Mixing
Calculus on Manifolds Theorem 3-14 (Sard's Theorem)
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I have a question regarding Theorem 3-14 in Spivak's Calculus on Manifolds:
The proof starts by considering a closed rectangle $Usubset A$ such that all the sides of $U$ are of the same length $l$. Let $epsilon>0$. Spivak claims that if $N$ is sufficiently large and $U$ is divided into $N^n$ rectangles, with sides of length $l/N$, then for each of these rectangles $S$, if $xin S$ we have
$$|Dg(x)(y-x)-(g(y)-g(x))|<epsilon|x-y|leq epsilonsqrt{n}(l/N)$$
for all $yin S$. I don't get why this is necessarily true. My attempt to prove this is as follows. Take any $epsilon>0$. Since $g$ is differentiable on $U$, for all $xin U$ there exists $delta'_x >0$ such that for all $y$,
$$|x-y|<delta'_ximplies|Dg(x)(y-x)-(g(y)-g(x))|<epsilon|x-y|.$$
Since $g$ is continuously differentiable, for all $xin U$ there exists $delta''_x$ such that for all $y$ and $v$,
$$|x-y|<delta''_ximplies|Dg(x)(v)-Dg(y)(v)|<epsilon|v|.$$
Let $delta_x=min{delta'_x,, delta''_x}$. Then ${B(x;delta_x/2),|,xin U}$ is an open cover of $U$. Since $U$ is compact, a finite number of these balls cover $U$. Say these balls are centered around $x_1,,x_2,,ldots,,x_k$. Let $delta=min{delta_{x_1}/2,,delta_{x_2}/2,ldots,,delta_{x_k}/2}$. Pick any arbitrary $xin U$. Then $xin B(x_i,delta_{x_i}/2)$ for some $1leq i leq k$. Thus
$$|Dg(x_i)(x-x_i)-(g(x)-g(x_i))|<epsilon|x_i-x|.$$
For any $yin B(x;delta)$ we have $yin B(x_i,delta_{x_i})$, ergo
$$|Dg(x_i)(y-x_i)-(g(y)-g(x_i))|<epsilon|x_i-y|.$$
Combining these two statements gives
$$|Dg(x_i)(x-y)-(g(x)-g(y))|<epsilon(|x_i-x|+|x_i-y|).$$
Again using the fact that $xin B(x_i,delta_{x_i}/2)$, we have
$$|Dg(x_i)(x-y)-Dg(x)(x-y)|<epsilon|x-y|,$$
hence
$$|Dg(x)(y-x)-(g(y)-g(x))|<epsilon(|x-y|+|x_i-x|+|x_i-y|).$$
Unfortunately, I don't know how to progress from here. Any help would very much be appreciated.
multivariable-calculus
asked Jan 20 at 20:52
HrhmHrhm
2,178417
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add a comment |
$begingroup$
I have a question regarding Theorem 3-14 in Spivak's Calculus on Manifolds:
The proof starts by considering a closed rectangle $Usubset A$ such that all the sides of $U$ are of the same length $l$. Let $epsilon>0$. Spivak claims that if $N$ is sufficiently large and $U$ is divided into $N^n$ rectangles, with sides of length $l/N$, then for each of these rectangles $S$, if $xin S$ we have
$$|Dg(x)(y-x)-(g(y)-g(x))|<epsilon|x-y|leq epsilonsqrt{n}(l/N)$$
for all $yin S$. I don't get why this is necessarily true. My attempt to prove this is as follows. Take any $epsilon>0$. Since $g$ is differentiable on $U$, for all $xin U$ there exists $delta'_x >0$ such that for all $y$,
$$|x-y|<delta'_ximplies|Dg(x)(y-x)-(g(y)-g(x))|<epsilon|x-y|.$$
Since $g$ is continuously differentiable, for all $xin U$ there exists $delta''_x$ such that for all $y$ and $v$,
$$|x-y|<delta''_ximplies|Dg(x)(v)-Dg(y)(v)|<epsilon|v|.$$
Let $delta_x=min{delta'_x,, delta''_x}$. Then ${B(x;delta_x/2),|,xin U}$ is an open cover of $U$. Since $U$ is compact, a finite number of these balls cover $U$. Say these balls are centered around $x_1,,x_2,,ldots,,x_k$. Let $delta=min{delta_{x_1}/2,,delta_{x_2}/2,ldots,,delta_{x_k}/2}$. Pick any arbitrary $xin U$. Then $xin B(x_i,delta_{x_i}/2)$ for some $1leq i leq k$. Thus
$$|Dg(x_i)(x-x_i)-(g(x)-g(x_i))|<epsilon|x_i-x|.$$
For any $yin B(x;delta)$ we have $yin B(x_i,delta_{x_i})$, ergo
$$|Dg(x_i)(y-x_i)-(g(y)-g(x_i))|<epsilon|x_i-y|.$$
Combining these two statements gives
$$|Dg(x_i)(x-y)-(g(x)-g(y))|<epsilon(|x_i-x|+|x_i-y|).$$
Again using the fact that $xin B(x_i,delta_{x_i}/2)$, we have
$$|Dg(x_i)(x-y)-Dg(x)(x-y)|<epsilon|x-y|,$$
hence
$$|Dg(x)(y-x)-(g(y)-g(x))|<epsilon(|x-y|+|x_i-x|+|x_i-y|).$$
Unfortunately, I don't know how to progress from here. Any help would very much be appreciated.
