$L^p$ convergence of composition
$begingroup$
Let $f in L^{p}(U)$ and $(t_{n})$ be a sequence of functions from $U$ to $U$ such that $$t_{n} rightarrow operatorname{id},$$
in the space $L^{infty}(U)$.
Does the following hold:
$$f circ t_{n} rightarrow f,$$
in the space $L^{p}(U)$?
I tried using the dominated convergence theorem for $p_{n}(x)=left|f(x) - f(t_{n}(x))right|^{p}$ to show that the statement is true, but I couldn't find the "dominating" function.
sequences-and-series analysis measure-theory convergence lp-spaces
edited Dec 11 '18 at 20:52
Davide Giraudo
128k17156268
asked Apr 30 '18 at 8:33
user490393user490393
265
$endgroup$
|
show 3 more comments
$begingroup$
Let $f in L^{p}(U)$ and $(t_{n})$ be a sequence of functions from $U$ to $U$ such that $$t_{n} rightarrow operatorname{id},$$
in the space $L^{infty}(U)$.
Does the following hold:
$$f circ t_{n} rightarrow f,$$
in the space $L^{p}(U)$?
I tried using the dominated convergence theorem for $p_{n}(x)=left|f(x) - f(t_{n}(x))right|^{p}$ to show that the statement is true, but I couldn't find the "dominating" function.
sequences-and-series analysis measure-theory convergence lp-spaces
edited Dec 11 '18 at 20:52
Davide Giraudo
128k17156268
asked Apr 30 '18 at 8:33
user490393user490393
265
$endgroup$
$begingroup$
Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
$endgroup$
– humanStampedist
Apr 30 '18 at 9:08
$begingroup$
please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
$endgroup$
– JustDroppedIn
Apr 30 '18 at 18:19
$begingroup$
Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
$endgroup$
– user490393
May 1 '18 at 7:16
$begingroup$
Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
$endgroup$
– JustDroppedIn
May 1 '18 at 14:54
$begingroup$
Is $U$ of finite measure?
$endgroup$
– JustDroppedIn
May 1 '18 at 14:55
|
show 3 more comments
$begingroup$
Let $f in L^{p}(U)$ and $(t_{n})$ be a sequence of functions from $U$ to $U$ such that $$t_{n} rightarrow operatorname{id},$$
in the space $L^{infty}(U)$.
Does the following hold:
$$f circ t_{n} rightarrow f,$$
in the space $L^{p}(U)$?
I tried using the dominated convergence theorem for $p_{n}(x)=left|f(x) - f(t_{n}(x))right|^{p}$ to show that the statement is true, but I couldn't find the "dominating" function.
sequences-and-series analysis measure-theory convergence lp-spaces
edited Dec 11 '18 at 20:52
Davide Giraudo
128k17156268
asked Apr 30 '18 at 8:33
user490393user490393
265
$endgroup$
Let $f in L^{p}(U)$ and $(t_{n})$ be a sequence of functions from $U$ to $U$ such that $$t_{n} rightarrow operatorname{id},$$
in the space $L^{infty}(U)$.
Does the following hold:
$$f circ t_{n} rightarrow f,$$
in the space $L^{p}(U)$?
I tried using the dominated convergence theorem for $p_{n}(x)=left|f(x) - f(t_{n}(x))right|^{p}$ to show that the statement is true, but I couldn't find the "dominating" function.
