Example of convergence under $|cdot|_2$, but not pointwise.
$begingroup$
Let $(f_n)_n$ be a sequence of continuous functions on $[0,1]$ and $f$ a continuous function on $[0,1]$. Assume that $|f_n-f|_2rightarrow 0$. Does it follow that $f_n(x)rightarrow f(x)$ for some $xin [0,1]$? Give a proof or counterexample. Also, can NOT use Lebesgue here.
$text{Proof}$
Let $f(x)=0$ and $I_{n,k}(x)={chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}:nin mathbb{N}, k=0,...,2^n}$ where $chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}$ is defined as
[begin{cases}
f_{n,k}(x) & text{if}space xin left[frac{k-1}{2^n},frac{k}{2^n}right] \
0 & text{if}space xnotin left[frac{k-1}{2^n},frac{k}{2^n}right] \
end{cases}
and $f_{n,k}(x)$ defined as
begin{cases}
(2^{n+1})x + (2-2k) & text{if}space xin left[ frac{k-1}{2^n},frac{2k-1}{2^{n+1}} right] \
-(2^{n+1})x + 2k & text{if}space xin left[ frac{2k-1}{2^{n+1}},frac{k}{2^n} right] \
end{cases}
Each "bump" is a triangle over the interval $left[frac{k-1}{2^n},frac{k}{2^n}right]$ with height $1$, so the area would be $frac{1}{2^{n+1}}$.
Hence, $largeint^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=frac{1}{2^{n+1}}$ and, therefore, we have that
$$ underset{nrightarrow infty}{lim}int^{1}_{0}I_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}int^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}frac{1}{2^{n+1}}=sum^{infty}_{k=1}0=0 $$
Hence, the sequence converges under $|cdot|_2$, but for any $xin [0,1]$ we can choose $Nin mathbb{N}$ so that
$$ |f_{N,k}(x)|>frac{1}{2} $$
Therefore, $I_{n,k}(x)$ converges no-where to $f(x)$.
Can anyone please check if my reasoning is correct? I know I'm probably not very convincing at the end since I do not know what value to make $N$, but I know that the interval can be divided up enough so that it isn't $0$. I'm also not sure if I have shown convergence under $|cdot|_2$ here either.
real-analysis integration
edited Dec 9 '18 at 9:16
Gaby Alfonso
1,1791318
asked Oct 1 '18 at 3:39
Joe Man AnalysisJoe Man Analysis
56419
$endgroup$
add a comment |
$begingroup$
Let $(f_n)_n$ be a sequence of continuous functions on $[0,1]$ and $f$ a continuous function on $[0,1]$. Assume that $|f_n-f|_2rightarrow 0$. Does it follow that $f_n(x)rightarrow f(x)$ for some $xin [0,1]$? Give a proof or counterexample. Also, can NOT use Lebesgue here.
$text{Proof}$
Let $f(x)=0$ and $I_{n,k}(x)={chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}:nin mathbb{N}, k=0,...,2^n}$ where $chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}$ is defined as
[begin{cases}
f_{n,k}(x) & text{if}space xin left[frac{k-1}{2^n},frac{k}{2^n}right] \
0 & text{if}space xnotin left[frac{k-1}{2^n},frac{k}{2^n}right] \
end{cases}
and $f_{n,k}(x)$ defined as
begin{cases}
(2^{n+1})x + (2-2k) & text{if}space xin left[ frac{k-1}{2^n},frac{2k-1}{2^{n+1}} right] \
-(2^{n+1})x + 2k & text{if}space xin left[ frac{2k-1}{2^{n+1}},frac{k}{2^n} right] \
end{cases}
Each "bump" is a triangle over the interval $left[frac{k-1}{2^n},frac{k}{2^n}right]$ with height $1$, so the area would be $frac{1}{2^{n+1}}$.
Hence, $largeint^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=frac{1}{2^{n+1}}$ and, therefore, we have that
$$ underset{nrightarrow infty}{lim}int^{1}_{0}I_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}int^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}frac{1}{2^{n+1}}=sum^{infty}_{k=1}0=0 $$
Hence, the sequence converges under $|cdot|_2$, but for any $xin [0,1]$ we can choose $Nin mathbb{N}$ so that
$$ |f_{N,k}(x)|>frac{1}{2} $$
Therefore, $I_{n,k}(x)$ converges no-where to $f(x)$.
Can anyone please check if my reasoning is correct? I know I'm probably not very convincing at the end since I do not know what value to make $N$, but I know that the interval can be divided up enough so that it isn't $0$. I'm also not sure if I have shown convergence under $|cdot|_2$ here either.
real-analysis integration
edited Dec 9 '18 at 9:16
Gaby Alfonso
1,1791318
asked Oct 1 '18 at 3:39
Joe Man AnalysisJoe Man Analysis
56419
$endgroup$
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
add a comment |
$begingroup$
Let $(f_n)_n$ be a sequence of continuous functions on $[0,1]$ and $f$ a continuous function on $[0,1]$. Assume that $|f_n-f|_2rightarrow 0$. Does it follow that $f_n(x)rightarrow f(x)$ for some $xin [0,1]$? Give a proof or counterexample. Also, can NOT use Lebesgue here.
