Example for a differentiable function that is not absolutely continuous.
$begingroup$
$mathbf{Theorem:}$
Let $f$ be defined on $I=[a,b]$ and absolutely continuous on $I$. Then $f$ is differentiable a.e in $(a,b)$, and its derivative is integrable over $[a,b]$ with $$int_{[a,b]}f'=f(b)-f(a).$$
I am ok with this theorem. But I need an example to show the converse is not true, $i.e$, we are looking for a function that is differentiable a.e on $(a,b)$ with $$int_{[a,b]}f'=f(b)-f(a),$$ but it is not absolutely continuous on its domain, which we want it to be compact.
I have an example which i am not sure about it:
Define $f(x)=x^{frac{1}{3}}$ for $x in [-1,1]$ and $g(x)=begin{cases} x^2 cos(pi/2x) & x in [-1,1]sim {0}\
0 & x =0end{cases}$.
Clearly, $f$ is increaing on $[-1,1]$, and we have that for any $1>epsilon>0$, $|f'(x)|=|frac{1}{3 sqrt[3]{x^2}}|leq |frac{1}{3sqrt[3]{epsilon^2}}|$ for all $x in [epsilon,1]cup [-1,-epsilon]$, thus $f$ is Lipschitz on $[epsilon,1]cup [-1,-epsilon]$, hence $f$ is absolutely continuos on $[epsilon,1]cup [-1,-epsilon]$. Thus, $f$ is absolutely continuous on $[0,1]cup [-1,0]=[-1,1]$ since it is increaing.
Now, we see obviously, $g$ is differentiable on $[-1,1]$ with $g'(x)=2x cos(pi/2x)+frac{pi}{2}sin(pi/2x)$, and we have $|g'(x)|leq 2+1=3$. Therefore, $g$ is Lipschitz on $[-1,1]$, thus it is absolutely continuous on $[-1,1].$
Now, consider the partition $mathcal{P}_n={-1,0,frac{1}{2n},frac{1}{2n-1},...,frac{1}{2},1}$ of $[-1,1]$. We have $(f circ g)(x)=x^{^{frac{2}{3}}}cos^{^{frac{1}{3}}}(frac{pi}{2x})$, so $$sum_{i=1}^{2n-1}|f(x_{i})-f(x_{i-1})|=|0-0|+|frac{cos^{1/3}(n pi)}{2n^{2/3}}-0|+|frac{cos^{1/3}(frac{(2n-1)pi}{2})}{(2n-1)^{2/3}}-frac{cos^{1/3}(n pi)}{2n^{2/3}}|+...$$
$$+ |frac{cos^{1/3}( pi)}{2^{2/3}}-frac{cos^{1/3}(pi/2)}{1^{2/3}}|=sum_{i=0}^n2^{^{1/3}}frac{1}{i^{2/3}}=2^{^{1/3}}sum_{i=0}^nfrac{1}{i^{2/3}}.$$
since $2/3<1$, then the sum diverges as $n rightarrow infty$, thus $fcirc g$ fails to be of a bounded variation on $[-1,1]$, and hence it is not absolutely continuous on $[-1,1].$
But I can not see that $$int_{[0,1]}(f circ g)'=f(g(b))-f(g(a))$$
Thanks for any help.
measure-theory lebesgue-measure
asked Nov 28 '18 at 21:43
Ahmed Ahmed
29519
$endgroup$
add a comment |
$begingroup$
$mathbf{Theorem:}$
Let $f$ be defined on $I=[a,b]$ and absolutely continuous on $I$. Then $f$ is differentiable a.e in $(a,b)$, and its derivative is integrable over $[a,b]$ with $$int_{[a,b]}f'=f(b)-f(a).$$
I am ok with this theorem. But I need an example to show the converse is not true, $i.e$, we are looking for a function that is differentiable a.e on $(a,b)$ with $$int_{[a,b]}f'=f(b)-f(a),$$ but it is not absolutely continuous on its domain, which we want it to be compact.
I have an example which i am not sure about it:
Define $f(x)=x^{frac{1}{3}}$ for $x in [-1,1]$ and $g(x)=begin{cases} x^2 cos(pi/2x) & x in [-1,1]sim {0}\
0 & x =0end{cases}$.
Clearly, $f$ is increaing on $[-1,1]$, and we have that for any $1>epsilon>0$, $|f'(x)|=|frac{1}{3 sqrt[3]{x^2}}|leq |frac{1}{3sqrt[3]{epsilon^2}}|$ for all $x in [epsilon,1]cup [-1,-epsilon]$, thus $f$ is Lipschitz on $[epsilon,1]cup [-1,-epsilon]$, hence $f$ is absolutely continuos on $[epsilon,1]cup [-1,-epsilon]$. Thus, $f$ is absolutely continuous on $[0,1]cup [-1,0]=[-1,1]$ since it is increaing.
