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Why in Euclidean space the distance function (say $f$) has the property that its $|\nabla f|=1$.

While looking for the reason I got referred to "Eikonal equation". Which I found more like a definition other than a proof, for which Euclidean distance is an especial case.

I am looking more for a sort of mathematical proof, either through discretization or calculation or a general proof to understand the reason.

Another relevant question, when talking about signed distance, is it really a distance? Can its value go till large negative or large positive values? or its value should be normalized between -1 and +1?

Soyol
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  • See page 2 here: {https://books.google.com/books?id=jewlBQAAQBAJ&pg=PR3&dq=arnold+lectures+on+partial+differential+equations&hl=en&sa=X&ved=0ahUKEwj4nuLGiL7UAhUH4D4KHRO7CjkQ6AEIKzAA#v=onepage&q=eikonal&f=false} – avs Jun 14 '17 at 19:28
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    $|\nabla f|$ is the greatest rate of increase of $f$ along a unit vector. For the function of distance from the origin, it's intuitively clear that the greatest rate of increase happens by going directly away from the origin (hence that's the direction of $\nabla f$), and that rate of increase is 1. The argument could be made rigorous using the triangle inequality. – Daniel Schepler Jun 14 '17 at 19:32

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Since $f$ is supposedly the distance function (I assume what is meant here is the function $x \mapsto |x|$), we have:

$f: (x_1,...,x_n) \mapsto \sqrt{x_1^2+...+x_n^2}$

So just taking partials, we have: $\nabla f= \Big(\frac{x_1}{\sqrt{x_1^2+...+x_n^2}}, ..., \frac{x_n}{\sqrt{x_1^2+...+x_n^2}}\Big)$

Taking the norm immediately gives $|\nabla f|=1$.

M10687
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    Thanks for the reply, it seems that it has been more straightforward to show it than I thought – Soyol Jun 14 '17 at 20:00