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Differentiability on the boundary
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Equivalent condition for differentiability on partial derivativesWill the rules of calculus stay the same when a real-valued function is defined over infinite number of variables?Which is the correct definition of stationary point for real-valued functions in Euclidean space?Differentiability conditions for a piecewise functionDifferentiability: Partially Defined FunctionsWhy is continuous differentiability required?The set of differentiability of an extension from half-plane to the planeContinuously differentiable function on the closed interval and Newton-Leibniz formulaQuestion regarding complex differentiability and vector differentiability from $R rightarrow R^2$existence of a non-negative smooth function on a neighborhood of point on a boundary of a smooth manifold.
$begingroup$
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?
real-analysis multivariable-calculus
asked Aug 31 '12 at 5:29
KumaraKumara
221118
$endgroup$
add a comment |
$begingroup$
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?
real-analysis multivariable-calculus
asked Aug 31 '12 at 5:29
KumaraKumara
221118
$endgroup$
$begingroup$
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
$endgroup$
– user8268
Aug 31 '12 at 7:13
$begingroup$
@user8268:Can $f$ be extended to an open set containing $[0,+infty)^n$ such that their gradients agree on the boundary, i.e., if $textbfv_0$ is a point on the boundary, $nabla g(textbfv_0)=lim_textbfvto textbfv_0nabla f(textbfv)$?
$endgroup$
– Kumara
Aug 31 '12 at 9:49
add a comment |
$begingroup$
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?
real-analysis multivariable-calculus
asked Aug 31 '12 at 5:29
KumaraKumara
221118
$endgroup$
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?
real-analysis multivariable-calculus
real-analysis multivariable-calculus
asked Aug 31 '12 at 5:29
KumaraKumara
221118
asked Aug 31 '12 at 5:29
KumaraKumara
221118
asked Aug 31 '12 at 5:29
KumaraKumara
221118
asked Aug 31 '12 at 5:29
KumaraKumara
221118
asked Aug 31 '12 at 5:29
KumaraKumara
221118
221118
$begingroup$
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
$endgroup$
– user8268
Aug 31 '12 at 7:13
$begingroup$
@user8268:Can $f$ be extended to an open set containing $[0,+infty)^n$ such that their gradients agree on the boundary, i.e., if $textbfv_0$ is a point on the boundary, $nabla g(textbfv_0)=lim_textbfvto textbfv_0nabla f(textbfv)$?
$endgroup$
– Kumara
Aug 31 '12 at 9:49
add a comment |
$begingroup$
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
$endgroup$
– user8268
Aug 31 '12 at 7:13
$begingroup$
@user8268:Can $f$ be extended to an open set containing $[0,+infty)^n$ such that their gradients agree on the boundary, i.e., if $textbfv_0$ is a point on the boundary, $nabla g(textbfv_0)=lim_textbfvto textbfv_0nabla f(textbfv)$?
$endgroup$
– Kumara
Aug 31 '12 at 9:49
$begingroup$
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
$endgroup$
– user8268
Aug 31 '12 at 7:13
$begingroup$
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
$endgroup$
– user8268
Aug 31 '12 at 7:13
$begingroup$
@user8268:Can $f$ be extended to an open set containing $[0,+infty)^n$ such that their gradients agree on the boundary, i.e., if $textbfv_0$ is a point on the boundary, $nabla g(textbfv_0)=lim_textbfvto textbfv_0nabla f(textbfv)$?
$endgroup$
– Kumara
Aug 31 '12 at 9:49
$begingroup$
@user8268:Can $f$ be extended to an open set containing $[0,+infty)^n$ such that their gradients agree on the boundary, i.e., if $textbfv_0$ is a point on the boundary, $nabla g(textbfv_0)=lim_textbfvto textbfv_0nabla f(textbfv)$?