Now, we see obviously, $g$ is differentiable on $[-1,1]$ with $g'(x)=2x cos(pi/2x)+frac{pi}{2}sin(pi/2x)$, and we have $|g'(x)|leq 2+1=3$. Therefore, $g$ is Lipschitz on $[-1,1]$, thus it is absolutely continuous on $[-1,1].$
Now, consider the partition $mathcal{P}_n={-1,0,frac{1}{2n},frac{1}{2n-1},...,frac{1}{2},1}$ of $[-1,1]$. We have $(f circ g)(x)=x^{^{frac{2}{3}}}cos^{^{frac{1}{3}}}(frac{pi}{2x})$, so $$sum_{i=1}^{2n-1}|f(x_{i})-f(x_{i-1})|=|0-0|+|frac{cos^{1/3}(n pi)}{2n^{2/3}}-0|+|frac{cos^{1/3}(frac{(2n-1)pi}{2})}{(2n-1)^{2/3}}-frac{cos^{1/3}(n pi)}{2n^{2/3}}|+...$$
$$+ |frac{cos^{1/3}( pi)}{2^{2/3}}-frac{cos^{1/3}(pi/2)}{1^{2/3}}|=sum_{i=0}^n2^{^{1/3}}frac{1}{i^{2/3}}=2^{^{1/3}}sum_{i=0}^nfrac{1}{i^{2/3}}.$$
since $2/3<1$, then the sum diverges as $n rightarrow infty$, thus $fcirc g$ fails to be of a bounded variation on $[-1,1]$, and hence it is not absolutely continuous on $[-1,1].$
But I can not see that $$int_{[0,1]}(f circ g)'=f(g(b))-f(g(a))$$
Thanks for any help.
measure-theory lebesgue-measure
asked Nov 28 '18 at 21:43
Ahmed Ahmed
29519
$endgroup$
1
$begingroup$
Looking under en.wikipedia.org/wiki/Absolute_continuity in the "Equivalent Definitions" section it seems those 2 conditions are indeed equivalent. But I'm no expert in this, please check this yourself.
$endgroup$
– Ingix
Nov 28 '18 at 23:47
add a comment |
$begingroup$
$mathbf{Theorem:}$
Let $f$ be defined on $I=[a,b]$ and absolutely continuous on $I$. Then $f$ is differentiable a.e in $(a,b)$, and its derivative is integrable over $[a,b]$ with $$int_{[a,b]}f'=f(b)-f(a).$$
I am ok with this theorem. But I need an example to show the converse is not true, $i.e$, we are looking for a function that is differentiable a.e on $(a,b)$ with $$int_{[a,b]}f'=f(b)-f(a),$$ but it is not absolutely continuous on its domain, which we want it to be compact.
I have an example which i am not sure about it:
Define $f(x)=x^{frac{1}{3}}$ for $x in [-1,1]$ and $g(x)=begin{cases} x^2 cos(pi/2x) & x in [-1,1]sim {0}\
0 & x =0end{cases}$.
Clearly, $f$ is increaing on $[-1,1]$, and we have that for any $1>epsilon>0$, $|f'(x)|=|frac{1}{3 sqrt[3]{x^2}}|leq |frac{1}{3sqrt[3]{epsilon^2}}|$ for all $x in [epsilon,1]cup [-1,-epsilon]$, thus $f$ is Lipschitz on $[epsilon,1]cup [-1,-epsilon]$, hence $f$ is absolutely continuos on $[epsilon,1]cup [-1,-epsilon]$. Thus, $f$ is absolutely continuous on $[0,1]cup [-1,0]=[-1,1]$ since it is increaing.
Now, we see obviously, $g$ is differentiable on $[-1,1]$ with $g'(x)=2x cos(pi/2x)+frac{pi}{2}sin(pi/2x)$, and we have $|g'(x)|leq 2+1=3$. Therefore, $g$ is Lipschitz on $[-1,1]$, thus it is absolutely continuous on $[-1,1].$
Now, consider the partition $mathcal{P}_n={-1,0,frac{1}{2n},frac{1}{2n-1},...,frac{1}{2},1}$ of $[-1,1]$. We have $(f circ g)(x)=x^{^{frac{2}{3}}}cos^{^{frac{1}{3}}}(frac{pi}{2x})$, so $$sum_{i=1}^{2n-1}|f(x_{i})-f(x_{i-1})|=|0-0|+|frac{cos^{1/3}(n pi)}{2n^{2/3}}-0|+|frac{cos^{1/3}(frac{(2n-1)pi}{2})}{(2n-1)^{2/3}}-frac{cos^{1/3}(n pi)}{2n^{2/3}}|+...$$
$$+ |frac{cos^{1/3}( pi)}{2^{2/3}}-frac{cos^{1/3}(pi/2)}{1^{2/3}}|=sum_{i=0}^n2^{^{1/3}}frac{1}{i^{2/3}}=2^{^{1/3}}sum_{i=0}^nfrac{1}{i^{2/3}}.$$
since $2/3<1$, then the sum diverges as $n rightarrow infty$, thus $fcirc g$ fails to be of a bounded variation on $[-1,1]$, and hence it is not absolutely continuous on $[-1,1].$
But I can not see that $$int_{[0,1]}(f circ g)'=f(g(b))-f(g(a))$$
Thanks for any help.
measure-theory lebesgue-measure
asked Nov 28 '18 at 21:43
Ahmed Ahmed
29519
$endgroup$
$mathbf{Theorem:}$
Let $f$ be defined on $I=[a,b]$ and absolutely continuous on $I$. Then $f$ is differentiable a.e in $(a,b)$, and its derivative is integrable over $[a,b]$ with $$int_{[a,b]}f'=f(b)-f(a).$$
I am ok with this theorem. But I need an example to show the converse is not true, $i.e$, we are looking for a function that is differentiable a.e on $(a,b)$ with $$int_{[a,b]}f'=f(b)-f(a),$$ but it is not absolutely continuous on its domain, which we want it to be compact.
I have an example which i am not sure about it:
Define $f(x)=x^{frac{1}{3}}$ for $x in [-1,1]$ and $g(x)=begin{cases} x^2 cos(pi/2x) & x in [-1,1]sim {0}\
0 & x =0end{cases}$.
Clearly, $f$ is increaing on $[-1,1]$, and we have that for any $1>epsilon>0$, $|f'(x)|=|frac{1}{3 sqrt[3]{x^2}}|leq |frac{1}{3sqrt[3]{epsilon^2}}|$ for all $x in [epsilon,1]cup [-1,-epsilon]$, thus $f$ is Lipschitz on $[epsilon,1]cup [-1,-epsilon]$, hence $f$ is absolutely continuos on $[epsilon,1]cup [-1,-epsilon]$. Thus, $f$ is absolutely continuous on $[0,1]cup [-1,0]=[-1,1]$ since it is increaing.
Now, we see obviously, $g$ is differentiable on $[-1,1]$ with $g'(x)=2x cos(pi/2x)+frac{pi}{2}sin(pi/2x)$, and we have $|g'(x)|leq 2+1=3$. Therefore, $g$ is Lipschitz on $[-1,1]$, thus it is absolutely continuous on $[-1,1].$
Now, consider the partition $mathcal{P}_n={-1,0,frac{1}{2n},frac{1}{2n-1},...,frac{1}{2},1}$ of $[-1,1]$. We have $(f circ g)(x)=x^{^{frac{2}{3}}}cos^{^{frac{1}{3}}}(frac{pi}{2x})$, so $$sum_{i=1}^{2n-1}|f(x_{i})-f(x_{i-1})|=|0-0|+|frac{cos^{1/3}(n pi)}{2n^{2/3}}-0|+|frac{cos^{1/3}(frac{(2n-1)pi}{2})}{(2n-1)^{2/3}}-frac{cos^{1/3}(n pi)}{2n^{2/3}}|+...$$
$$+ |frac{cos^{1/3}( pi)}{2^{2/3}}-frac{cos^{1/3}(pi/2)}{1^{2/3}}|=sum_{i=0}^n2^{^{1/3}}frac{1}{i^{2/3}}=2^{^{1/3}}sum_{i=0}^nfrac{1}{i^{2/3}}.$$
since $2/3<1$, then the sum diverges as $n rightarrow infty$, thus $fcirc g$ fails to be of a bounded variation on $[-1,1]$, and hence it is not absolutely continuous on $[-1,1].$
But I can not see that $$int_{[0,1]}(f circ g)'=f(g(b))-f(g(a))$$
Thanks for any help.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Nov 28 '18 at 21:43
Ahmed Ahmed
29519
asked Nov 28 '18 at 21:43
Ahmed Ahmed
29519
edited Nov 28 '18 at 22:12
Ahmed
asked Nov 28 '18 at 21:43
Ahmed Ahmed
29519
asked Nov 28 '18 at 21:43
Ahmed Ahmed
29519
asked Nov 28 '18 at 21:43
Ahmed Ahmed
29519
29519
1
$begingroup$
Looking under en.wikipedia.org/wiki/Absolute_continuity in the "Equivalent Definitions" section it seems those 2 conditions are indeed equivalent. But I'm no expert in this, please check this yourself.
$endgroup$
– Ingix
Nov 28 '18 at 23:47
add a comment |
1
$begingroup$
Looking under en.wikipedia.org/wiki/Absolute_continuity in the "Equivalent Definitions" section it seems those 2 conditions are indeed equivalent. But I'm no expert in this, please check this yourself.
$endgroup$
– Ingix
Nov 28 '18 at 23:47
1
1
$begingroup$
Looking under en.wikipedia.org/wiki/Absolute_continuity in the "Equivalent Definitions" section it seems those 2 conditions are indeed equivalent. But I'm no expert in this, please check this yourself.
$endgroup$
– Ingix
Nov 28 '18 at 23:47
$begingroup$
Looking under en.wikipedia.org/wiki/Absolute_continuity in the "Equivalent Definitions" section it seems those 2 conditions are indeed equivalent. But I'm no expert in this, please check this yourself.
$endgroup$
– Ingix
Nov 28 '18 at 23:47
add a comment |
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$begingroup$
Looking under en.wikipedia.org/wiki/Absolute_continuity in the "Equivalent Definitions" section it seems those 2 conditions are indeed equivalent. But I'm no expert in this, please check this yourself.
$endgroup$
– Ingix
Nov 28 '18 at 23:47