Showing an Integrable function is everywhere discontinuous
$begingroup$
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
edited Dec 25 '18 at 3:31
Math1000
19.4k31746
asked Dec 25 '18 at 3:24
Jo'Jo'
306110
$endgroup$
add a comment |
$begingroup$
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
edited Dec 25 '18 at 3:31
Math1000
19.4k31746
asked Dec 25 '18 at 3:24
Jo'Jo'
306110
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Dec 25 '18 at 3:31
Math1000
19.4k31746
asked Dec 25 '18 at 3:24
Jo'Jo'
306110
$endgroup$
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
edited Dec 25 '18 at 3:31
Math1000
19.4k31746
asked Dec 25 '18 at 3:24
Jo'Jo'
306110
edited Dec 25 '18 at 3:31
Math1000
19.4k31746
asked Dec 25 '18 at 3:24
Jo'Jo'
306110
edited Dec 25 '18 at 3:31
Math1000
19.4k31746
edited Dec 25 '18 at 3:31
Math1000
19.4k31746
edited Dec 25 '18 at 3:31
Math1000
19.4k31746
19.4k31746
asked Dec 25 '18 at 3:24
Jo'Jo'
306110
asked Dec 25 '18 at 3:24
Jo'Jo'
306110
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
add a comment |
$begingroup$
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
add a comment |
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 ...)
answered Dec 25 '18 at 3:47
Henning MakholmHenning Makholm
244k17313557
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add a comment |
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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 ...)
answered Dec 25 '18 at 3:47
Henning MakholmHenning Makholm
244k17313557
$endgroup$
add a comment |
$begingroup$
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 ...)
answered Dec 25 '18 at 3:47
Henning MakholmHenning Makholm
244k17313557
$endgroup$
add a comment |
$begingroup$
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 ...)
answered Dec 25 '18 at 3:47
Henning MakholmHenning Makholm
244k17313557
$endgroup$
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 ...)
answered Dec 25 '18 at 3:47
Henning MakholmHenning Makholm
244k17313557
answered Dec 25 '18 at 3:47
Henning MakholmHenning Makholm
244k17313557
answered Dec 25 '18 at 3:47
Henning MakholmHenning Makholm
244k17313557
answered Dec 25 '18 at 3:47
Henning MakholmHenning Makholm
244k17313557
244k17313557
add a comment |
add a comment |
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$begingroup$
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