Frechet differentiable implies continuty in $mathbb{R} ^n$
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Need help completing a proof. The statement is as follows: Let $E subset mathbb{R} ^m$ be an open set. If $f:E longrightarrow mathbb{R}^n $ is Frechet differentiable at $a in E$, then it is continuous at a.
Proof: Let $f:E longrightarrow mathbb{R}^n $ be Frechet differentiable. Then for $a in E$ , $exists$ a linear map $A:mathbb{R}^m longrightarrow mathbb{R}^n $ s.t. $forall epsilon>0$, $existsdelta >0$, if $h in mathbb{R}^m$ and $0<||h||<delta$, then $$||f(a+h)-f(a)-Ah||<epsilon||h||$$ We then have for $$||f(a+h)-f(a)||leq||h||frac{||f(a+h)-f(a)-Ah||}{||h||}+||Ah||$$ by the triangle inequality. But i get stuck here i feel that im very close to finishing it but just cant see it. Any hints or help will be appreciated!
ordinary-differential-equations
asked Sep 5 '17 at 4:52
Rami2314Rami2314
404
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add a comment |
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Need help completing a proof. The statement is as follows: Let $E subset mathbb{R} ^m$ be an open set. If $f:E longrightarrow mathbb{R}^n $ is Frechet differentiable at $a in E$, then it is continuous at a.
Proof: Let $f:E longrightarrow mathbb{R}^n $ be Frechet differentiable. Then for $a in E$ , $exists$ a linear map $A:mathbb{R}^m longrightarrow mathbb{R}^n $ s.t. $forall epsilon>0$, $existsdelta >0$, if $h in mathbb{R}^m$ and $0<||h||<delta$, then $$||f(a+h)-f(a)-Ah||<epsilon||h||$$ We then have for $$||f(a+h)-f(a)||leq||h||frac{||f(a+h)-f(a)-Ah||}{||h||}+||Ah||$$ by the triangle inequality. But i get stuck here i feel that im very close to finishing it but just cant see it. Any hints or help will be appreciated!
ordinary-differential-equations
asked Sep 5 '17 at 4:52
Rami2314Rami2314
404
$endgroup$
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What is the meaning of $frac 1 h$ for $h in mathbb{R^m}$? And $f(a+h)$ is not well defined here since $a in E subset R^n$. Perhaps you mean $E subset R^m$ ?
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– nicomezi
Sep 5 '17 at 5:03
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Sorry yes you are right $E subset mathbb{R^m}$. And the $frac{1}{h}$ is suppose to be $frac{1}{||h||}$
$endgroup$
– Rami2314
Sep 5 '17 at 13:08
add a comment |
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Need help completing a proof. The statement is as follows: Let $E subset mathbb{R} ^m$ be an open set. If $f:E longrightarrow mathbb{R}^n $ is Frechet differentiable at $a in E$, then it is continuous at a.
Proof: Let $f:E longrightarrow mathbb{R}^n $ be Frechet differentiable. Then for $a in E$ , $exists$ a linear map $A:mathbb{R}^m longrightarrow mathbb{R}^n $ s.t. $forall epsilon>0$, $existsdelta >0$, if $h in mathbb{R}^m$ and $0<||h||<delta$, then $$||f(a+h)-f(a)-Ah||<epsilon||h||$$ We then have for $$||f(a+h)-f(a)||leq||h||frac{||f(a+h)-f(a)-Ah||}{||h||}+||Ah||$$ by the triangle inequality. But i get stuck here i feel that im very close to finishing it but just cant see it. Any hints or help will be appreciated!
ordinary-differential-equations
asked Sep 5 '17 at 4:52
Rami2314Rami2314
404
$endgroup$
Need help completing a proof. The statement is as follows: Let $E subset mathbb{R} ^m$ be an open set. If $f:E longrightarrow mathbb{R}^n $ is Frechet differentiable at $a in E$, then it is continuous at a.
