathbb R^n\rightarrow \mathbb R$ be a smooth function such that $\|\nabla f(x)\|=1$ for all $x\in \mathbb R^n$ and $f(0)=0$.

I would like to prove that $f$ is linear.

I first looked at the solution of the O.D.E. $$\dfrac{d\gamma}{dt}(t)=\nabla f(\gamma(t))$$ I noticed that there exists only one solution $\gamma_x$ passing through $x\in \mathbb R^n$ at $t=0$, and such a solution is defined over $\mathbb R$. I also proved that $$\gamma_x(t)=x+t\nabla f(x)$$ and $$f(\gamma_x(t))=f(x)+t.$$

Furthermore, the following hint is given :

Show that if $f(x)=f(y)$ then $\langle \nabla f(x),x-y\rangle=\langle \nabla f(y),x-y\rangle=0$.

I did it, but now, I don't understand why it follows that $f$ is linear.

Edit: I write here the proof of the hint;

Let $c:[0,1]\rightarrow \mathbb R^n$ be the (usual) parametrization of the segment $[x+t\nabla f(x),y]$. Since $f(x)=f(y)$, we get $$\begin{align*}|t|=|f(\gamma_x(t))-f(x)| & =|f(x+t\nabla f(x))-f(y)|\\ & =|\int_0^1 \langle \nabla f(c(s)),c'(s)\rangle ds| \\ & \leq \int_0^1 \|c'(s)\|ds \\ & =\|y-x-t\nabla f(x)\|\end{align*}$$ From this inequality, which is true for any $t\in \mathbb R$ and using the fact that $\nabla f(x)$ is a unitary vector, it follows that : $$t^2\leq \|x-y\|^2 + t^2 +2t\langle x-y,\nabla f(x) \rangle$$ Dividing by $t$ and taking limits to $\pm \infty$ gives the hint.

Answer 1

From your tags, I suppose I can use some geometric argument here.

Because of the condition $\|\nabla f\| = 1$, the level set

$$M = M_c = \{x\in \mathbb R^n: f(x) = c\}$$

is a smooth hypersurface for all $c\in \mathbb R$ and for all $x\in M$,

$$T_xM = \{ y + x \in \mathbb R^n : \langle y, \nabla f (x) \rangle = 0\}.$$

Let $x\in M$ and $y \in M$ be closed to $x$. The condition

$$\langle \nabla f(x), x-y \rangle = 0$$

says that $x-y$ lie in the tangent space of $T_x M$ for all $y\in M$. Then $y\in T_xM$ for all $y$ closed to $x$ (Some arguments are needed here). With some more argument, one can show that $M = T_xM$ for all $x\in M$. This mean that

• For all $c$, $M_c = V + cw$, where $w\in \mathbb R^n$ is a vector with norm one and $V = \{ x: \langle x, w\rangle = 0\}$,
• $f(u+ tw) = g(t)$, where $u\in V$ and $g:\mathbb R\to \mathbb R$ and $g'(t)=1$ for all $t$.

The second claim show that $f$ is linear.

Answer 2

To begin with, the fact that $\gamma$ is solution of the above ODE and is given by $\gamma(t)=x+t\nabla f(t)$ provides for all $t$: $$ \nabla f(x+t\nabla f(x)) = \nabla f(\gamma(t)) = \gamma'(t)=\nabla f(x).$$

We now use the hint. For arbitray $x,y\in\mathbb{R}^d$, with $t$ such that $$f(x+t\nabla f(x))=f(x)+t=f(y),$$ we deduce using the above property: $$ <\nabla f(x),x+t\nabla f(x) -y> =0,$$ hence when re-arranging: $$ f(y) = f(x) + <\nabla f(x) , y-x >,$$ and the function is affine.

Another way to conclude is to consider that for all $s\in\mathbb{R}$, $$ f(x+(t+s)\nabla f(x))=f(y+s\nabla f(y)),$$ hence $$ <\nabla f(x),x+t\nabla f(x) -y> + s <\nabla f(x), \nabla f(x)-\nabla f(y)> =0,$$ and by the equality case of Cauchy-Schwartz' inequality, $$ <\nabla f(x), \nabla f(x)> \ = \ <\nabla f(x),\nabla f(y)> \quad \implies\quad \nabla f(x)=\nabla f(y).$$

Answer 3

I know this is a very old question, but the claim is not true. Take $f(x) = \|x\|$, which has gradient $\nabla f(x) = x/\|x\|$. Obviously, $\|\nabla f(x)\|=1$, but the function does not satisfy the properties of a linear function.