Definition of smooth functions on arbitrary subsets of $mathbb{R}^n$ and partial derivatives
$begingroup$
Let $A$ be an arbitrary subset of $mathbb{R}^n$, and let $f:Ato mathbb{R}$ be a function.
We say that $f$ is smooth if for each point $p$ in $A$ there exists an open subset $U$ of $mathbb{R}^n$ containing $p$ and a smooth map $g:Uto mathbb{R}$ such that $f$ and $g$ agrees on $A cap U$.
Now let $U$ be an open subset of $mathbb{H}^n$, with $mathbb{H}^n={xin mathbb{R}^n:x^nge0 }$
Let $f:Uto mathbb{R}$ be a smooth function in the above sense.
Let $pin U cap partialmathbb{H}^n$. Thus there are $tilde U$ open subset of $mathbb{R}^n$ and a smooth map $g:tilde Uto mathbb{R}$ such that $g$ and $f$ agrees on $U cap tilde U$.
I want to show that the partial derivatives of $f$ at $p$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension i.e. of the choice of $g$.
Here is my argument.
$$lim_{tto0^+}frac{f(p+te_i)-f(p)}{t}=lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}qquad [1]$$ beacuse $f(p)=g(p)$ and for sufficient positive small $t$ we have $p+te_i in U cap tilde U$ and so $f(p+te_i)=g(p+te_i)$.
Now, since $g$ is smooth at $p$ by hypothesis, we have that the right hand side of the above equation exists, and we have
$$lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}=lim_{tto0^-}frac{g(p+te_i)-g(p)}{t}=:partial_i|_p g$$
So I can define $partial_i|_p f:=partial_i|_p g$ and this value is indipendent of the choice of $g$ by $[1]$.
So I have shown the claim in the above yellow box.
Is my argument right? In John M. Lee Textbook Introduction to smooth manifolds, on page 27, he says that By continuity, all partial derivatives of $f$ at points of $Ucap partialmathbb{H}^n$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension.
But it seems I dont'use the continuity of any function, so I'm not convinced of having understood the matter.
multivariable-calculus differential-geometry smooth-manifolds smooth-functions
asked Jan 8 at 17:31
MinatoMinato
595314
$endgroup$
add a comment |
$begingroup$
Let $A$ be an arbitrary subset of $mathbb{R}^n$, and let $f:Ato mathbb{R}$ be a function.
We say that $f$ is smooth if for each point $p$ in $A$ there exists an open subset $U$ of $mathbb{R}^n$ containing $p$ and a smooth map $g:Uto mathbb{R}$ such that $f$ and $g$ agrees on $A cap U$.
Now let $U$ be an open subset of $mathbb{H}^n$, with $mathbb{H}^n={xin mathbb{R}^n:x^nge0 }$
Let $f:Uto mathbb{R}$ be a smooth function in the above sense.
Let $pin U cap partialmathbb{H}^n$. Thus there are $tilde U$ open subset of $mathbb{R}^n$ and a smooth map $g:tilde Uto mathbb{R}$ such that $g$ and $f$ agrees on $U cap tilde U$.
I want to show that the partial derivatives of $f$ at $p$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension i.e. of the choice of $g$.
Here is my argument.
$$lim_{tto0^+}frac{f(p+te_i)-f(p)}{t}=lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}qquad [1]$$ beacuse $f(p)=g(p)$ and for sufficient positive small $t$ we have $p+te_i in U cap tilde U$ and so $f(p+te_i)=g(p+te_i)$.
Now, since $g$ is smooth at $p$ by hypothesis, we have that the right hand side of the above equation exists, and we have
$$lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}=lim_{tto0^-}frac{g(p+te_i)-g(p)}{t}=:partial_i|_p g$$
So I can define $partial_i|_p f:=partial_i|_p g$ and this value is indipendent of the choice of $g$ by $[1]$.
So I have shown the claim in the above yellow box.
Is my argument right? In John M. Lee Textbook Introduction to smooth manifolds, on page 27, he says that By continuity, all partial derivatives of $f$ at points of $Ucap partialmathbb{H}^n$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension.
But it seems I dont'use the continuity of any function, so I'm not convinced of having understood the matter.
multivariable-calculus differential-geometry smooth-manifolds smooth-functions
asked Jan 8 at 17:31
MinatoMinato
595314
$endgroup$
add a comment |
$begingroup$
Let $A$ be an arbitrary subset of $mathbb{R}^n$, and let $f:Ato mathbb{R}$ be a function.