Proof: Let $f:E longrightarrow mathbb{R}^n $ be Frechet differentiable. Then for $a in E$ , $exists$ a linear map $A:mathbb{R}^m longrightarrow mathbb{R}^n $ s.t. $forall epsilon>0$, $existsdelta >0$, if $h in mathbb{R}^m$ and $0<||h||<delta$, then $$||f(a+h)-f(a)-Ah||<epsilon||h||$$ We then have for $$||f(a+h)-f(a)||leq||h||frac{||f(a+h)-f(a)-Ah||}{||h||}+||Ah||$$ by the triangle inequality. But i get stuck here i feel that im very close to finishing it but just cant see it. Any hints or help will be appreciated!
ordinary-differential-equations
ordinary-differential-equations
asked Sep 5 '17 at 4:52
Rami2314Rami2314
404
asked Sep 5 '17 at 4:52
Rami2314Rami2314
404
edited Sep 5 '17 at 13:11
Rami2314
asked Sep 5 '17 at 4:52
Rami2314Rami2314
404
asked Sep 5 '17 at 4:52
Rami2314Rami2314
404
asked Sep 5 '17 at 4:52
Rami2314Rami2314
404
404
$begingroup$
What is the meaning of $frac 1 h$ for $h in mathbb{R^m}$? And $f(a+h)$ is not well defined here since $a in E subset R^n$. Perhaps you mean $E subset R^m$ ?
$endgroup$
– nicomezi
Sep 5 '17 at 5:03
$begingroup$
Sorry yes you are right $E subset mathbb{R^m}$. And the $frac{1}{h}$ is suppose to be $frac{1}{||h||}$
$endgroup$
– Rami2314
Sep 5 '17 at 13:08
add a comment |
$begingroup$
What is the meaning of $frac 1 h$ for $h in mathbb{R^m}$? And $f(a+h)$ is not well defined here since $a in E subset R^n$. Perhaps you mean $E subset R^m$ ?
$endgroup$
– nicomezi
Sep 5 '17 at 5:03
$begingroup$
Sorry yes you are right $E subset mathbb{R^m}$. And the $frac{1}{h}$ is suppose to be $frac{1}{||h||}$
$endgroup$
– Rami2314
Sep 5 '17 at 13:08
$begingroup$
What is the meaning of $frac 1 h$ for $h in mathbb{R^m}$? And $f(a+h)$ is not well defined here since $a in E subset R^n$. Perhaps you mean $E subset R^m$ ?
$endgroup$
– nicomezi
Sep 5 '17 at 5:03
$begingroup$
What is the meaning of $frac 1 h$ for $h in mathbb{R^m}$? And $f(a+h)$ is not well defined here since $a in E subset R^n$. Perhaps you mean $E subset R^m$ ?
$endgroup$
– nicomezi
Sep 5 '17 at 5:03
$begingroup$
Sorry yes you are right $E subset mathbb{R^m}$. And the $frac{1}{h}$ is suppose to be $frac{1}{||h||}$
$endgroup$
– Rami2314
Sep 5 '17 at 13:08
$begingroup$
Sorry yes you are right $E subset mathbb{R^m}$. And the $frac{1}{h}$ is suppose to be $frac{1}{||h||}$
$endgroup$
– Rami2314
Sep 5 '17 at 13:08
add a comment |
1 Answer
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Suppose $f$ is Frechet differentiable at $a$, with derivative $A$. Then we know that $$lim_{hto0} frac{|f(a+h)-f(a)-Ah|}{|h|} = 0.$$ Define a function $eta$ by $eta(h) = {|f(a+h)-f(a)-Ah|}big/{|h|} $ if $hne0$ and $eta(0)=0$. By the differentiability of $f$ at $a$ the function $eta(h)$ is continuous at $h=0$. (This is only a restatement of the differentiablilty of $f$, not a deep theorem.) So we know $$tag{*}f(a+h) = f(a) + Ah +|h|eta(h).$$ What we want is $lim_{hto0}f(a+h)=f(a)$. But this can be read off of (*): the right hand side of (*) is $f(a)$ plus terms that manifestly converge to $0$ as $hto0$: $Ahto0$ as $hto0$, and $|h|to0$ and $eta(h)to0$ as $hto 0$, and so does their product $|h|eta(h)$, too, etc.
answered Sep 5 '17 at 5:35
kimchi loverkimchi lover
10.4k31128
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$begingroup$
Suppose $f$ is Frechet differentiable at $a$, with derivative $A$. Then we know that $$lim_{hto0} frac{|f(a+h)-f(a)-Ah|}{|h|} = 0.$$ Define a function $eta$ by $eta(h) = {|f(a+h)-f(a)-Ah|}big/{|h|} $ if $hne0$ and $eta(0)=0$. By the differentiability of $f$ at $a$ the function $eta(h)$ is continuous at $h=0$. (This is only a restatement of the differentiablilty of $f$, not a deep theorem.) So we know $$tag{*}f(a+h) = f(a) + Ah +|h|eta(h).$$ What we want is $lim_{hto0}f(a+h)=f(a)$. But this can be read off of (*): the right hand side of (*) is $f(a)$ plus terms that manifestly converge to $0$ as $hto0$: $Ahto0$ as $hto0$, and $|h|to0$ and $eta(h)to0$ as $hto 0$, and so does their product $|h|eta(h)$, too, etc.
