Is this function continuous/smooth?












0












$begingroup$


Suppose $A subset mathbb{T}_2$ is a measurable set, where $mathbb{T}_2$ is the torus group.



For a fixed $n$, define the function $f_A:mathbb{T}_2^n rightarrow mathbb{R}$ as



$$f_A(x_1, ..., x_n) = begin{cases} 1 text{ if } x_1 in A, x_2 in A ... ,x_n in A \ 0 text{ otherwise}end{cases}.$$



Now with the diagonal action of $mathbb{T}_2$ on $mathbb{T}_2^n$ (written multiplicatively), define the function $g_A:mathbb{T}_2^nrightarrow mathbb{R}$ as



$$g_A(x) = int_{mathbb{T}_2}f_A(gx) dg $$



Is the function $g_A$ continous? It is smooth? What if I consider general a Lie group $G$ acting on a manifold $M$, and consider $A subset M$ to make a similar construction?






functional-analysis analysis lie-groups geometric-measure-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's certainly not continuous in general: when $A={0}$, then $g_A(x)$ equals $1$ if $x=(0,dots,0)$ but equals $0$ otherwise.
    $endgroup$
    – Greg Martin
    Dec 28 '18 at 21:32










  • $begingroup$
    I don't think so. For $A= { 0 }$, $f_A$ is the function you described. $g_A(0,0...)$ is the integral of $f_A(t,t,t.... )$ where $t in mathbb{T}_2$ and is therefore zero when $x=(0,0,0...)$.
    $endgroup$
    – Breakfastisready
    Dec 28 '18 at 22:21












  • $begingroup$
    I think you're right; I got confused between additive and multiplicative representations of the torus operation.
    $endgroup$
    – Greg Martin
    Dec 28 '18 at 23:54
















0












$begingroup$


Suppose $A subset mathbb{T}_2$ is a measurable set, where $mathbb{T}_2$ is the torus group.



For a fixed $n$, define the function $f_A:mathbb{T}_2^n rightarrow mathbb{R}$ as



$$f_A(x_1, ..., x_n) = begin{cases} 1 text{ if } x_1 in A, x_2 in A ... ,x_n in A \ 0 text{ otherwise}end{cases}.$$



Now with the diagonal action of $mathbb{T}_2$ on $mathbb{T}_2^n$ (written multiplicatively), define the function $g_A:mathbb{T}_2^nrightarrow mathbb{R}$ as



$$g_A(x) = int_{mathbb{T}_2}f_A(gx) dg $$



Is the function $g_A$ continous? It is smooth? What if I consider general a Lie group $G$ acting on a manifold $M$, and consider $A subset M$ to make a similar construction?






functional-analysis analysis lie-groups geometric-measure-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's certainly not continuous in general: when $A={0}$, then $g_A(x)$ equals $1$ if $x=(0,dots,0)$ but equals $0$ otherwise.
    $endgroup$
    – Greg Martin
    Dec 28 '18 at 21:32










  • $begingroup$
    I don't think so. For $A= { 0 }$, $f_A$ is the function you described. $g_A(0,0...)$ is the integral of $f_A(t,t,t.... )$ where $t in mathbb{T}_2$ and is therefore zero when $x=(0,0,0...)$.
    $endgroup$
    – Breakfastisready
    Dec 28 '18 at 22:21












  • $begingroup$
    I think you're right; I got confused between additive and multiplicative representations of the torus operation.
    $endgroup$
    – Greg Martin
    Dec 28 '18 at 23:54














0












0








0


1



$begingroup$


Suppose $A subset mathbb{T}_2$ is a measurable set, where $mathbb{T}_2$ is the torus group.