We say that $f$ is smooth if for each point $p$ in $A$ there exists an open subset $U$ of $mathbb{R}^n$ containing $p$ and a smooth map $g:Uto mathbb{R}$ such that $f$ and $g$ agrees on $A cap U$.
Now let $U$ be an open subset of $mathbb{H}^n$, with $mathbb{H}^n={xin mathbb{R}^n:x^nge0 }$
Let $f:Uto mathbb{R}$ be a smooth function in the above sense.
Let $pin U cap partialmathbb{H}^n$. Thus there are $tilde U$ open subset of $mathbb{R}^n$ and a smooth map $g:tilde Uto mathbb{R}$ such that $g$ and $f$ agrees on $U cap tilde U$.
I want to show that the partial derivatives of $f$ at $p$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension i.e. of the choice of $g$.
Here is my argument.
$$lim_{tto0^+}frac{f(p+te_i)-f(p)}{t}=lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}qquad [1]$$ beacuse $f(p)=g(p)$ and for sufficient positive small $t$ we have $p+te_i in U cap tilde U$ and so $f(p+te_i)=g(p+te_i)$.
Now, since $g$ is smooth at $p$ by hypothesis, we have that the right hand side of the above equation exists, and we have
$$lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}=lim_{tto0^-}frac{g(p+te_i)-g(p)}{t}=:partial_i|_p g$$
So I can define $partial_i|_p f:=partial_i|_p g$ and this value is indipendent of the choice of $g$ by $[1]$.
So I have shown the claim in the above yellow box.
Is my argument right? In John M. Lee Textbook Introduction to smooth manifolds, on page 27, he says that By continuity, all partial derivatives of $f$ at points of $Ucap partialmathbb{H}^n$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension.
But it seems I dont'use the continuity of any function, so I'm not convinced of having understood the matter.
multivariable-calculus differential-geometry smooth-manifolds smooth-functions
asked Jan 8 at 17:31
MinatoMinato
595314
$endgroup$
Let $A$ be an arbitrary subset of $mathbb{R}^n$, and let $f:Ato mathbb{R}$ be a function.
We say that $f$ is smooth if for each point $p$ in $A$ there exists an open subset $U$ of $mathbb{R}^n$ containing $p$ and a smooth map $g:Uto mathbb{R}$ such that $f$ and $g$ agrees on $A cap U$.
Now let $U$ be an open subset of $mathbb{H}^n$, with $mathbb{H}^n={xin mathbb{R}^n:x^nge0 }$
Let $f:Uto mathbb{R}$ be a smooth function in the above sense.
Let $pin U cap partialmathbb{H}^n$. Thus there are $tilde U$ open subset of $mathbb{R}^n$ and a smooth map $g:tilde Uto mathbb{R}$ such that $g$ and $f$ agrees on $U cap tilde U$.
I want to show that the partial derivatives of $f$ at $p$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension i.e. of the choice of $g$.
Here is my argument.
$$lim_{tto0^+}frac{f(p+te_i)-f(p)}{t}=lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}qquad [1]$$ beacuse $f(p)=g(p)$ and for sufficient positive small $t$ we have $p+te_i in U cap tilde U$ and so $f(p+te_i)=g(p+te_i)$.
Now, since $g$ is smooth at $p$ by hypothesis, we have that the right hand side of the above equation exists, and we have
$$lim_{tto0^+}frac{g(p+te_i)-g(p)}{t}=lim_{tto0^-}frac{g(p+te_i)-g(p)}{t}=:partial_i|_p g$$
So I can define $partial_i|_p f:=partial_i|_p g$ and this value is indipendent of the choice of $g$ by $[1]$.
So I have shown the claim in the above yellow box.
Is my argument right? In John M. Lee Textbook Introduction to smooth manifolds, on page 27, he says that By continuity, all partial derivatives of $f$ at points of $Ucap partialmathbb{H}^n$ are determined by their values in Int$mathbb{H}^n$, and therefore in particular are indipendent of the choice of exstension.
But it seems I dont'use the continuity of any function, so I'm not convinced of having understood the matter.
multivariable-calculus differential-geometry smooth-manifolds smooth-functions
multivariable-calculus differential-geometry smooth-manifolds smooth-functions
asked Jan 8 at 17:31
MinatoMinato
595314
asked Jan 8 at 17:31
MinatoMinato
595314
edited Jan 8 at 22:50
Minato
asked Jan 8 at 17:31
MinatoMinato
595314
asked Jan 8 at 17:31
MinatoMinato
595314
asked Jan 8 at 17:31
MinatoMinato
595314
595314
add a comment |
add a comment |
1 Answer
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Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
$$
frac{partial g}{partial x^i}(p)
=
lim_{jto infty} frac{partial g}{partial x^i}(q_j)
=
lim_{jto infty} frac{partial f}{partial x^i}(q_j),
$$
and this last expression manifestly depends only on the original function $f$.