sequences-and-series analysis measure-theory convergence lp-spaces
sequences-and-series analysis measure-theory convergence lp-spaces
edited Dec 11 '18 at 20:52
Davide Giraudo
128k17156268
asked Apr 30 '18 at 8:33
user490393user490393
265
edited Dec 11 '18 at 20:52
Davide Giraudo
128k17156268
asked Apr 30 '18 at 8:33
user490393user490393
265
edited Dec 11 '18 at 20:52
Davide Giraudo
128k17156268
edited Dec 11 '18 at 20:52
Davide Giraudo
128k17156268
edited Dec 11 '18 at 20:52
Davide Giraudo
128k17156268
128k17156268
asked Apr 30 '18 at 8:33
user490393user490393
265
asked Apr 30 '18 at 8:33
user490393user490393
265
asked Apr 30 '18 at 8:33
user490393user490393
265
265
$begingroup$
Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
$endgroup$
– humanStampedist
Apr 30 '18 at 9:08
$begingroup$
please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
$endgroup$
– JustDroppedIn
Apr 30 '18 at 18:19
$begingroup$
Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
$endgroup$
– user490393
May 1 '18 at 7:16
$begingroup$
Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
$endgroup$
– JustDroppedIn
May 1 '18 at 14:54
$begingroup$
Is $U$ of finite measure?
$endgroup$
– JustDroppedIn
May 1 '18 at 14:55
|
show 3 more comments
$begingroup$
Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
$endgroup$
– humanStampedist
Apr 30 '18 at 9:08
$begingroup$
please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
$endgroup$
– JustDroppedIn
Apr 30 '18 at 18:19
$begingroup$
Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
$endgroup$
– user490393
May 1 '18 at 7:16
$begingroup$
Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
$endgroup$
– JustDroppedIn
May 1 '18 at 14:54
$begingroup$
Is $U$ of finite measure?
$endgroup$
– JustDroppedIn
May 1 '18 at 14:55
$begingroup$
Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
$endgroup$
– humanStampedist
Apr 30 '18 at 9:08
$begingroup$
Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
$endgroup$
– humanStampedist
Apr 30 '18 at 9:08
$begingroup$
please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
$endgroup$
– JustDroppedIn
Apr 30 '18 at 18:19
$begingroup$
please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
$endgroup$
– JustDroppedIn
Apr 30 '18 at 18:19
$begingroup$
Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
$endgroup$
– user490393
May 1 '18 at 7:16
$begingroup$
Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
$endgroup$
– user490393
May 1 '18 at 7:16
$begingroup$
Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
$endgroup$
– JustDroppedIn
May 1 '18 at 14:54
$begingroup$
Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
$endgroup$
– JustDroppedIn
May 1 '18 at 14:54
$begingroup$
Is $U$ of finite measure?
$endgroup$
– JustDroppedIn
May 1 '18 at 14:55
$begingroup$
Is $U$ of finite measure?
$endgroup$
– JustDroppedIn
May 1 '18 at 14:55
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
$$
t_n(x) = frac{k}{2^n}
$$ for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
$$
fmapsto fcirc t_n
$$ is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
$$
1_mathbb{Q}circ t_n (x) = 1,
$$ while
$$
0circ t_n(x) = 0.
$$ Therefore, we see that composition may not be well-defined in general.
Instead of $L^p$ of equivalence classes, we may adopt the space
$$
mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
$$ of actual functions. However, we can see that the convergence can also fail in this case by the example
$$
1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
$$ Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.
answered Dec 11 '18 at 21:34
SongSong
18.5k21651
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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oldest
votes
$begingroup$
For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
$$
t_n(x) = frac{k}{2^n}
$$ for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
$$
fmapsto fcirc t_n
$$ is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
$$
1_mathbb{Q}circ t_n (x) = 1,
$$ while
$$
0circ t_n(x) = 0.
$$ Therefore, we see that composition may not be well-defined in general.
Instead of $L^p$ of equivalence classes, we may adopt the space
$$
mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
$$ of actual functions. However, we can see that the convergence can also fail in this case by the example
$$
1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
$$ Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.
answered Dec 11 '18 at 21:34
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
$$
t_n(x) = frac{k}{2^n}
$$ for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
$$
fmapsto fcirc t_n
$$ is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
$$
1_mathbb{Q}circ t_n (x) = 1,
$$ while
$$
0circ t_n(x) = 0.
$$ Therefore, we see that composition may not be well-defined in general.