For a fixed $n$, define the function $f_A:mathbb{T}_2^n rightarrow mathbb{R}$ as



$$f_A(x_1, ..., x_n) = begin{cases} 1 text{ if } x_1 in A, x_2 in A ... ,x_n in A \ 0 text{ otherwise}end{cases}.$$



Now with the diagonal action of $mathbb{T}_2$ on $mathbb{T}_2^n$ (written multiplicatively), define the function $g_A:mathbb{T}_2^nrightarrow mathbb{R}$ as



$$g_A(x) = int_{mathbb{T}_2}f_A(gx) dg $$



Is the function $g_A$ continous? It is smooth? What if I consider general a Lie group $G$ acting on a manifold $M$, and consider $A subset M$ to make a similar construction?






functional-analysis analysis lie-groups geometric-measure-theory







share|cite|improve this question











$endgroup$




Suppose $A subset mathbb{T}_2$ is a measurable set, where $mathbb{T}_2$ is the torus group.



For a fixed $n$, define the function $f_A:mathbb{T}_2^n rightarrow mathbb{R}$ as



$$f_A(x_1, ..., x_n) = begin{cases} 1 text{ if } x_1 in A, x_2 in A ... ,x_n in A \ 0 text{ otherwise}end{cases}.$$



Now with the diagonal action of $mathbb{T}_2$ on $mathbb{T}_2^n$ (written multiplicatively), define the function $g_A:mathbb{T}_2^nrightarrow mathbb{R}$ as



$$g_A(x) = int_{mathbb{T}_2}f_A(gx) dg $$



Is the function $g_A$ continous? It is smooth? What if I consider general a Lie group $G$ acting on a manifold $M$, and consider $A subset M$ to make a similar construction?






functional-analysis analysis lie-groups geometric-measure-theory




functional-analysis analysis lie-groups geometric-measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 2:05







Breakfastisready

















asked Dec 28 '18 at 21:09









BreakfastisreadyBreakfastisready

929




929












  • $begingroup$
    It's certainly not continuous in general: when $A={0}$, then $g_A(x)$ equals $1$ if $x=(0,dots,0)$ but equals $0$ otherwise.
    $endgroup$
    – Greg Martin
    Dec 28 '18 at 21:32










  • $begingroup$
    I don't think so. For $A= { 0 }$, $f_A$ is the function you described. $g_A(0,0...)$ is the integral of $f_A(t,t,t.... )$ where $t in mathbb{T}_2$ and is therefore zero when $x=(0,0,0...)$.
    $endgroup$
    – Breakfastisready
    Dec 28 '18 at 22:21












  • $begingroup$
    I think you're right; I got confused between additive and multiplicative representations of the torus operation.
    $endgroup$
    – Greg Martin
    Dec 28 '18 at 23:54


















  • $begingroup$
    It's certainly not continuous in general: when $A={0}$, then $g_A(x)$ equals $1$ if $x=(0,dots,0)$ but equals $0$ otherwise.
    $endgroup$
    – Greg Martin
    Dec 28 '18 at 21:32










  • $begingroup$
    I don't think so. For $A= { 0 }$, $f_A$ is the function you described. $g_A(0,0...)$ is the integral of $f_A(t,t,t.... )$ where $t in mathbb{T}_2$ and is therefore zero when $x=(0,0,0...)$.
    $endgroup$
    – Breakfastisready
    Dec 28 '18 at 22:21












  • $begingroup$
    I think you're right; I got confused between additive and multiplicative representations of the torus operation.
    $endgroup$
    – Greg Martin
    Dec 28 '18 at 23:54
















$begingroup$
It's certainly not continuous in general: when $A={0}$, then $g_A(x)$ equals $1$ if $x=(0,dots,0)$ but equals $0$ otherwise.
$endgroup$
– Greg Martin
Dec 28 '18 at 21:32




$begingroup$
It's certainly not continuous in general: when $A={0}$, then $g_A(x)$ equals $1$ if $x=(0,dots,0)$ but equals $0$ otherwise.
$endgroup$
– Greg Martin
Dec 28 '18 at 21:32












$begingroup$
I don't think so. For $A= { 0 }$, $f_A$ is the function you described. $g_A(0,0...)$ is the integral of $f_A(t,t,t.... )$ where $t in mathbb{T}_2$ and is therefore zero when $x=(0,0,0...)$.
$endgroup$
– Breakfastisready
Dec 28 '18 at 22:21






