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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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    @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, $\nabla g(\textbf{v}_0)=\lim_{\textbf{v}\to \textbf{v}_0}\nabla f(\textbf{v})$?2012-08-31

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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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    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. http://en.wikipedia.org/wiki/Partition_of_unity2012-08-31