multivariable-calculus
asked Jan 20 at 20:52
HrhmHrhm
2,178417
$endgroup$
add a comment |
$begingroup$
I have a question regarding Theorem 3-14 in Spivak's Calculus on Manifolds:
The proof starts by considering a closed rectangle $Usubset A$ such that all the sides of $U$ are of the same length $l$. Let $epsilon>0$. Spivak claims that if $N$ is sufficiently large and $U$ is divided into $N^n$ rectangles, with sides of length $l/N$, then for each of these rectangles $S$, if $xin S$ we have
$$|Dg(x)(y-x)-(g(y)-g(x))|<epsilon|x-y|leq epsilonsqrt{n}(l/N)$$
for all $yin S$. I don't get why this is necessarily true. My attempt to prove this is as follows. Take any $epsilon>0$. Since $g$ is differentiable on $U$, for all $xin U$ there exists $delta'_x >0$ such that for all $y$,
$$|x-y|<delta'_ximplies|Dg(x)(y-x)-(g(y)-g(x))|<epsilon|x-y|.$$
Since $g$ is continuously differentiable, for all $xin U$ there exists $delta''_x$ such that for all $y$ and $v$,
$$|x-y|<delta''_ximplies|Dg(x)(v)-Dg(y)(v)|<epsilon|v|.$$
Let $delta_x=min{delta'_x,, delta''_x}$. Then ${B(x;delta_x/2),|,xin U}$ is an open cover of $U$. Since $U$ is compact, a finite number of these balls cover $U$. Say these balls are centered around $x_1,,x_2,,ldots,,x_k$. Let $delta=min{delta_{x_1}/2,,delta_{x_2}/2,ldots,,delta_{x_k}/2}$. Pick any arbitrary $xin U$. Then $xin B(x_i,delta_{x_i}/2)$ for some $1leq i leq k$. Thus
$$|Dg(x_i)(x-x_i)-(g(x)-g(x_i))|<epsilon|x_i-x|.$$
For any $yin B(x;delta)$ we have $yin B(x_i,delta_{x_i})$, ergo
$$|Dg(x_i)(y-x_i)-(g(y)-g(x_i))|<epsilon|x_i-y|.$$
Combining these two statements gives
$$|Dg(x_i)(x-y)-(g(x)-g(y))|<epsilon(|x_i-x|+|x_i-y|).$$
Again using the fact that $xin B(x_i,delta_{x_i}/2)$, we have
$$|Dg(x_i)(x-y)-Dg(x)(x-y)|<epsilon|x-y|,$$
hence
$$|Dg(x)(y-x)-(g(y)-g(x))|<epsilon(|x-y|+|x_i-x|+|x_i-y|).$$
Unfortunately, I don't know how to progress from here. Any help would very much be appreciated.
multivariable-calculus
asked Jan 20 at 20:52
HrhmHrhm
2,178417
$endgroup$
I have a question regarding Theorem 3-14 in Spivak's Calculus on Manifolds:
The proof starts by considering a closed rectangle $Usubset A$ such that all the sides of $U$ are of the same length $l$. Let $epsilon>0$. Spivak claims that if $N$ is sufficiently large and $U$ is divided into $N^n$ rectangles, with sides of length $l/N$, then for each of these rectangles $S$, if $xin S$ we have
$$|Dg(x)(y-x)-(g(y)-g(x))|<epsilon|x-y|leq epsilonsqrt{n}(l/N)$$
for all $yin S$. I don't get why this is necessarily true. My attempt to prove this is as follows. Take any $epsilon>0$. Since $g$ is differentiable on $U$, for all $xin U$ there exists $delta'_x >0$ such that for all $y$,
$$|x-y|<delta'_ximplies|Dg(x)(y-x)-(g(y)-g(x))|<epsilon|x-y|.$$
Since $g$ is continuously differentiable, for all $xin U$ there exists $delta''_x$ such that for all $y$ and $v$,
$$|x-y|<delta''_ximplies|Dg(x)(v)-Dg(y)(v)|<epsilon|v|.$$
Let $delta_x=min{delta'_x,, delta''_x}$. Then ${B(x;delta_x/2),|,xin U}$ is an open cover of $U$. Since $U$ is compact, a finite number of these balls cover $U$. Say these balls are centered around $x_1,,x_2,,ldots,,x_k$. Let $delta=min{delta_{x_1}/2,,delta_{x_2}/2,ldots,,delta_{x_k}/2}$. Pick any arbitrary $xin U$. Then $xin B(x_i,delta_{x_i}/2)$ for some $1leq i leq k$. Thus
$$|Dg(x_i)(x-x_i)-(g(x)-g(x_i))|<epsilon|x_i-x|.$$
For any $yin B(x;delta)$ we have $yin B(x_i,delta_{x_i})$, ergo
$$|Dg(x_i)(y-x_i)-(g(y)-g(x_i))|<epsilon|x_i-y|.$$
Combining these two statements gives
$$|Dg(x_i)(x-y)-(g(x)-g(y))|<epsilon(|x_i-x|+|x_i-y|).$$
Again using the fact that $xin B(x_i,delta_{x_i}/2)$, we have
$$|Dg(x_i)(x-y)-Dg(x)(x-y)|<epsilon|x-y|,$$
hence
$$|Dg(x)(y-x)-(g(y)-g(x))|<epsilon(|x-y|+|x_i-x|+|x_i-y|).$$
Unfortunately, I don't know how to progress from here. Any help would very much be appreciated.
