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Suppose we have a function $f : \Re^N \rightarrow \Re$ which, given a vector, returns the value of its smallest element. How can I approximate $f$ with a differentiable function(s)?

flag	asked Aug 11 2010 at 5:33 eakbas 121●6

The tag of the question is probably wrong. Sorry for that. – eakbas Aug 11 2010 at 5:45

Aaron Bergman · Accepted Answer · 2010-08-11 05:54:43Z UTC

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If signs aren't a big deal, use the generalized mean formula

$$ \left(\frac{1}{n}\sum x_i^k\right)^{1/k} $$

for $k\to -\infty$.

edited Aug 11 2010 at 5:54

answered Aug 11 2010 at 5:47

Aaron Bergman
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Nice, but you probably want to get rid of 1/n so that the result is as close to the smallest x_i as possible? – Neil Aug 11 2010 at 7:42

@Neil: The $1/n$ normalization is the standard one since it gives exactly what you want when all the $x_i$ are equal. But in fact if $n$ is fixed and $k$ approaches negative infinity, it doesn't matter asymptotically what constant you put inside the parentheses. – Tracy Hall Aug 11 2010 at 8:28

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Since the function is piecewise linear, probably the most efficient smooth approximation is just by convolution with standard mollifiers $\epsilon^-n \rho(x/\epsilon)$, this simply 'rounds the corners' and leaves the function unchanged at most points. But the OP should really clarify what he needs. – Piero D'Ancona Aug 11 2010 at 11:10

Neil · Answer 2 · 2010-08-11 05:41:59Z UTC

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What about $f: [x_1 \dots\ x_n] \mapsto \frac{1}{\sum_i x_i^{-1}}$?

answered Aug 11 2010 at 5:41

Neil
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dmuir · Answer 3 · 2012-12-20 15:52:23Z UTC

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For two dimensions, we have min(x,y) = (x+y-|x-y|)/2, so you just need a differentiable approximation to x -> |x|. Then for higher dimesions we have min(x,y,z) = min(x, min(y,z)) etc.

answered Dec 20 at 15:52

dmuir
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A differentiable approximation to the minimum function

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