Showing an Integrable function is everywhere discontinuous












1












$begingroup$



Question

Let $f(x) = x^{-1/2}$ for $0<x<1$ and $0$ otherwise, $r_n$ be an
enumeration of rationals. Define
begin{equation}
g(x) = sum_n g_n(x) quad text{where} quad g_n(x) = 2^{-n}f(x-r_n)
end{equation}
Show that $g$ is discontinuous everywhere.




Attempt

I claim $g$ is unbounded for any interval $I$, which
implies it cannot be continuous anywhere. Note that if $(x-r_n) leq 2^{-2n}$
then $g_n(x) geq 1$. Construction relies on the fact that
any interval contains infinitely many $r_n$.



For $I$, there exists $r_kin I$ with a corresponding interval $I_k
subset I$ of size $2^{-2k}$ with left endpoint $r_k$. Repeat for
$I_k$, there exists $r_{k'}in I_k$ with a corresponding interval
$I_{k'}subset I_k$ of size $2^{-2k'}$ with left end point $r_{k'}$.
Therefore, there exists an increasing sequence of $r_k$'s
and they converge, say to $x$. Note that $(x - r_k) leq 2^{-2k}$
for all $k$ in this construction. It follows, $g(x) = sum_n g_n(x)
geq sum_k g_k(x) geq sum_k 1= infty$.



Appreciate if one can verify/point out the mistake/suggest another approach. Notation is not very precise, but hopefully explains it.






real-analysis integration proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    I just wanted to comment that your interval $I$ should be closed. Certainly $f(x)=x$ is unbounded on $mathbb{R}$, but it’s continuous everywhere.
    $endgroup$
    – Live Free or π Hard
    Dec 25 '18 at 3:33










  • $begingroup$
    Why would $g$ be unbounded? Note that $|f(x)|leq 1$ for all $x$. So $|g_n(x)|leq 2^{-n}$ meaning $|g(x)|leq sumlimits_n 2^{-n}=1$
    $endgroup$
    – J. Pistachio
    Dec 25 '18 at 3:40










  • $begingroup$
    @J.Pistachio: $|f(x)|le 1$ would be the case if $f(x)$ was $x^{1/2}$, but it is $x^{-1/2}$.
    $endgroup$
    – Henning Makholm
    Dec 25 '18 at 3:44












  • $begingroup$
    Ah, didn't see the negative. Sorry about that
    $endgroup$
    – J. Pistachio
    Dec 25 '18 at 3:49
















1












$begingroup$



Question

Let $f(x) = x^{-1/2}$ for $0<x<1$ and $0$ otherwise, $r_n$ be an
enumeration of rationals. Define
begin{equation}
g(x) = sum_n g_n(x) quad text{where} quad g_n(x) = 2^{-n}f(x-r_n)
end{equation}
Show that $g$ is discontinuous everywhere.




Attempt

I claim $g$ is unbounded for any interval $I$, which
implies it cannot be continuous anywhere. Note that if $(x-r_n) leq 2^{-2n}$
then $g_n(x) geq 1$. Construction relies on the fact that
any interval contains infinitely many $r_n$.



For $I$, there exists $r_kin I$ with a corresponding interval $I_k
subset I$ of size $2^{-2k}$ with left endpoint $r_k$. Repeat for
$I_k$, there exists $r_{k'}in I_k$ with a corresponding interval
$I_{k'}subset I_k$ of size $2^{-2k'}$ with left end point $r_{k'}$.
Therefore, there exists an increasing sequence of $r_k$'s
and they converge, say to $x$. Note that $(x - r_k) leq 2^{-2k}$
for all $k$ in this construction. It follows, $g(x) = sum_n g_n(x)
geq sum_k g_k(x) geq sum_k 1= infty$.



Appreciate if one can verify/point out the mistake/suggest another approach. Notation is not very precise, but hopefully explains it.






real-analysis integration proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    I just wanted to comment that your interval $I$ should be closed. Certainly $f(x)=x$ is unbounded on $mathbb{R}$, but it’s continuous everywhere.
    $endgroup$
    – Live Free or π Hard
    Dec 25 '18 at 3:33










  • $begingroup$
    Why would $g$ be unbounded? Note that $|f(x)|leq 1$ for all $x$. So $|g_n(x)|leq 2^{-n}$ meaning $|g(x)|leq sumlimits_n 2^{-n}=1$
    $endgroup$
    – J. Pistachio
    Dec 25 '18 at 3:40










  • $begingroup$
    @J.Pistachio: $|f(x)|le 1$ would be the case if $f(x)$ was $x^{1/2}$, but it is $x^{-1/2}$.
    $endgroup$
    – Henning Makholm
    Dec 25 '18 at 3:44












  • $begingroup$
    Ah, didn't see the negative. Sorry about that
    $endgroup$
    – J. Pistachio
    Dec 25 '18 at 3:49














1












1








1





$begingroup$



Question

Let $f(x) = x^{-1/2}$ for $0<x<1$ and $0$ otherwise, $r_n$ be an
enumeration of rationals. Define
begin{equation}
g(x) = sum_n g_n(x) quad text{where} quad g_n(x) = 2^{-n}f(x-r_n)
end{equation}
Show that $g$ is discontinuous everywhere.




