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1

$begingroup$

Let $U$ be an open subset of some finite dimensional Euclidean space $mathbb{R}^n$.
Suppose $U$ has a $C^1$ boundary.
This means that for each $x in partial U$, there exists an $r>0$ and a $C^1$ function defined on some open subset of $mathbb{R}^{n-1}$ so that $B(x,r) cap U = B(x,r) cap {(x,y):y>f(x)}$ and $B(x,r) cap partial U = B(x,r) cap {(x,y):y=f(x)} $ after "changing the order of coordinates" if needed.

Now, Let $f: U rightarrow mathbb{R}$ be a $C^1$ map where $f$ and each of its partial derivatives may be extended continuously to $bar{U}$. Let $c:Irightarrow mathbb{R}^n$ be a $C^1$ map where $I$ is an open interval and $c(I) subseteq bar{U}$. Then can we say that the chain rule holds? i.e. $ frac{d}{dt} f(c(t)) = nabla f (c(t))c' (t)$ where $nabla(f)$ is (uniquely) extended to $bar{U}$?

I think that this problem is related to some kind of "smooth extension theorem"... maybe?

analysis

share|cite|improve this question

edited Mar 13 at 7:04

J. Doe

asked Mar 13 at 6:49

J. DoeJ. Doe

888

$endgroup$

add a comment | 

0

1

$begingroup$

Let $U$ be an open subset of some finite dimensional Euclidean space $mathbb{R}^n$.
Suppose $U$ has a $C^1$ boundary.
This means that for each $x in partial U$, there exists an $r>0$ and a $C^1$ function defined on some open subset of $mathbb{R}^{n-1}$ so that $B(x,r) cap U = B(x,r) cap {(x,y):y>f(x)}$ and $B(x,r) cap partial U = B(x,r) cap {(x,y):y=f(x)} $ after "changing the order of coordinates" if needed.

Now, Let $f: U rightarrow mathbb{R}$ be a $C^1$ map where $f$ and each of its partial derivatives may be extended continuously to $bar{U}$. Let $c:Irightarrow mathbb{R}^n$ be a $C^1$ map where $I$ is an open interval and $c(I) subseteq bar{U}$. Then can we say that the chain rule holds? i.e. $ frac{d}{dt} f(c(t)) = nabla f (c(t))c' (t)$ where $nabla(f)$ is (uniquely) extended to $bar{U}$?

I think that this problem is related to some kind of "smooth extension theorem"... maybe?

analysis

share|cite|improve this question

edited Mar 13 at 7:04

J. Doe

asked Mar 13 at 6:49

J. DoeJ. Doe

888

$endgroup$

add a comment | 

$begingroup$

Let $U$ be an open subset of some finite dimensional Euclidean space $mathbb{R}^n$.
Suppose $U$ has a $C^1$ boundary.
This means that for each $x in partial U$, there exists an $r>0$ and a $C^1$ function defined on some open subset of $mathbb{R}^{n-1}$ so that $B(x,r) cap U = B(x,r) cap {(x,y):y>f(x)}$ and $B(x,r) cap partial U = B(x,r) cap {(x,y):y=f(x)} $ after "changing the order of coordinates" if needed.

Now, Let $f: U rightarrow mathbb{R}$ be a $C^1$ map where $f$ and each of its partial derivatives may be extended continuously to $bar{U}$. Let $c:Irightarrow mathbb{R}^n$ be a $C^1$ map where $I$ is an open interval and $c(I) subseteq bar{U}$. Then can we say that the chain rule holds? i.e. $ frac{d}{dt} f(c(t)) = nabla f (c(t))c' (t)$ where $nabla(f)$ is (uniquely) extended to $bar{U}$?

I think that this problem is related to some kind of "smooth extension theorem"... maybe?

analysis

share|cite|improve this question

edited Mar 13 at 7:04

J. Doe

asked Mar 13 at 6:49

J. DoeJ. Doe

888

$endgroup$

share|cite|improve this question

edited Mar 13 at 7:04

J. Doe

asked Mar 13 at 6:49

J. DoeJ. Doe

888

share|cite|improve this question

edited Mar 13 at 7:04

J. Doe

asked Mar 13 at 6:49

J. DoeJ. Doe

888

asked Mar 13 at 6:49

J. DoeJ. Doe

888

asked Mar 13 at 6:49

J. DoeJ. Doe

888