Sequence of functions on $mathcal{L}^1([0,1])$ with $lim_{nrightarrow infty}||f_n||_1=0$ but $sup{f_n(x):...












1












$begingroup$


Problem Statement.



Good evening. As the title suggests, I am having a hard time with the following exercise:




Prove that there exists a sequence $f_1,f_2,ldots$ of functions on
$mathcal{L}^1([0,1])$ such that $lim_{nrightarrow
infty}||f_n||_1=0$
but $$sup{f_n(x): ninmathbf{Z}^+}=infty$$
for every $xin [0,1].$






Notation.



Here $mathcal{L}^1([0,1])$ means $mathcal{L}^1(lambda_{[0,1]})$ where $lambda_{[0,1]}$ represents Lebesgue measure restricted to the Borel subsets of $mathbf{R}$ that are contained in $[0,1]$.





Questions.



Ok, so my first question is how do I interpret $lim_{nrightarrow
infty}||f_n||_1=0$
? I know that by definition $$||f||_1=int |f|,dmu$$ so I think this statement means $$lim_{nrightarrow infty}||f_n||=lim_{nrightarrowinfty}int|f_n|,dmu=0.$$ But I haven't been able to get anywhere with this, possibly because of my confusion with the other part of the question, which is even finding a sequence of functions that have the property $sup{f_n(x): ninmathbf{Z}^+}=infty$. I initially tried playing around with functions like $$f_n(x)=frac{1}{sqrt{nx}}$$ but this is undefined at $x=0$. So to summarize, my two questions are:




  1. Understanding the limit notation I pointed out.

  2. Finding a sequence of functions with an undefined supremum at every $x$.


Any suggestions towards either of these points of confusion would be wonderful.



Thank you for your time!










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Try a sequence of functions where the height of the functions grows, but the size of the interval shrinks quickly. This will bound the area, but also make the height(and thus the sup) grow.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:21










  • $begingroup$
    @rubikscube09 Hmmm, ok. Would something like $f_n(x)=e^{nx}$ work?
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:24










  • $begingroup$
    @rubikscube09 I think the integral on that would grow instead of decay though hmmm... I'll keep trying! Thanks for the suggestion!
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:26








  • 1




    $begingroup$
    Do you know about indicator functions? They might be useful for something like this.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:28






  • 2




    $begingroup$
    Try to see how you could represent thin spikes that keep getting thinner and taller with indicator functions.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:34
















1












$begingroup$


Problem Statement.



Good evening. As the title suggests, I am having a hard time with the following exercise:




Prove that there exists a sequence $f_1,f_2,ldots$ of functions on
$mathcal{L}^1([0,1])$ such that $lim_{nrightarrow
infty}||f_n||_1=0$
but $$sup{f_n(x): ninmathbf{Z}^+}=infty$$
for every $xin [0,1].$






Notation.



Here $mathcal{L}^1([0,1])$ means $mathcal{L}^1(lambda_{[0,1]})$ where $lambda_{[0,1]}$ represents Lebesgue measure restricted to the Borel subsets of $mathbf{R}$ that are contained in $[0,1]$.





Questions.



Ok, so my first question is how do I interpret $lim_{nrightarrow
infty}||f_n||_1=0$
? I know that by definition $$||f||_1=int |f|,dmu$$ so I think this statement means $$lim_{nrightarrow infty}||f_n||=lim_{nrightarrowinfty}int|f_n|,dmu=0.$$ But I haven't been able to get anywhere with this, possibly because of my confusion with the other part of the question, which is even finding a sequence of functions that have the property $sup{f_n(x): ninmathbf{Z}^+}=infty$. I initially tried playing around with functions like $$f_n(x)=frac{1}{sqrt{nx}}$$ but this is undefined at $x=0$. So to summarize, my two questions are:




  1. Understanding the limit notation I pointed out.

  2. Finding a sequence of functions with an undefined supremum at every $x$.


Any suggestions towards either of these points of confusion would be wonderful.



Thank you for your time!










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Try a sequence of functions where the height of the functions grows, but the size of the interval shrinks quickly. This will bound the area, but also make the height(and thus the sup) grow.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:21










  • $begingroup$
    @rubikscube09 Hmmm, ok. Would something like $f_n(x)=e^{nx}$ work?
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:24










  • $begingroup$
    @rubikscube09 I think the integral on that would grow instead of decay though hmmm... I'll keep trying! Thanks for the suggestion!
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:26








  • 1




    $begingroup$
    Do you know about indicator functions? They might be useful for something like this.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:28






  • 2




    $begingroup$
    Try to see how you could represent thin spikes that keep getting thinner and taller with indicator functions.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:34














1












1








1





$begingroup$


Problem Statement.



Good evening. As the title suggests, I am having a hard time with the following exercise:




Prove that there exists a sequence $f_1,f_2,ldots$ of functions on
$mathcal{L}^1([0,1])$ such that $lim_{nrightarrow
infty}||f_n||_1=0$
but $$sup{f_n(x): ninmathbf{Z}^+}=infty$$
for every $xin [0,1].$






Notation.



