Closure of $C^1[0,1]$ functions under the Lipschitz norm
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I have been trying to prove that $C^1[0,1]$, i.e the space of continuously differentiable functions is closed under the $C^{0,1}[0,1]$ norm.
$$||f||_{C^{0,1}}=||f||_{C^0}+sup_{xne y}frac{|f(x)-f(y)|}{|x-y|}$$
So what I have been able to see is that if $f_n to f$ in the $text{Lip}$ norm for $f_n in C^1$ then $f$ is differentiable with bounded derivative.
I cannot see why this $f$ must have a continuous derivative. More precisely
$$left|frac{f(x+h)-f(x)}{h}-frac{f(y+h)-f(y)}{h}right|lesssim ||f_n-f||_{C^{0,1}}+|f_n'(x)-f'_n(y)|$$
But of course we just have uniform convergence for $f_n$ in $C^0$. So I can't see why we can make the second term small independent of $n$. Perhaps I'm missing something.
Any help is appreciated
real-analysis functional-analysis banach-spaces
asked Jan 21 at 23:58
SanchitSanchit
8710
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add a comment |
$begingroup$
I have been trying to prove that $C^1[0,1]$, i.e the space of continuously differentiable functions is closed under the $C^{0,1}[0,1]$ norm.
$$||f||_{C^{0,1}}=||f||_{C^0}+sup_{xne y}frac{|f(x)-f(y)|}{|x-y|}$$
So what I have been able to see is that if $f_n to f$ in the $text{Lip}$ norm for $f_n in C^1$ then $f$ is differentiable with bounded derivative.
I cannot see why this $f$ must have a continuous derivative. More precisely
$$left|frac{f(x+h)-f(x)}{h}-frac{f(y+h)-f(y)}{h}right|lesssim ||f_n-f||_{C^{0,1}}+|f_n'(x)-f'_n(y)|$$
But of course we just have uniform convergence for $f_n$ in $C^0$. So I can't see why we can make the second term small independent of $n$. Perhaps I'm missing something.
Any help is appreciated
real-analysis functional-analysis banach-spaces
asked Jan 21 at 23:58
SanchitSanchit
8710
$endgroup$
add a comment |
$begingroup$
I have been trying to prove that $C^1[0,1]$, i.e the space of continuously differentiable functions is closed under the $C^{0,1}[0,1]$ norm.
$$||f||_{C^{0,1}}=||f||_{C^0}+sup_{xne y}frac{|f(x)-f(y)|}{|x-y|}$$
So what I have been able to see is that if $f_n to f$ in the $text{Lip}$ norm for $f_n in C^1$ then $f$ is differentiable with bounded derivative.
I cannot see why this $f$ must have a continuous derivative. More precisely
$$left|frac{f(x+h)-f(x)}{h}-frac{f(y+h)-f(y)}{h}right|lesssim ||f_n-f||_{C^{0,1}}+|f_n'(x)-f'_n(y)|$$
But of course we just have uniform convergence for $f_n$ in $C^0$. So I can't see why we can make the second term small independent of $n$. Perhaps I'm missing something.
Any help is appreciated
real-analysis functional-analysis banach-spaces
asked Jan 21 at 23:58
SanchitSanchit
8710
$endgroup$
I have been trying to prove that $C^1[0,1]$, i.e the space of continuously differentiable functions is closed under the $C^{0,1}[0,1]$ norm.
$$||f||_{C^{0,1}}=||f||_{C^0}+sup_{xne y}frac{|f(x)-f(y)|}{|x-y|}$$
So what I have been able to see is that if $f_n to f$ in the $text{Lip}$ norm for $f_n in C^1$ then $f$ is differentiable with bounded derivative.
I cannot see why this $f$ must have a continuous derivative. More precisely
$$left|frac{f(x+h)-f(x)}{h}-frac{f(y+h)-f(y)}{h}right|lesssim ||f_n-f||_{C^{0,1}}+|f_n'(x)-f'_n(y)|$$
But of course we just have uniform convergence for $f_n$ in $C^0$. So I can't see why we can make the second term small independent of $n$. Perhaps I'm missing something.
Any help is appreciated
real-analysis functional-analysis banach-spaces
real-analysis functional-analysis banach-spaces
asked Jan 21 at 23:58
SanchitSanchit
8710
asked Jan 21 at 23:58
SanchitSanchit
8710
asked Jan 21 at 23:58
SanchitSanchit
8710
asked Jan 21 at 23:58
SanchitSanchit
8710
asked Jan 21 at 23:58
SanchitSanchit
8710
8710
add a comment |
add a comment |
1 Answer
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If $f_n to f$ uniformly and $f_n' to g$ uniformly then $f$ is differentiable and $g=f'$. It follows immediately that $f$ is continuously differentiable because $g$ is continuous. [ $|f_n'(x)-f_m'(x)| leq |f_n-f_m||_{C^{0,1}} to 0$ so ${f_n'}$ converges uniformly].
answered Jan 22 at 0:06
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
$begingroup$
Thanks for the help! Somehow I missed this line of thinking
$endgroup$
– Sanchit
Jan 22 at 0:16
add a comment |
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$begingroup$
If $f_n to f$ uniformly and $f_n' to g$ uniformly then $f$ is differentiable and $g=f'$. It follows immediately that $f$ is continuously differentiable because $g$ is continuous. [ $|f_n'(x)-f_m'(x)| leq |f_n-f_m||_{C^{0,1}} to 0$ so ${f_n'}$ converges uniformly].
answered Jan 22 at 0:06
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
$begingroup$
Thanks for the help! Somehow I missed this line of thinking
$endgroup$
– Sanchit
Jan 22 at 0:16
add a comment |
$begingroup$
If $f_n to f$ uniformly and $f_n' to g$ uniformly then $f$ is differentiable and $g=f'$. It follows immediately that $f$ is continuously differentiable because $g$ is continuous. [ $|f_n'(x)-f_m'(x)| leq |f_n-f_m||_{C^{0,1}} to 0$ so ${f_n'}$ converges uniformly].
answered Jan 22 at 0:06
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
$begingroup$
Thanks for the help! Somehow I missed this line of thinking
$endgroup$
– Sanchit
Jan 22 at 0:16
add a comment |
$begingroup$
If $f_n to f$ uniformly and $f_n' to g$ uniformly then $f$ is differentiable and $g=f'$. It follows immediately that $f$ is continuously differentiable because $g$ is continuous. [ $|f_n'(x)-f_m'(x)| leq |f_n-f_m||_{C^{0,1}} to 0$ so ${f_n'}$ converges uniformly].
answered Jan 22 at 0:06
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
If $f_n to f$ uniformly and $f_n' to g$ uniformly then $f$ is differentiable and $g=f'$. It follows immediately that $f$ is continuously differentiable because $g$ is continuous. [ $|f_n'(x)-f_m'(x)| leq |f_n-f_m||_{C^{0,1}} to 0$ so ${f_n'}$ converges uniformly].
answered Jan 22 at 0:06
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 22 at 0:06
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 22 at 0:06
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Jan 22 at 0:06
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
$begingroup$
Thanks for the help! Somehow I missed this line of thinking
$endgroup$
– Sanchit
Jan 22 at 0:16
add a comment |
$begingroup$
Thanks for the help! Somehow I missed this line of thinking
$endgroup$
– Sanchit
Jan 22 at 0:16
$begingroup$
Thanks for the help! Somehow I missed this line of thinking
$endgroup$
– Sanchit
Jan 22 at 0:16
$begingroup$
Thanks for the help! Somehow I missed this line of thinking
$endgroup$
– Sanchit
Jan 22 at 0:16
add a comment |
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