$endgroup$
– Kumara
Aug 31 '12 at 9:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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:
$$ lim_h to 0^+ fracf(x+h)-f(x)h$$
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 $(0,+infty)^n$.
answered Aug 31 '12 at 8:40
Ilies ZidaneIlies Zidane
87145
$endgroup$
$begingroup$
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 $limtextbfvto (0.5,0,0,...,0)nabla f(textbfv)$? (I am talking only about real valued functions here)
$endgroup$
– Kumara
Aug 31 '12 at 9:19
$begingroup$
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.
$endgroup$
– Ilies Zidane
Aug 31 '12 at 11:27
$begingroup$
Can you point some reference?
$endgroup$
– Kumara
Aug 31 '12 at 12:59
$begingroup$
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
$endgroup$
– Ilies Zidane
Aug 31 '12 at 19:15
add a comment |
$begingroup$
I think the answer by Ilies Zidane above does not meet the OP question. There are two different settings here:
$(1)$ To extend the derivative $ f': , ]0, infty[ to mathbbR $ of a continuous function $ f: [0, infty[ , to mathbbR $ differentiable on the open interval $ ]0, infty[ $ and defined only for nonnegative real numbers (hence nothing is assumed in the negative part of the real line) and
$(2)$ To talk about the derivative of a function $ g: , ] varepsilon , infty [ , to mathbbR $ (where $ varepsilon > 0 $) whose restriction $ g|_[0, infty[ : [0, infty[ , to mathbbR $ is continuous and differentiable on the open interval $ ]0, infty[ $.
In the first case we can extend the function $f$ in any way that suit us. In the second case there is an unique way we can extend $ g|_[0, infty[ $ in order to recover the already defined function $g , $. As an example, in (1) we begin with $f$ defined as $ f(x)=x=|x| $ and we can extend it as $ x mapsto x $ for nonnegative numbers. But in (2), if $ g(x)=|x| $, the same extension applied to $ g|_[0, infty[ $ does not recover $g , $.
answered Mar 27 at 14:43
GustavoGustavo
866621
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
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:
$$ lim_h to 0^+ fracf(x+h)-f(x)h$$
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 $(0,+infty)^n$.
answered Aug 31 '12 at 8:40
Ilies ZidaneIlies Zidane
87145
$endgroup$
$begingroup$
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 $limtextbfvto (0.5,0,0,...,0)nabla f(textbfv)$? (I am talking only about real valued functions here)
$endgroup$
– Kumara
Aug 31 '12 at 9:19
$begingroup$
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.
$endgroup$
– Ilies Zidane
Aug 31 '12 at 11:27
$begingroup$
Can you point some reference?
$endgroup$
– Kumara
Aug 31 '12 at 12:59
$begingroup$
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
$endgroup$
– Ilies Zidane
Aug 31 '12 at 19:15
add a comment |
$begingroup$
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:
$$ lim_h to 0^+ fracf(x+h)-f(x)h$$
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 $(0,+infty)^n$.
answered Aug 31 '12 at 8:40
Ilies ZidaneIlies Zidane
87145
$endgroup$
$begingroup$
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 $limtextbfvto (0.5,0,0,...,0)nabla f(textbfv)$? (I am talking only about real valued functions here)
$endgroup$
– Kumara
Aug 31 '12 at 9:19
$begingroup$
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.
$endgroup$
– Ilies Zidane
Aug 31 '12 at 11:27
$begingroup$
Can you point some reference?
$endgroup$
– Kumara
Aug 31 '12 at 12:59
$begingroup$
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
$endgroup$
– Ilies Zidane
Aug 31 '12 at 19:15
add a comment |
$begingroup$
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:
$$ lim_h to 0^+ fracf(x+h)-f(x)h$$
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 $(0,+infty)^n$.
answered Aug 31 '12 at 8:40
Ilies ZidaneIlies Zidane
87145
$endgroup$
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:
$$ lim_h to 0^+ fracf(x+h)-f(x)h$$
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 $(0,+infty)^n$.
answered Aug 31 '12 at 8:40
Ilies ZidaneIlies Zidane
87145
answered Aug 31 '12 at 8:40
Ilies ZidaneIlies Zidane
87145
answered Aug 31 '12 at 8:40
Ilies ZidaneIlies Zidane
87145
answered Aug 31 '12 at 8:40
Ilies ZidaneIlies Zidane
87145
87145
$begingroup$
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 $limtextbfvto (0.5,0,0,...,0)nabla f(textbfv)$? (I am talking only about real valued functions here)
$endgroup$
– Kumara
Aug 31 '12 at 9:19
$begingroup$
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.