Instead of $L^p$ of equivalence classes, we may adopt the space
$$
mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
$$ of actual functions. However, we can see that the convergence can also fail in this case by the example
$$
1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
$$ Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.
answered Dec 11 '18 at 21:34
SongSong
18.5k21651
$endgroup$
add a comment |
$begingroup$
For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
$$
t_n(x) = frac{k}{2^n}
$$ for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
$$
fmapsto fcirc t_n
$$ is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
$$
1_mathbb{Q}circ t_n (x) = 1,
$$ while
$$
0circ t_n(x) = 0.
$$ Therefore, we see that composition may not be well-defined in general.
Instead of $L^p$ of equivalence classes, we may adopt the space
$$
mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
$$ of actual functions. However, we can see that the convergence can also fail in this case by the example
$$
1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
$$ Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.
answered Dec 11 '18 at 21:34
SongSong
18.5k21651
$endgroup$
For the convergence to happen, it is important to know how $t_n$ mixes the domain $U$. And there is a subtle technical issue associated with the definition of $L^p$. Let $U=[0,1]$ equipped with the Lebesgue measure and $t_n$ be
$$
t_n(x) = frac{k}{2^n}
$$ for $xin [frac{k}{2^n},frac{k+1}{2^n})$ for $k=0,1,ldots,2^n$. We can easily see that $t_n to operatorname{id}$ in $L^infty$. An issue here is that
$$
fmapsto fcirc t_n
$$ is not well-defined on $L^p([0,1])$. To see this, recall that $L^p([0,1])$ space is not a family of functions, but a family of equivalence classes of functions. Pick $0 = 1_mathbb{Q} $ in $L^p([0,1])$. The composition with $t_n$ gives incompatible results that
$$
1_mathbb{Q}circ t_n (x) = 1,
$$ while
$$
0circ t_n(x) = 0.
$$ Therefore, we see that composition may not be well-defined in general.
Instead of $L^p$ of equivalence classes, we may adopt the space
$$
mathcal{L}^p = {f:Uto mathbb{C};|;ftext{ is measurable},int_U|f|^p dx <infty}
$$ of actual functions. However, we can see that the convergence can also fail in this case by the example
$$
1_mathbb{Q}circ t_n = 1 notto 1_mathbb{Q}.
$$ Thus, unfortunately, we conclude that mixing up the domain does not work properly in general. To properly define composition, some additional conditions such as preserving measure should be imposed on $t_n$.
answered Dec 11 '18 at 21:34
SongSong
18.5k21651
edited Dec 12 '18 at 6:12
answered Dec 11 '18 at 21:34
SongSong
18.5k21651
answered Dec 11 '18 at 21:34
SongSong
18.5k21651
answered Dec 11 '18 at 21:34
SongSong
18.5k21651
18.5k21651
add a comment |
add a comment |
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$begingroup$
Maybe another convergence theorem is more appropriate. Have you tried applying the Vitali convergence theorem? See en.wikipedia.org/wiki/Vitali_convergence_theorem
$endgroup$
– humanStampedist
Apr 30 '18 at 9:08
$begingroup$
please add some context, is $p>1$? $is Usubsetmathbb{R}^n$? is the measure the Lebesgue measure? are $t_n$ continuous? In your attempt there is not even pointwise convergence.
$endgroup$
– JustDroppedIn
Apr 30 '18 at 18:19
$begingroup$
Yes, Lebesgue measure is used and $U subset mathbb{R}^{n}$, but $p$ is not restricted (in order to see when the statement holds and when it does not). Doesn't the uniform convergence imply pointwise convergence, for almost all points? Thank you for the "Vitali convergence", I'll take a look.
$endgroup$
– user490393
May 1 '18 at 7:16
$begingroup$
Yes, but you cant know that if $x_nto x$ then $ f(x_n)to f(x)$ if $f$ is not continuous
$endgroup$
– JustDroppedIn
May 1 '18 at 14:54
$begingroup$
Is $U$ of finite measure?
$endgroup$
– JustDroppedIn
May 1 '18 at 14:55