Attempt

I claim $g$ is unbounded for any interval $I$, which
implies it cannot be continuous anywhere. Note that if $(x-r_n) leq 2^{-2n}$
then $g_n(x) geq 1$. Construction relies on the fact that
any interval contains infinitely many $r_n$.



For $I$, there exists $r_kin I$ with a corresponding interval $I_k
subset I$ of size $2^{-2k}$ with left endpoint $r_k$. Repeat for
$I_k$, there exists $r_{k'}in I_k$ with a corresponding interval
$I_{k'}subset I_k$ of size $2^{-2k'}$ with left end point $r_{k'}$.
Therefore, there exists an increasing sequence of $r_k$'s
and they converge, say to $x$. Note that $(x - r_k) leq 2^{-2k}$
for all $k$ in this construction. It follows, $g(x) = sum_n g_n(x)
geq sum_k g_k(x) geq sum_k 1= infty$.



Appreciate if one can verify/point out the mistake/suggest another approach. Notation is not very precise, but hopefully explains it.






real-analysis integration proof-verification







share|cite|improve this question











$endgroup$





Question

Let $f(x) = x^{-1/2}$ for $0<x<1$ and $0$ otherwise, $r_n$ be an
enumeration of rationals. Define
begin{equation}
g(x) = sum_n g_n(x) quad text{where} quad g_n(x) = 2^{-n}f(x-r_n)
end{equation}
Show that $g$ is discontinuous everywhere.




Attempt

I claim $g$ is unbounded for any interval $I$, which
implies it cannot be continuous anywhere. Note that if $(x-r_n) leq 2^{-2n}$
then $g_n(x) geq 1$. Construction relies on the fact that
any interval contains infinitely many $r_n$.



For $I$, there exists $r_kin I$ with a corresponding interval $I_k
subset I$ of size $2^{-2k}$ with left endpoint $r_k$. Repeat for
$I_k$, there exists $r_{k'}in I_k$ with a corresponding interval
$I_{k'}subset I_k$ of size $2^{-2k'}$ with left end point $r_{k'}$.
Therefore, there exists an increasing sequence of $r_k$'s
and they converge, say to $x$. Note that $(x - r_k) leq 2^{-2k}$
for all $k$ in this construction. It follows, $g(x) = sum_n g_n(x)
geq sum_k g_k(x) geq sum_k 1= infty$.



Appreciate if one can verify/point out the mistake/suggest another approach. Notation is not very precise, but hopefully explains it.






real-analysis integration proof-verification




real-analysis integration proof-verification






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 3:31









Math1000

19.4k31746




19.4k31746










asked Dec 25 '18 at 3:24









Jo'Jo'

306110




306110












  • $begingroup$
    I just wanted to comment that your interval $I$ should be closed. Certainly $f(x)=x$ is unbounded on $mathbb{R}$, but it’s continuous everywhere.
    $endgroup$
    – Live Free or π Hard
    Dec 25 '18 at 3:33










  • $begingroup$
    Why would $g$ be unbounded? Note that $|f(x)|leq 1$ for all $x$. So $|g_n(x)|leq 2^{-n}$ meaning $|g(x)|leq sumlimits_n 2^{-n}=1$
    $endgroup$
    – J. Pistachio
    Dec 25 '18 at 3:40










  • $begingroup$
    @J.Pistachio: $|f(x)|le 1$ would be the case if $f(x)$ was $x^{1/2}$, but it is $x^{-1/2}$.
    $endgroup$
    – Henning Makholm
    Dec 25 '18 at 3:44












  • $begingroup$
    Ah, didn't see the negative. Sorry about that
    $endgroup$
    – J. Pistachio
    Dec 25 '18 at 3:49


















  • $begingroup$
    I just wanted to comment that your interval $I$ should be closed. Certainly $f(x)=x$ is unbounded on $mathbb{R}$, but it’s continuous everywhere.
    $endgroup$
    – Live Free or π Hard
    Dec 25 '18 at 3:33










  • $begingroup$
    Why would $g$ be unbounded? Note that $|f(x)|leq 1$ for all $x$. So $|g_n(x)|leq 2^{-n}$ meaning $|g(x)|leq sumlimits_n 2^{-n}=1$
    $endgroup$
    – J. Pistachio
    Dec 25 '18 at 3:40










  • $begingroup$
    @J.Pistachio: $|f(x)|le 1$ would be the case if $f(x)$ was $x^{1/2}$, but it is $x^{-1/2}$.
    $endgroup$
    – Henning Makholm
    Dec 25 '18 at 3:44