answered Sep 5 '17 at 5:35
kimchi loverkimchi lover
10.4k31128
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add a comment |
$begingroup$
Suppose $f$ is Frechet differentiable at $a$, with derivative $A$. Then we know that $$lim_{hto0} frac{|f(a+h)-f(a)-Ah|}{|h|} = 0.$$ Define a function $eta$ by $eta(h) = {|f(a+h)-f(a)-Ah|}big/{|h|} $ if $hne0$ and $eta(0)=0$. By the differentiability of $f$ at $a$ the function $eta(h)$ is continuous at $h=0$. (This is only a restatement of the differentiablilty of $f$, not a deep theorem.) So we know $$tag{*}f(a+h) = f(a) + Ah +|h|eta(h).$$ What we want is $lim_{hto0}f(a+h)=f(a)$. But this can be read off of (*): the right hand side of (*) is $f(a)$ plus terms that manifestly converge to $0$ as $hto0$: $Ahto0$ as $hto0$, and $|h|to0$ and $eta(h)to0$ as $hto 0$, and so does their product $|h|eta(h)$, too, etc.
answered Sep 5 '17 at 5:35
kimchi loverkimchi lover
10.4k31128
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add a comment |
$begingroup$
Suppose $f$ is Frechet differentiable at $a$, with derivative $A$. Then we know that $$lim_{hto0} frac{|f(a+h)-f(a)-Ah|}{|h|} = 0.$$ Define a function $eta$ by $eta(h) = {|f(a+h)-f(a)-Ah|}big/{|h|} $ if $hne0$ and $eta(0)=0$. By the differentiability of $f$ at $a$ the function $eta(h)$ is continuous at $h=0$. (This is only a restatement of the differentiablilty of $f$, not a deep theorem.) So we know $$tag{*}f(a+h) = f(a) + Ah +|h|eta(h).$$ What we want is $lim_{hto0}f(a+h)=f(a)$. But this can be read off of (*): the right hand side of (*) is $f(a)$ plus terms that manifestly converge to $0$ as $hto0$: $Ahto0$ as $hto0$, and $|h|to0$ and $eta(h)to0$ as $hto 0$, and so does their product $|h|eta(h)$, too, etc.
answered Sep 5 '17 at 5:35
kimchi loverkimchi lover
10.4k31128
$endgroup$
Suppose $f$ is Frechet differentiable at $a$, with derivative $A$. Then we know that $$lim_{hto0} frac{|f(a+h)-f(a)-Ah|}{|h|} = 0.$$ Define a function $eta$ by $eta(h) = {|f(a+h)-f(a)-Ah|}big/{|h|} $ if $hne0$ and $eta(0)=0$. By the differentiability of $f$ at $a$ the function $eta(h)$ is continuous at $h=0$. (This is only a restatement of the differentiablilty of $f$, not a deep theorem.) So we know $$tag{*}f(a+h) = f(a) + Ah +|h|eta(h).$$ What we want is $lim_{hto0}f(a+h)=f(a)$. But this can be read off of (*): the right hand side of (*) is $f(a)$ plus terms that manifestly converge to $0$ as $hto0$: $Ahto0$ as $hto0$, and $|h|to0$ and $eta(h)to0$ as $hto 0$, and so does their product $|h|eta(h)$, too, etc.
answered Sep 5 '17 at 5:35
kimchi loverkimchi lover
10.4k31128
edited Dec 13 '18 at 16:44
answered Sep 5 '17 at 5:35
kimchi loverkimchi lover
10.4k31128
answered Sep 5 '17 at 5:35
kimchi loverkimchi lover
10.4k31128
answered Sep 5 '17 at 5:35
kimchi loverkimchi lover
10.4k31128
10.4k31128
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$begingroup$
What is the meaning of $frac 1 h$ for $h in mathbb{R^m}$? And $f(a+h)$ is not well defined here since $a in E subset R^n$. Perhaps you mean $E subset R^m$ ?
$endgroup$
– nicomezi
Sep 5 '17 at 5:03
$begingroup$
Sorry yes you are right $E subset mathbb{R^m}$. And the $frac{1}{h}$ is suppose to be $frac{1}{||h||}$
$endgroup$
– Rami2314
Sep 5 '17 at 13:08