Here $mathcal{L}^1([0,1])$ means $mathcal{L}^1(lambda_{[0,1]})$ where $lambda_{[0,1]}$ represents Lebesgue measure restricted to the Borel subsets of $mathbf{R}$ that are contained in $[0,1]$.





Questions.



Ok, so my first question is how do I interpret $lim_{nrightarrow
infty}||f_n||_1=0$
? I know that by definition $$||f||_1=int |f|,dmu$$ so I think this statement means $$lim_{nrightarrow infty}||f_n||=lim_{nrightarrowinfty}int|f_n|,dmu=0.$$ But I haven't been able to get anywhere with this, possibly because of my confusion with the other part of the question, which is even finding a sequence of functions that have the property $sup{f_n(x): ninmathbf{Z}^+}=infty$. I initially tried playing around with functions like $$f_n(x)=frac{1}{sqrt{nx}}$$ but this is undefined at $x=0$. So to summarize, my two questions are:




  1. Understanding the limit notation I pointed out.

  2. Finding a sequence of functions with an undefined supremum at every $x$.


Any suggestions towards either of these points of confusion would be wonderful.



Thank you for your time!










share|cite|improve this question









$endgroup$




Problem Statement.



Good evening. As the title suggests, I am having a hard time with the following exercise:




Prove that there exists a sequence $f_1,f_2,ldots$ of functions on
$mathcal{L}^1([0,1])$ such that $lim_{nrightarrow
infty}||f_n||_1=0$
but $$sup{f_n(x): ninmathbf{Z}^+}=infty$$
for every $xin [0,1].$






Notation.



Here $mathcal{L}^1([0,1])$ means $mathcal{L}^1(lambda_{[0,1]})$ where $lambda_{[0,1]}$ represents Lebesgue measure restricted to the Borel subsets of $mathbf{R}$ that are contained in $[0,1]$.





Questions.



Ok, so my first question is how do I interpret $lim_{nrightarrow
infty}||f_n||_1=0$
? I know that by definition $$||f||_1=int |f|,dmu$$ so I think this statement means $$lim_{nrightarrow infty}||f_n||=lim_{nrightarrowinfty}int|f_n|,dmu=0.$$ But I haven't been able to get anywhere with this, possibly because of my confusion with the other part of the question, which is even finding a sequence of functions that have the property $sup{f_n(x): ninmathbf{Z}^+}=infty$. I initially tried playing around with functions like $$f_n(x)=frac{1}{sqrt{nx}}$$ but this is undefined at $x=0$. So to summarize, my two questions are:




  1. Understanding the limit notation I pointed out.

  2. Finding a sequence of functions with an undefined supremum at every $x$.


Any suggestions towards either of these points of confusion would be wonderful.



Thank you for your time!







real-analysis sequences-and-series measure-theory lp-spaces supremum-and-infimum






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 5:16









Thy Art is MathThy Art is Math

489211




489211








  • 1




    $begingroup$
    Try a sequence of functions where the height of the functions grows, but the size of the interval shrinks quickly. This will bound the area, but also make the height(and thus the sup) grow.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:21










  • $begingroup$
    @rubikscube09 Hmmm, ok. Would something like $f_n(x)=e^{nx}$ work?
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:24










  • $begingroup$
    @rubikscube09 I think the integral on that would grow instead of decay though hmmm... I'll keep trying! Thanks for the suggestion!
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:26








  • 1




    $begingroup$
    Do you know about indicator functions? They might be useful for something like this.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:28






  • 2




    $begingroup$
    Try to see how you could represent thin spikes that keep getting thinner and taller with indicator functions.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:34














  • 1




    $begingroup$
    Try a sequence of functions where the height of the functions grows, but the size of the interval shrinks quickly. This will bound the area, but also make the height(and thus the sup) grow.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:21










  • $begingroup$
    @rubikscube09 Hmmm, ok. Would something like $f_n(x)=e^{nx}$ work?
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:24










  • $begingroup$
    @rubikscube09 I think the integral on that would grow instead of decay though hmmm... I'll keep trying! Thanks for the suggestion!
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:26








  • 1




    $begingroup$
    Do you know about indicator functions? They might be useful for something like this.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:28






  • 2




    $begingroup$
    Try to see how you could represent thin spikes that keep getting thinner and taller with indicator functions.
    $endgroup$
    – rubikscube09
    Dec 2 '18 at 5:34








1




1




$begingroup$
Try a sequence of functions where the height of the functions grows, but the size of the interval shrinks quickly. This will bound the area, but also make the height(and thus the sup) grow.
$endgroup$
– rubikscube09
Dec 2 '18 at 5:21




$begingroup$
Try a sequence of functions where the height of the functions grows, but the size of the interval shrinks quickly. This will bound the area, but also make the height(and thus the sup) grow.
$endgroup$
– rubikscube09
Dec 2 '18 at 5:21