$endgroup$
– Ilies Zidane
Aug 31 '12 at 11:27
$begingroup$
Can you point some reference?
$endgroup$
– Kumara
Aug 31 '12 at 12:59
$begingroup$
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
$endgroup$
– Ilies Zidane
Aug 31 '12 at 19:15
add a comment |
$begingroup$
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 $limtextbfvto (0.5,0,0,...,0)nabla f(textbfv)$? (I am talking only about real valued functions here)
$endgroup$
– Kumara
Aug 31 '12 at 9:19
$begingroup$
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.
$endgroup$
– Ilies Zidane
Aug 31 '12 at 11:27
$begingroup$
Can you point some reference?
$endgroup$
– Kumara
Aug 31 '12 at 12:59
$begingroup$
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
$endgroup$
– Ilies Zidane
Aug 31 '12 at 19:15
$begingroup$
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 $limtextbfvto (0.5,0,0,...,0)nabla f(textbfv)$? (I am talking only about real valued functions here)
$endgroup$
– Kumara
Aug 31 '12 at 9:19
$begingroup$
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 $limtextbfvto (0.5,0,0,...,0)nabla f(textbfv)$? (I am talking only about real valued functions here)
$endgroup$
– Kumara
Aug 31 '12 at 9:19
$begingroup$
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.
$endgroup$
– Ilies Zidane
Aug 31 '12 at 11:27
$begingroup$
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.
$endgroup$
– Ilies Zidane
Aug 31 '12 at 11:27
$begingroup$
Can you point some reference?
$endgroup$
– Kumara
Aug 31 '12 at 12:59
$begingroup$
Can you point some reference?
$endgroup$
– Kumara
Aug 31 '12 at 12:59
$begingroup$
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
$endgroup$
– Ilies Zidane
Aug 31 '12 at 19:15
$begingroup$
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
$endgroup$
– Ilies Zidane
Aug 31 '12 at 19:15
add a comment |
$begingroup$
I think the answer by Ilies Zidane above does not meet the OP question. There are two different settings here:
$(1)$ To extend the derivative $ f': , ]0, infty[ to mathbbR $ of a continuous function $ f: [0, infty[ , to mathbbR $ differentiable on the open interval $ ]0, infty[ $ and defined only for nonnegative real numbers (hence nothing is assumed in the negative part of the real line) and
$(2)$ To talk about the derivative of a function $ g: , ] varepsilon , infty [ , to mathbbR $ (where $ varepsilon > 0 $) whose restriction $ g|_[0, infty[ : [0, infty[ , to mathbbR $ is continuous and differentiable on the open interval $ ]0, infty[ $.
In the first case we can extend the function $f$ in any way that suit us. In the second case there is an unique way we can extend $ g|_[0, infty[ $ in order to recover the already defined function $g , $. As an example, in (1) we begin with $f$ defined as $ f(x)=x=|x| $ and we can extend it as $ x mapsto x $ for nonnegative numbers. But in (2), if $ g(x)=|x| $, the same extension applied to $ g|_[0, infty[ $ does not recover $g , $.
answered Mar 27 at 14:43
GustavoGustavo
866621
$endgroup$
add a comment |
$begingroup$
I think the answer by Ilies Zidane above does not meet the OP question. There are two different settings here:
$(1)$ To extend the derivative $ f': , ]0, infty[ to mathbbR $ of a continuous function $ f: [0, infty[ , to mathbbR $ differentiable on the open interval $ ]0, infty[ $ and defined only for nonnegative real numbers (hence nothing is assumed in the negative part of the real line) and
$(2)$ To talk about the derivative of a function $ g: , ] varepsilon , infty [ , to mathbbR $ (where $ varepsilon > 0 $) whose restriction $ g|_[0, infty[ : [0, infty[ , to mathbbR $ is continuous and differentiable on the open interval $ ]0, infty[ $.
