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?
Differentiability on the boundary
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real-analysis
multivariable-calculus
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0the usual definition is: there exists an extension of $f$ to an open set containing $[0,\infty)^n$, which is as differentiable as you require – 2012-08-31
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0@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