multivariable-calculus
multivariable-calculus
asked Jan 20 at 20:52
HrhmHrhm
2,178417
asked Jan 20 at 20:52
HrhmHrhm
2,178417
asked Jan 20 at 20:52
HrhmHrhm
2,178417
asked Jan 20 at 20:52
HrhmHrhm
2,178417
asked Jan 20 at 20:52
HrhmHrhm
2,178417
2,178417
add a comment |
add a comment |
1 Answer
1
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Hint :
You cannot show what you want like this : the $(x_i)$ introduced prevents you to have a bound with $|x-y|$.
The reason is you must show that the inequality :
$|Dg(x)(y-x) - (g (y) - g (x)) | < epsilon |x - y|$
is true for both $x$ and $y$ variyng (not only for a $x $ fixed and a $y $ varying). You need the continuity of $Dg$ to show this.
Then you use the compacity of $U times U $ to conclude.
answered Jan 20 at 22:34
DLeMeurDLeMeur
3148
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add a comment |
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$begingroup$
Hint :
You cannot show what you want like this : the $(x_i)$ introduced prevents you to have a bound with $|x-y|$.
The reason is you must show that the inequality :
$|Dg(x)(y-x) - (g (y) - g (x)) | < epsilon |x - y|$
is true for both $x$ and $y$ variyng (not only for a $x $ fixed and a $y $ varying). You need the continuity of $Dg$ to show this.
Then you use the compacity of $U times U $ to conclude.
answered Jan 20 at 22:34
DLeMeurDLeMeur
3148
$endgroup$
add a comment |
$begingroup$
Hint :
You cannot show what you want like this : the $(x_i)$ introduced prevents you to have a bound with $|x-y|$.
The reason is you must show that the inequality :
$|Dg(x)(y-x) - (g (y) - g (x)) | < epsilon |x - y|$
is true for both $x$ and $y$ variyng (not only for a $x $ fixed and a $y $ varying). You need the continuity of $Dg$ to show this.
Then you use the compacity of $U times U $ to conclude.
answered Jan 20 at 22:34
DLeMeurDLeMeur
3148
$endgroup$
add a comment |
$begingroup$
Hint :
You cannot show what you want like this : the $(x_i)$ introduced prevents you to have a bound with $|x-y|$.
The reason is you must show that the inequality :
$|Dg(x)(y-x) - (g (y) - g (x)) | < epsilon |x - y|$
is true for both $x$ and $y$ variyng (not only for a $x $ fixed and a $y $ varying). You need the continuity of $Dg$ to show this.
Then you use the compacity of $U times U $ to conclude.
answered Jan 20 at 22:34
DLeMeurDLeMeur
3148
$endgroup$
Hint :
You cannot show what you want like this : the $(x_i)$ introduced prevents you to have a bound with $|x-y|$.
The reason is you must show that the inequality :
$|Dg(x)(y-x) - (g (y) - g (x)) | < epsilon |x - y|$
is true for both $x$ and $y$ variyng (not only for a $x $ fixed and a $y $ varying). You need the continuity of $Dg$ to show this.
Then you use the compacity of $U times U $ to conclude.
answered Jan 20 at 22:34
DLeMeurDLeMeur
3148
edited Jan 20 at 22:44
answered Jan 20 at 22:34
DLeMeurDLeMeur
3148
answered Jan 20 at 22:34
DLeMeurDLeMeur
3148
answered Jan 20 at 22:34
DLeMeurDLeMeur
3148
3148
add a comment |
add a comment |
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