In the first case we can extend the function $f$ in any way that suit us. In the second case there is an unique way we can extend $ g|_[0, infty[ $ in order to recover the already defined function $g , $. As an example, in (1) we begin with $f$ defined as $ f(x)=x=|x| $ and we can extend it as $ x mapsto x $ for nonnegative numbers. But in (2), if $ g(x)=|x| $, the same extension applied to $ g|_[0, infty[ $ does not recover $g , $.
answered Mar 27 at 14:43
GustavoGustavo
866621
$endgroup$
add a comment |
$begingroup$
I think the answer by Ilies Zidane above does not meet the OP question. There are two different settings here:
$(1)$ To extend the derivative $ f': , ]0, infty[ to mathbbR $ of a continuous function $ f: [0, infty[ , to mathbbR $ differentiable on the open interval $ ]0, infty[ $ and defined only for nonnegative real numbers (hence nothing is assumed in the negative part of the real line) and
$(2)$ To talk about the derivative of a function $ g: , ] varepsilon , infty [ , to mathbbR $ (where $ varepsilon > 0 $) whose restriction $ g|_[0, infty[ : [0, infty[ , to mathbbR $ is continuous and differentiable on the open interval $ ]0, infty[ $.
In the first case we can extend the function $f$ in any way that suit us. In the second case there is an unique way we can extend $ g|_[0, infty[ $ in order to recover the already defined function $g , $. As an example, in (1) we begin with $f$ defined as $ f(x)=x=|x| $ and we can extend it as $ x mapsto x $ for nonnegative numbers. But in (2), if $ g(x)=|x| $, the same extension applied to $ g|_[0, infty[ $ does not recover $g , $.
answered Mar 27 at 14:43
GustavoGustavo
866621
$endgroup$
I think the answer by Ilies Zidane above does not meet the OP question. There are two different settings here:
$(1)$ To extend the derivative $ f': , ]0, infty[ to mathbbR $ of a continuous function $ f: [0, infty[ , to mathbbR $ differentiable on the open interval $ ]0, infty[ $ and defined only for nonnegative real numbers (hence nothing is assumed in the negative part of the real line) and
$(2)$ To talk about the derivative of a function $ g: , ] varepsilon , infty [ , to mathbbR $ (where $ varepsilon > 0 $) whose restriction $ g|_[0, infty[ : [0, infty[ , to mathbbR $ is continuous and differentiable on the open interval $ ]0, infty[ $.
In the first case we can extend the function $f$ in any way that suit us. In the second case there is an unique way we can extend $ g|_[0, infty[ $ in order to recover the already defined function $g , $. As an example, in (1) we begin with $f$ defined as $ f(x)=x=|x| $ and we can extend it as $ x mapsto x $ for nonnegative numbers. But in (2), if $ g(x)=|x| $, the same extension applied to $ g|_[0, infty[ $ does not recover $g , $.
answered Mar 27 at 14:43
GustavoGustavo
866621
answered Mar 27 at 14:43
GustavoGustavo
866621
answered Mar 27 at 14:43
GustavoGustavo
866621
answered Mar 27 at 14:43
GustavoGustavo
866621
866621
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$begingroup$
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
$endgroup$
– user8268
Aug 31 '12 at 7:13
$begingroup$
@user8268:Can $f$ be extended to an open set containing $[0,+infty)^n$ such that their gradients agree on the boundary, i.e., if $textbfv_0$ is a point on the boundary, $nabla g(textbfv_0)=lim_textbfvto textbfv_0nabla f(textbfv)$?
$endgroup$
– Kumara
Aug 31 '12 at 9:49