$begingroup$
I don't think so. For $A= { 0 }$, $f_A$ is the function you described. $g_A(0,0...)$ is the integral of $f_A(t,t,t.... )$ where $t in mathbb{T}_2$ and is therefore zero when $x=(0,0,0...)$.
$endgroup$
– Breakfastisready
Dec 28 '18 at 22:21














$begingroup$
I think you're right; I got confused between additive and multiplicative representations of the torus operation.
$endgroup$
– Greg Martin
Dec 28 '18 at 23:54




$begingroup$
I think you're right; I got confused between additive and multiplicative representations of the torus operation.
$endgroup$
– Greg Martin
Dec 28 '18 at 23:54










1 Answer
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$begingroup$

Let $f_n$ be a sequence of continuous functions converging to $f_A$ (one can think of many such constructions).



Then $f_n$ are uniformly continuous because the domain is compact. One can easily prove that $g_n: x mapsto int_{g in mathbb{T}_2} f_n(g x) dx$ uniformly converge to $g_A$, and therefore $g_A$ must be continuous. A similar proof should generalize to the case of a compact group acting on a (compact?) manifold.



Smoothness is too much to ask for. We can take $n=2$ and $A$ to be a horizontal strip. Then the value of $g_A(x,y)$ is just a piecewise linear function only depending on the vertical distance of the points $x,y in mathbb{T}_2$






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    $begingroup$

    Let $f_n$ be a sequence of continuous functions converging to $f_A$ (one can think of many such constructions).



    Then $f_n$ are uniformly continuous because the domain is compact. One can easily prove that $g_n: x mapsto int_{g in mathbb{T}_2} f_n(g x) dx$ uniformly converge to $g_A$, and therefore $g_A$ must be continuous. A similar proof should generalize to the case of a compact group acting on a (compact?) manifold.



    Smoothness is too much to ask for. We can take $n=2$ and $A$ to be a horizontal strip. Then the value of $g_A(x,y)$ is just a piecewise linear function only depending on the vertical distance of the points $x,y in mathbb{T}_2$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Let $f_n$ be a sequence of continuous functions converging to $f_A$ (one can think of many such constructions).



      Then $f_n$ are uniformly continuous because the domain is compact. One can easily prove that $g_n: x mapsto int_{g in mathbb{T}_2} f_n(g x) dx$ uniformly converge to $g_A$, and therefore $g_A$ must be continuous. A similar proof should generalize to the case of a compact group acting on a (compact?) manifold.



      Smoothness is too much to ask for. We can take $n=2$ and $A$ to be a horizontal strip. Then the value of $g_A(x,y)$ is just a piecewise linear function only depending on the vertical distance of the points $x,y in mathbb{T}_2$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $f_n$ be a sequence of continuous functions converging to $f_A$ (one can think of many such constructions).



        Then $f_n$ are uniformly continuous because the domain is compact. One can easily prove that $g_n: x mapsto int_{g in mathbb{T}_2} f_n(g x) dx$ uniformly converge to $g_A$, and therefore $g_A$ must be continuous. A similar proof should generalize to the case of a compact group acting on a (compact?) manifold.



        Smoothness is too much to ask for. We can take $n=2$ and $A$ to be a horizontal strip. Then the value of $g_A(x,y)$ is just a piecewise linear function only depending on the vertical distance of the points $x,y in mathbb{T}_2$






        share|cite|improve this answer









        $endgroup$



        Let $f_n$ be a sequence of continuous functions converging to $f_A$ (one can think of many such constructions).



        Then $f_n$ are uniformly continuous because the domain is compact. One can easily prove that $g_n: x mapsto int_{g in mathbb{T}_2} f_n(g x) dx$ uniformly converge to $g_A$, and therefore $g_A$ must be continuous. A similar proof should generalize to the case of a compact group acting on a (compact?) manifold.



        Smoothness is too much to ask for. We can take $n=2$ and $A$ to be a horizontal strip. Then the value of $g_A(x,y)$ is just a piecewise linear function only depending on the vertical distance of the points $x,y in mathbb{T}_2$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 5:03









        BreakfastisreadyBreakfastisready

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        929






























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