  • $begingroup$
    Ah, didn't see the negative. Sorry about that
    $endgroup$
    – J. Pistachio
    Dec 25 '18 at 3:49
















$begingroup$
I just wanted to comment that your interval $I$ should be closed. Certainly $f(x)=x$ is unbounded on $mathbb{R}$, but it’s continuous everywhere.
$endgroup$
– Live Free or π Hard
Dec 25 '18 at 3:33




$begingroup$
I just wanted to comment that your interval $I$ should be closed. Certainly $f(x)=x$ is unbounded on $mathbb{R}$, but it’s continuous everywhere.
$endgroup$
– Live Free or π Hard
Dec 25 '18 at 3:33












$begingroup$
Why would $g$ be unbounded? Note that $|f(x)|leq 1$ for all $x$. So $|g_n(x)|leq 2^{-n}$ meaning $|g(x)|leq sumlimits_n 2^{-n}=1$
$endgroup$
– J. Pistachio
Dec 25 '18 at 3:40




$begingroup$
Why would $g$ be unbounded? Note that $|f(x)|leq 1$ for all $x$. So $|g_n(x)|leq 2^{-n}$ meaning $|g(x)|leq sumlimits_n 2^{-n}=1$
$endgroup$
– J. Pistachio
Dec 25 '18 at 3:40












$begingroup$
@J.Pistachio: $|f(x)|le 1$ would be the case if $f(x)$ was $x^{1/2}$, but it is $x^{-1/2}$.
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:44






$begingroup$
@J.Pistachio: $|f(x)|le 1$ would be the case if $f(x)$ was $x^{1/2}$, but it is $x^{-1/2}$.
$endgroup$
– Henning Makholm
Dec 25 '18 at 3:44














$begingroup$
Ah, didn't see the negative. Sorry about that
$endgroup$
– J. Pistachio
Dec 25 '18 at 3:49




$begingroup$
Ah, didn't see the negative. Sorry about that
$endgroup$
– J. Pistachio
Dec 25 '18 at 3:49










1 Answer
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Apparently the codomain of $g$ includes $infty$. That can in itself be okay; there are good topologies on the extended real line.



Your argument seems to lead to the the fact that there is a dense set of points $x$ where $g(x)=infty$. However, that in itself does not make $g$ discontinuous everywhere -- as far as you have argued so far, there might be an entire interval somewhere where $g(x)=infty$. Then $g$ would be perfectly continuous in the interior of such interval.



So you need an explicit argument that this is not the case. (One is suggested by the title of your question, but you should write it down explicitly ...)






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    $begingroup$

    Apparently the codomain of $g$ includes $infty$. That can in itself be okay; there are good topologies on the extended real line.



    Your argument seems to lead to the the fact that there is a dense set of points $x$ where $g(x)=infty$. However, that in itself does not make $g$ discontinuous everywhere -- as far as you have argued so far, there might be an entire interval somewhere where $g(x)=infty$. Then $g$ would be perfectly continuous in the interior of such interval.



    So you need an explicit argument that this is not the case. (One is suggested by the title of your question, but you should write it down explicitly ...)






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Apparently the codomain of $g$ includes $infty$. That can in itself be okay; there are good topologies on the extended real line.



      Your argument seems to lead to the the fact that there is a dense set of points $x$ where $g(x)=infty$. However, that in itself does not make $g$ discontinuous everywhere -- as far as you have argued so far, there might be an entire interval somewhere where $g(x)=infty$. Then $g$ would be perfectly continuous in the interior of such interval.



      So you need an explicit argument that this is not the case. (One is suggested by the title of your question, but you should write it down explicitly ...)






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Apparently the codomain of $g$ includes $infty$. That can in itself be okay; there are good topologies on the extended real line.



        Your argument seems to lead to the the fact that there is a dense set of points $x$ where $g(x)=infty$. However, that in itself does not make $g$ discontinuous everywhere -- as far as you have argued so far, there might be an entire interval somewhere where $g(x)=infty$. Then $g$ would be perfectly continuous in the interior of such interval.



        So you need an explicit argument that this is not the case. (One is suggested by the title of your question, but you should write it down explicitly ...)






        share|cite|improve this answer









        $endgroup$



        Apparently the codomain of $g$ includes $infty$. That can in itself be okay; there are good topologies on the extended real line.



        Your argument seems to lead to the the fact that there is a dense set of points $x$ where $g(x)=infty$. However, that in itself does not make $g$ discontinuous everywhere -- as far as you have argued so far, there might be an entire interval somewhere where $g(x)=infty$. Then $g$ would be perfectly continuous in the interior of such interval.



        So you need an explicit argument that this is not the case. (One is suggested by the title of your question, but you should write it down explicitly ...)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 3:47









        Henning MakholmHenning Makholm

        244k17313557




        244k17313557






























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