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Suppose that we have a non negative real valued function $f$ defined only on $[0,\infty)^n$. Can one talk about the differentiability of such a function on the boundary? In the classical books on multivariable calculus, when they define differentiability of a multivariable function at a point, they always start with an assumption that the function is defined on an open neighborhood of the point. Can someone clear this for me?

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the usual definition is: there exists an extension of $f$ to an open set containing $[0,\infty)^n$, which is as differentiable as you require –  user8268 Aug 31 '12 at 7:13
    
@user8268:Can $f$ be extended to an open set containing $[0,+\infty)^n$ such that their gradients agree on the boundary, i.e., if $\textbf{v}_0$ is a point on the boundary, $\nabla g(\textbf{v}_0)=\lim_{\textbf{v}\to \textbf{v}_0}\nabla f(\textbf{v})$? –  Kumara Aug 31 '12 at 9:49

1 Answer 1

No you can't, at least it's not the usual differentiation.

It's like talking about the differentiation of $x \mapsto |x|$, for $x>0$ you can always define the derivative by:

$$ \lim_{h \to 0^+} \frac{f(x+h)-f(x)}{h}$$

In this case you get $1$. But if your map is defined in a neighborhood of $[0,+\infty)$: $(-\varepsilon, +\infty)$, for some $\varepsilon > 0$, this definition doesn't agree with the usual one.

In general, we just don't talk about differentiation in the boundary. It could be defined and continuous on $[0,+\infty)^n$ and differentiable on $(0,+\infty)^n$.

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If I know that my $f$ is defined in $[0,+\infty)^n$ but differentiable in the interior $(0,+\infty)^n$, can I get an extension $g$ such that $g$ is differentiable in an open set containing $[0,+\infty)^n$ and $f=g$ on $[0,+\infty)^n$ and the gradient of $g$ at a boundary point say $(0.5,0,0,...,0)$ is equal to $\lim{\textbf{v}\to (0.5,0,0,...,0)}\nabla f(\textbf{v})$? (I am talking only about real valued functions here) –  Kumara Aug 31 '12 at 9:19
    
Yes of course. Let's see how it works in one variable. You define $g$ on $(-\varepsilon,0)$: for $x \in (-\varepsilon,0)$ to be $f(-x)$. Then $g$ has the properties you want. –  Ilies Zidane Aug 31 '12 at 11:27
    
Can you point some reference? –  Kumara Aug 31 '12 at 12:59
    
I don't know reference about this, as you said, usually we work on open sets. Nevertheless you can also do the construction using partition of unity, and you can easily find reference on it: mainly any book on differential geometry. en.wikipedia.org/wiki/Partition_of_unity –  Ilies Zidane Aug 31 '12 at 19:15

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