This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.
answered Jan 9 at 5:39
Jack LeeJack Lee
27.7k54868
$endgroup$
$begingroup$
Thank you very much prof. Lee :)
$endgroup$
– Minato
Jan 9 at 17:55
$begingroup$
@Minato — you’re welcome!
$endgroup$
– Jack Lee
Jan 10 at 0:19
add a comment |
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1 Answer
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$begingroup$
Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
$$
frac{partial g}{partial x^i}(p)
=
lim_{jto infty} frac{partial g}{partial x^i}(q_j)
=
lim_{jto infty} frac{partial f}{partial x^i}(q_j),
$$
and this last expression manifestly depends only on the original function $f$.
This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.
answered Jan 9 at 5:39
Jack LeeJack Lee
27.7k54868
$endgroup$
$begingroup$
Thank you very much prof. Lee :)
$endgroup$
– Minato
Jan 9 at 17:55
$begingroup$
@Minato — you’re welcome!
$endgroup$
– Jack Lee
Jan 10 at 0:19
add a comment |
$begingroup$
Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
$$
frac{partial g}{partial x^i}(p)
=
lim_{jto infty} frac{partial g}{partial x^i}(q_j)
=
lim_{jto infty} frac{partial f}{partial x^i}(q_j),
$$
and this last expression manifestly depends only on the original function $f$.
This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.
answered Jan 9 at 5:39
Jack LeeJack Lee
27.7k54868
$endgroup$
$begingroup$
Thank you very much prof. Lee :)
$endgroup$
– Minato
Jan 9 at 17:55
$begingroup$
@Minato — you’re welcome!
$endgroup$
– Jack Lee
Jan 10 at 0:19
add a comment |
$begingroup$
Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
$$
frac{partial g}{partial x^i}(p)
=
lim_{jto infty} frac{partial g}{partial x^i}(q_j)
=
lim_{jto infty} frac{partial f}{partial x^i}(q_j),
$$
and this last expression manifestly depends only on the original function $f$.
This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.
answered Jan 9 at 5:39
Jack LeeJack Lee
27.7k54868
$endgroup$
Yes, your argument works. The argument I had in mind was this: Given $pin partial mathbb H^n$, there's a sequence of points $q_jin text{Int} mathbb H^n$ such that $q_jto p$. Because $partial g/partial x^i$ is continuous,
$$
frac{partial g}{partial x^i}(p)
=
lim_{jto infty} frac{partial g}{partial x^i}(q_j)
=
lim_{jto infty} frac{partial f}{partial x^i}(q_j),
$$
and this last expression manifestly depends only on the original function $f$.
This argument might be a little simpler, and it has the advantage that it applies to any subset of $mathbb R^n$ that is the closure of its interior, not just to $mathbb H^n$. But there's nothing wrong with your argument.
answered Jan 9 at 5:39
Jack LeeJack Lee
27.7k54868
answered Jan 9 at 5:39
Jack LeeJack Lee
27.7k54868
answered Jan 9 at 5:39
Jack LeeJack Lee
27.7k54868
answered Jan 9 at 5:39
Jack LeeJack Lee
27.7k54868
27.7k54868
$begingroup$
Thank you very much prof. Lee :)
$endgroup$
– Minato
Jan 9 at 17:55
$begingroup$
@Minato — you’re welcome!
$endgroup$
– Jack Lee
Jan 10 at 0:19
add a comment |
$begingroup$
Thank you very much prof. Lee :)
$endgroup$
– Minato
Jan 9 at 17:55
$begingroup$
@Minato — you’re welcome!
$endgroup$
– Jack Lee
Jan 10 at 0:19
$begingroup$
Thank you very much prof. Lee :)
$endgroup$
– Minato
Jan 9 at 17:55
$begingroup$
Thank you very much prof. Lee :)
$endgroup$
– Minato
Jan 9 at 17:55
$begingroup$
@Minato — you’re welcome!
$endgroup$
– Jack Lee
Jan 10 at 0:19
$begingroup$
@Minato — you’re welcome!
$endgroup$
– Jack Lee
Jan 10 at 0:19
add a comment |
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