Compactness of $A:=${$f in C[0,1], |f|_infty le K, |f'|_infty le M$} The 2019 Stack Overflow...
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Compactness of $A:=${$f in C[0,1], |f|_infty le K, |f'|_infty le M$}
The 2019 Stack Overflow Developer Survey Results Are InUniform convergence of subsequences implying uniform convergenceCompactness in $C([0,1])$Arzela-Ascoli Theorem in $L^p[0,1]$Compactness of the $K subset C[0,1]$About the proof of a corollary of Arzela-Ascoli Theorem.Folland Problem 4.64 relating to Hölder continuity and compactness.Arzela Ascoli counterexamplescharacterizing compactness in function space.For which n differential operator $C^{(n)}[0, 1] rightarrow C[0,1]$ is compactArzela-Ascoli Theorem for a subset of $C([0,1])$
$begingroup$
Here we use infinity norm as metrics for $C[0,1]$. The professor claims that this set is compact. I can show this set is relative compact by Arzela Ascoli, i.e., for each subsequence there is a further sub-subsequence that uniformly converges to a continuous function on $[0,1]$. But it is not intuitive to me why this set is compact, i.e. the limiting function also satisfies $|f'|_infty le M$. Is it true that $A$ is close and therefore compact?
real-analysis functional-analysis analysis
asked Mar 21 at 15:39
Daniel LiDaniel Li
787414
$endgroup$
add a comment |
$begingroup$
Here we use infinity norm as metrics for $C[0,1]$. The professor claims that this set is compact. I can show this set is relative compact by Arzela Ascoli, i.e., for each subsequence there is a further sub-subsequence that uniformly converges to a continuous function on $[0,1]$. But it is not intuitive to me why this set is compact, i.e. the limiting function also satisfies $|f'|_infty le M$. Is it true that $A$ is close and therefore compact?
real-analysis functional-analysis analysis
asked Mar 21 at 15:39
Daniel LiDaniel Li
787414
$endgroup$
add a comment |
$begingroup$
Here we use infinity norm as metrics for $C[0,1]$. The professor claims that this set is compact. I can show this set is relative compact by Arzela Ascoli, i.e., for each subsequence there is a further sub-subsequence that uniformly converges to a continuous function on $[0,1]$. But it is not intuitive to me why this set is compact, i.e. the limiting function also satisfies $|f'|_infty le M$. Is it true that $A$ is close and therefore compact?
real-analysis functional-analysis analysis
asked Mar 21 at 15:39
Daniel LiDaniel Li
787414
$endgroup$
Here we use infinity norm as metrics for $C[0,1]$. The professor claims that this set is compact. I can show this set is relative compact by Arzela Ascoli, i.e., for each subsequence there is a further sub-subsequence that uniformly converges to a continuous function on $[0,1]$. But it is not intuitive to me why this set is compact, i.e. the limiting function also satisfies $|f'|_infty le M$. Is it true that $A$ is close and therefore compact?
real-analysis functional-analysis analysis
real-analysis functional-analysis analysis
asked Mar 21 at 15:39
Daniel LiDaniel Li
787414
asked Mar 21 at 15:39
Daniel LiDaniel Li
787414
asked Mar 21 at 15:39
Daniel LiDaniel Li
787414
asked Mar 21 at 15:39
Daniel LiDaniel Li
787414
asked Mar 21 at 15:39
Daniel LiDaniel Li
787414
787414
add a comment |
add a comment |
1 Answer
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$begingroup$
No it is not compact. For example, $A$ contains
$$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$
since
$$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$
But $f_n$ converges to $|x-1/2|$ which is not differentiable.
Remark: If instead one consider
$$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$
here
$$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$
then $Asubset A'$ and $A'$ is compact.
answered Mar 21 at 17:19
Arctic CharArctic Char
471115
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No it is not compact. For example, $A$ contains
$$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$
since
$$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$
But $f_n$ converges to $|x-1/2|$ which is not differentiable.
Remark: If instead one consider
$$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$
here
$$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$
then $Asubset A'$ and $A'$ is compact.
answered Mar 21 at 17:19
Arctic CharArctic Char
471115
$endgroup$
add a comment |
$begingroup$
No it is not compact. For example, $A$ contains
$$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$
since
$$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$
But $f_n$ converges to $|x-1/2|$ which is not differentiable.
Remark: If instead one consider
$$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$
here
$$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$
then $Asubset A'$ and $A'$ is compact.
answered Mar 21 at 17:19
Arctic CharArctic Char
471115
$endgroup$
add a comment |
$begingroup$
No it is not compact. For example, $A$ contains
$$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$
since
$$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$
But $f_n$ converges to $|x-1/2|$ which is not differentiable.
Remark: If instead one consider
$$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$
here
$$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$
then $Asubset A'$ and $A'$ is compact.
answered Mar 21 at 17:19
Arctic CharArctic Char
471115
$endgroup$
No it is not compact. For example, $A$ contains
$$ f_n (x) = sqrt{(x-1/2)^2 + 1/n}.$$
since
$$ |f_n'(x)| = frac{2|x-1/2|}{sqrt{(x-1/2)^2 + 1/n}}le 2. $$
But $f_n$ converges to $|x-1/2|$ which is not differentiable.
Remark: If instead one consider
$$A' = { fin C[0,1] : |f|_inftyle K, operatorname{Lip}f le M},$$
here
$$operatorname{Lip}f := sup_{x, yin [0,1], xneq y} frac{|f(x) - f(y)|}{|x-y|},$$
then $Asubset A'$ and $A'$ is compact.
answered Mar 21 at 17:19
Arctic CharArctic Char
471115
answered Mar 21 at 17:19
Arctic CharArctic Char
471115
answered Mar 21 at 17:19
Arctic CharArctic Char
471115
answered Mar 21 at 17:19
Arctic CharArctic Char
471115
471115
add a comment |
add a comment |
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