$begingroup$
@rubikscube09 Hmmm, ok. Would something like $f_n(x)=e^{nx}$ work?
$endgroup$
– Thy Art is Math
Dec 2 '18 at 5:24




$begingroup$
@rubikscube09 Hmmm, ok. Would something like $f_n(x)=e^{nx}$ work?
$endgroup$
– Thy Art is Math
Dec 2 '18 at 5:24












$begingroup$
@rubikscube09 I think the integral on that would grow instead of decay though hmmm... I'll keep trying! Thanks for the suggestion!
$endgroup$
– Thy Art is Math
Dec 2 '18 at 5:26






$begingroup$
@rubikscube09 I think the integral on that would grow instead of decay though hmmm... I'll keep trying! Thanks for the suggestion!
$endgroup$
– Thy Art is Math
Dec 2 '18 at 5:26






1




1




$begingroup$
Do you know about indicator functions? They might be useful for something like this.
$endgroup$
– rubikscube09
Dec 2 '18 at 5:28




$begingroup$
Do you know about indicator functions? They might be useful for something like this.
$endgroup$
– rubikscube09
Dec 2 '18 at 5:28




2




2




$begingroup$
Try to see how you could represent thin spikes that keep getting thinner and taller with indicator functions.
$endgroup$
– rubikscube09
Dec 2 '18 at 5:34




$begingroup$
Try to see how you could represent thin spikes that keep getting thinner and taller with indicator functions.
$endgroup$
– rubikscube09
Dec 2 '18 at 5:34










1 Answer
1






active

oldest

votes


















2












$begingroup$

Here is an example of such a function:
$$
f_n:=ktimes 1_{[j/2^k,(j+1)/2^k]},
$$

where $n=2^k+j$ and $0le j< 2^k$.





$$
f_1equiv 0, f_2=1_{[0,1/2]}, f_3=1_{[1/2,1]} \
f_4=2times 1_{[0,1/4]}, f_5=2times 1_{[1/4,1/2]}, f_6=2times 1_{[1/2,3/4]}, f_7=2times 1_{[3/4,1]} \
ldots
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, the examples made that way clearer, thank you!
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:54











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Here is an example of such a function:
$$
f_n:=ktimes 1_{[j/2^k,(j+1)/2^k]},
$$

where $n=2^k+j$ and $0le j< 2^k$.





$$
f_1equiv 0, f_2=1_{[0,1/2]}, f_3=1_{[1/2,1]} \
f_4=2times 1_{[0,1/4]}, f_5=2times 1_{[1/4,1/2]}, f_6=2times 1_{[1/2,3/4]}, f_7=2times 1_{[3/4,1]} \
ldots
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, the examples made that way clearer, thank you!
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:54
















2












$begingroup$

Here is an example of such a function:
$$
f_n:=ktimes 1_{[j/2^k,(j+1)/2^k]},
$$

where $n=2^k+j$ and $0le j< 2^k$.





$$
f_1equiv 0, f_2=1_{[0,1/2]}, f_3=1_{[1/2,1]} \
f_4=2times 1_{[0,1/4]}, f_5=2times 1_{[1/4,1/2]}, f_6=2times 1_{[1/2,3/4]}, f_7=2times 1_{[3/4,1]} \
ldots
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Ok, the examples made that way clearer, thank you!
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:54














2












2








2





$begingroup$

Here is an example of such a function:
$$
f_n:=ktimes 1_{[j/2^k,(j+1)/2^k]},
$$

where $n=2^k+j$ and $0le j< 2^k$.





$$
f_1equiv 0, f_2=1_{[0,1/2]}, f_3=1_{[1/2,1]} \
f_4=2times 1_{[0,1/4]}, f_5=2times 1_{[1/4,1/2]}, f_6=2times 1_{[1/2,3/4]}, f_7=2times 1_{[3/4,1]} \
ldots
$$






share|cite|improve this answer











$endgroup$



Here is an example of such a function:
$$
f_n:=ktimes 1_{[j/2^k,(j+1)/2^k]},
$$

where $n=2^k+j$ and $0le j< 2^k$.





$$
f_1equiv 0, f_2=1_{[0,1/2]}, f_3=1_{[1/2,1]} \
f_4=2times 1_{[0,1/4]}, f_5=2times 1_{[1/4,1/2]}, f_6=2times 1_{[1/2,3/4]}, f_7=2times 1_{[3/4,1]} \
ldots
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 '18 at 5:53

























answered Dec 2 '18 at 5:38









d.k.o.d.k.o.

8,667528




8,667528












  • $begingroup$
    Ok, the examples made that way clearer, thank you!
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:54


















  • $begingroup$
    Ok, the examples made that way clearer, thank you!
    $endgroup$
    – Thy Art is Math
    Dec 2 '18 at 5:54
















$begingroup$
Ok, the examples made that way clearer, thank you!
$endgroup$
– Thy Art is Math
Dec 2 '18 at 5:54




$begingroup$
Ok, the examples made that way clearer, thank you!
$endgroup$
– Thy Art is Math
Dec 2 '18 at 5:54


















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