7
$\begingroup$

Let $F$ be the class of locally Lipschitz continuous functions $z=f(x,y)$, from $\mathbb R \times\mathbb R \to\mathbb R,$ such that the euclidean norm $|\ \mathrm{grad}\ f (x,y)\ |$ of its gradient vector is some function, call it $g,$ of its value, i.e. $|\ grad\ f (x,y)\ | = g(\ f(x, y)\ )\ a.e..$ Let $T$ be the subclass of those functions $f$ in $F$ which are a constant $c$ times the distance to some nonempty set, so that $|\ \mathrm{grad}\ f\ | = |\ c\ |\ a.e..$ I call $T$ the class of trivial functions in $F$. Question: are there nontrivial functions in $F$ ? That is, $F = T$ ? If yes, how does one prove it ? If not, can you give me a specific example of a nontrivial function ? Otherwise, how do you show that $F$ is larger than $T$ ? Or, more generally, what is the general form of functions in the class $F$ ?

$\endgroup$
6
  • 4
    $\begingroup$ The distance to a nonempty set doesn't have continuous partial derivatives. $(x,y) \mapsto ax+by$ does. $\endgroup$ Commented Mar 30, 2014 at 2:35
  • $\begingroup$ previous comment gives a non-trivial function, according to your definition. On the other hand distance to say two points is not differentiable, and distance to a non-empty set is 0 on this set, in particular if this set is open, then grad distance is also 0 on this set (but not outside this set). $\endgroup$
    – Mirko
    Commented Mar 30, 2014 at 3:09
  • $\begingroup$ We probably want to consider functions on open sets $U\subset\mathbb R^2$ here. $\endgroup$ Commented Mar 30, 2014 at 3:29
  • 2
    $\begingroup$ en.wikipedia.org/wiki/Eikonal_equation $\endgroup$
    – Will Jagy
    Commented Mar 30, 2014 at 4:44
  • $\begingroup$ mathoverflow.net/questions/82227/… $\endgroup$
    – Will Jagy
    Commented Mar 30, 2014 at 4:46

1 Answer 1

3
$\begingroup$

Functions in $F$ and $T$ are, at least around a level set where the gradient is nontrivial, the same up to "reparametrizations preserving level sets." To see this, assume that $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is in $F$, that $0$ is a nontrivial level set of $f$, and that $g > 0$ in a neighborhood of $0$. If we take a function $h: \mathbb{R} \rightarrow \mathbb{R}$ with $h(0) = 0$ and $h'(z) = \frac{1}{g(z)}$ then the function $\tilde{f}(x) = h(f(x))$ has the same level sets of $f$, and furthermore $$|\nabla\tilde{f}| = h'(f)|\nabla f| = 1,$$ so $\tilde{f}$ is the signed distance from its $0$ level set nearby this set. Heuristically, we "adjusted the heights of horizontal slices of $f$" so that it becomes the distance function.

As a simple example, take the simple example $f(x) = |x|^2$. Since $|\nabla f| = 2|x| = 2\sqrt{f}$ we have that $f \in F$. Then taking $h(z) = \sqrt{z}$ we transform this function to $|x|$, the distance from $0$. If we have any increasing radial function we can do the same. More generally, one can view the above discussion as saying that any $f \in F$ can, around "nondegenerate level sets," be written as some reparametrization of the distance function from this level set.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .