Limit of increasing piecewise function is increasing?
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Suppose we have a sequence of functions defined on some interval $[0,1]$ which are piecewise constant and each of the functions is also increasing. It converges (when you shrink the partitions of $[0,1]$ so as to get a non-piecewise function) in some $L^p$ space to a function. Is the limit also increasing?
Do we not need some convergence in $C^0$ spaces to say this?
functional-analysis
asked Dec 6 '18 at 14:19
StopUsingFacebookStopUsingFacebook
1908
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add a comment |
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Suppose we have a sequence of functions defined on some interval $[0,1]$ which are piecewise constant and each of the functions is also increasing. It converges (when you shrink the partitions of $[0,1]$ so as to get a non-piecewise function) in some $L^p$ space to a function. Is the limit also increasing?
Do we not need some convergence in $C^0$ spaces to say this?
functional-analysis
asked Dec 6 '18 at 14:19
StopUsingFacebookStopUsingFacebook
1908
$endgroup$
1
$begingroup$
How can one element of a space (in your case a function) converge to another? That's like saying that $2$ converges to $1$...
$endgroup$
– 5xum
Dec 6 '18 at 14:28
add a comment |
$begingroup$
Suppose we have a sequence of functions defined on some interval $[0,1]$ which are piecewise constant and each of the functions is also increasing. It converges (when you shrink the partitions of $[0,1]$ so as to get a non-piecewise function) in some $L^p$ space to a function. Is the limit also increasing?
Do we not need some convergence in $C^0$ spaces to say this?
functional-analysis
asked Dec 6 '18 at 14:19
StopUsingFacebookStopUsingFacebook
1908
$endgroup$
Suppose we have a sequence of functions defined on some interval $[0,1]$ which are piecewise constant and each of the functions is also increasing. It converges (when you shrink the partitions of $[0,1]$ so as to get a non-piecewise function) in some $L^p$ space to a function. Is the limit also increasing?
Do we not need some convergence in $C^0$ spaces to say this?
functional-analysis
functional-analysis
asked Dec 6 '18 at 14:19
StopUsingFacebookStopUsingFacebook
1908
asked Dec 6 '18 at 14:19
StopUsingFacebookStopUsingFacebook
1908
edited Dec 6 '18 at 14:33
StopUsingFacebook
asked Dec 6 '18 at 14:19
StopUsingFacebookStopUsingFacebook
1908
asked Dec 6 '18 at 14:19
StopUsingFacebookStopUsingFacebook
1908
asked Dec 6 '18 at 14:19
StopUsingFacebookStopUsingFacebook
1908
1908
1
$begingroup$
How can one element of a space (in your case a function) converge to another? That's like saying that $2$ converges to $1$...
$endgroup$
– 5xum
Dec 6 '18 at 14:28
add a comment |
1
$begingroup$
How can one element of a space (in your case a function) converge to another? That's like saying that $2$ converges to $1$...
$endgroup$
– 5xum
Dec 6 '18 at 14:28
1
1
$begingroup$
How can one element of a space (in your case a function) converge to another? That's like saying that $2$ converges to $1$...
$endgroup$
– 5xum
Dec 6 '18 at 14:28
$begingroup$
How can one element of a space (in your case a function) converge to another? That's like saying that $2$ converges to $1$...
$endgroup$
– 5xum
Dec 6 '18 at 14:28
add a comment |
1 Answer
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If we have $f_n rightarrow f$ in $L^p(lambda)$, then we get for some subsequence also that $f_{n_k} rightarrow f$ almost everywhere. Let $N$ be the (measurable) nullset such that $f_{n_k}(x) rightarrow f(x)$ for all $x notin N$ For any $x le y$ with $x,y, notin N$, we get that $$f(x) = lim_{k rightarrow infty} f_{n_k}(x) lelim_{k rightarrow infty} f_{n_k}(y) = f(y).$$
Thus $f$ is monotone on $N^c$. Set now
$$widetilde{f}(x) := max_{y le x, y in N^c} f(x).$$
By definition $widetilde{f}(x)$ is a monotone function with $widetilde{f}(x) = f(x)$ for any $x in N^c$. Any monotone function on $mathbb{R}$ has at most countable (jump) discontinuities and is measurable. Thus in $L^p$ we have that $[widetilde{f}] = [f]$, i.e. $f = widetilde{f}$ almost everywhere. Of course, we cannot expect more in $L^p$.
answered Dec 6 '18 at 14:48
p4schp4sch
5,430318
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$begingroup$
If we have $f_n rightarrow f$ in $L^p(lambda)$, then we get for some subsequence also that $f_{n_k} rightarrow f$ almost everywhere. Let $N$ be the (measurable) nullset such that $f_{n_k}(x) rightarrow f(x)$ for all $x notin N$ For any $x le y$ with $x,y, notin N$, we get that $$f(x) = lim_{k rightarrow infty} f_{n_k}(x) lelim_{k rightarrow infty} f_{n_k}(y) = f(y).$$
Thus $f$ is monotone on $N^c$. Set now
$$widetilde{f}(x) := max_{y le x, y in N^c} f(x).$$
By definition $widetilde{f}(x)$ is a monotone function with $widetilde{f}(x) = f(x)$ for any $x in N^c$. Any monotone function on $mathbb{R}$ has at most countable (jump) discontinuities and is measurable. Thus in $L^p$ we have that $[widetilde{f}] = [f]$, i.e. $f = widetilde{f}$ almost everywhere. Of course, we cannot expect more in $L^p$.
answered Dec 6 '18 at 14:48
p4schp4sch
5,430318
$endgroup$
add a comment |
$begingroup$
If we have $f_n rightarrow f$ in $L^p(lambda)$, then we get for some subsequence also that $f_{n_k} rightarrow f$ almost everywhere. Let $N$ be the (measurable) nullset such that $f_{n_k}(x) rightarrow f(x)$ for all $x notin N$ For any $x le y$ with $x,y, notin N$, we get that $$f(x) = lim_{k rightarrow infty} f_{n_k}(x) lelim_{k rightarrow infty} f_{n_k}(y) = f(y).$$
Thus $f$ is monotone on $N^c$. Set now
$$widetilde{f}(x) := max_{y le x, y in N^c} f(x).$$
By definition $widetilde{f}(x)$ is a monotone function with $widetilde{f}(x) = f(x)$ for any $x in N^c$. Any monotone function on $mathbb{R}$ has at most countable (jump) discontinuities and is measurable. Thus in $L^p$ we have that $[widetilde{f}] = [f]$, i.e. $f = widetilde{f}$ almost everywhere. Of course, we cannot expect more in $L^p$.
answered Dec 6 '18 at 14:48
p4schp4sch
5,430318
$endgroup$
add a comment |
$begingroup$
If we have $f_n rightarrow f$ in $L^p(lambda)$, then we get for some subsequence also that $f_{n_k} rightarrow f$ almost everywhere. Let $N$ be the (measurable) nullset such that $f_{n_k}(x) rightarrow f(x)$ for all $x notin N$ For any $x le y$ with $x,y, notin N$, we get that $$f(x) = lim_{k rightarrow infty} f_{n_k}(x) lelim_{k rightarrow infty} f_{n_k}(y) = f(y).$$
Thus $f$ is monotone on $N^c$. Set now
$$widetilde{f}(x) := max_{y le x, y in N^c} f(x).$$
By definition $widetilde{f}(x)$ is a monotone function with $widetilde{f}(x) = f(x)$ for any $x in N^c$. Any monotone function on $mathbb{R}$ has at most countable (jump) discontinuities and is measurable. Thus in $L^p$ we have that $[widetilde{f}] = [f]$, i.e. $f = widetilde{f}$ almost everywhere. Of course, we cannot expect more in $L^p$.
answered Dec 6 '18 at 14:48
p4schp4sch
5,430318
$endgroup$
If we have $f_n rightarrow f$ in $L^p(lambda)$, then we get for some subsequence also that $f_{n_k} rightarrow f$ almost everywhere. Let $N$ be the (measurable) nullset such that $f_{n_k}(x) rightarrow f(x)$ for all $x notin N$ For any $x le y$ with $x,y, notin N$, we get that $$f(x) = lim_{k rightarrow infty} f_{n_k}(x) lelim_{k rightarrow infty} f_{n_k}(y) = f(y).$$
Thus $f$ is monotone on $N^c$. Set now
$$widetilde{f}(x) := max_{y le x, y in N^c} f(x).$$
By definition $widetilde{f}(x)$ is a monotone function with $widetilde{f}(x) = f(x)$ for any $x in N^c$. Any monotone function on $mathbb{R}$ has at most countable (jump) discontinuities and is measurable. Thus in $L^p$ we have that $[widetilde{f}] = [f]$, i.e. $f = widetilde{f}$ almost everywhere. Of course, we cannot expect more in $L^p$.
answered Dec 6 '18 at 14:48
p4schp4sch
5,430318
answered Dec 6 '18 at 14:48
p4schp4sch
5,430318
answered Dec 6 '18 at 14:48
p4schp4sch
5,430318
answered Dec 6 '18 at 14:48
p4schp4sch
5,430318
5,430318
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How can one element of a space (in your case a function) converge to another? That's like saying that $2$ converges to $1$...
$endgroup$
– 5xum
Dec 6 '18 at 14:28