$text{Proof}$
Let $f(x)=0$ and $I_{n,k}(x)={chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}:nin mathbb{N}, k=0,...,2^n}$ where $chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}$ is defined as
[begin{cases}
f_{n,k}(x) & text{if}space xin left[frac{k-1}{2^n},frac{k}{2^n}right] \
0 & text{if}space xnotin left[frac{k-1}{2^n},frac{k}{2^n}right] \
end{cases}
and $f_{n,k}(x)$ defined as
begin{cases}
(2^{n+1})x + (2-2k) & text{if}space xin left[ frac{k-1}{2^n},frac{2k-1}{2^{n+1}} right] \
-(2^{n+1})x + 2k & text{if}space xin left[ frac{2k-1}{2^{n+1}},frac{k}{2^n} right] \
end{cases}
Each "bump" is a triangle over the interval $left[frac{k-1}{2^n},frac{k}{2^n}right]$ with height $1$, so the area would be $frac{1}{2^{n+1}}$.
Hence, $largeint^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=frac{1}{2^{n+1}}$ and, therefore, we have that
$$ underset{nrightarrow infty}{lim}int^{1}_{0}I_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}int^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}frac{1}{2^{n+1}}=sum^{infty}_{k=1}0=0 $$
Hence, the sequence converges under $|cdot|_2$, but for any $xin [0,1]$ we can choose $Nin mathbb{N}$ so that
$$ |f_{N,k}(x)|>frac{1}{2} $$
Therefore, $I_{n,k}(x)$ converges no-where to $f(x)$.
Can anyone please check if my reasoning is correct? I know I'm probably not very convincing at the end since I do not know what value to make $N$, but I know that the interval can be divided up enough so that it isn't $0$. I'm also not sure if I have shown convergence under $|cdot|_2$ here either.
real-analysis integration
edited Dec 9 '18 at 9:16
Gaby Alfonso
1,1791318
asked Oct 1 '18 at 3:39
Joe Man AnalysisJoe Man Analysis
56419
$endgroup$
Let $(f_n)_n$ be a sequence of continuous functions on $[0,1]$ and $f$ a continuous function on $[0,1]$. Assume that $|f_n-f|_2rightarrow 0$. Does it follow that $f_n(x)rightarrow f(x)$ for some $xin [0,1]$? Give a proof or counterexample. Also, can NOT use Lebesgue here.
$text{Proof}$
Let $f(x)=0$ and $I_{n,k}(x)={chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}:nin mathbb{N}, k=0,...,2^n}$ where $chi_{left[frac{k-1}{2^n},frac{k}{2^n}right]}$ is defined as
[begin{cases}
f_{n,k}(x) & text{if}space xin left[frac{k-1}{2^n},frac{k}{2^n}right] \
0 & text{if}space xnotin left[frac{k-1}{2^n},frac{k}{2^n}right] \
end{cases}
and $f_{n,k}(x)$ defined as
begin{cases}
(2^{n+1})x + (2-2k) & text{if}space xin left[ frac{k-1}{2^n},frac{2k-1}{2^{n+1}} right] \
-(2^{n+1})x + 2k & text{if}space xin left[ frac{2k-1}{2^{n+1}},frac{k}{2^n} right] \
end{cases}
Each "bump" is a triangle over the interval $left[frac{k-1}{2^n},frac{k}{2^n}right]$ with height $1$, so the area would be $frac{1}{2^{n+1}}$.
Hence, $largeint^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=frac{1}{2^{n+1}}$ and, therefore, we have that
$$ underset{nrightarrow infty}{lim}int^{1}_{0}I_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}int^{frac{k-1}{2^n}}_{frac{k}{2^n}}f_{n,k}(x)dx=underset{nrightarrow infty}{lim}sum^{2^n}_{k=1}frac{1}{2^{n+1}}=sum^{infty}_{k=1}0=0 $$
Hence, the sequence converges under $|cdot|_2$, but for any $xin [0,1]$ we can choose $Nin mathbb{N}$ so that
$$ |f_{N,k}(x)|>frac{1}{2} $$
Therefore, $I_{n,k}(x)$ converges no-where to $f(x)$.
Can anyone please check if my reasoning is correct? I know I'm probably not very convincing at the end since I do not know what value to make $N$, but I know that the interval can be divided up enough so that it isn't $0$. I'm also not sure if I have shown convergence under $|cdot|_2$ here either.
real-analysis integration
real-analysis integration
edited Dec 9 '18 at 9:16
Gaby Alfonso
1,1791318
asked Oct 1 '18 at 3:39
Joe Man AnalysisJoe Man Analysis
56419
edited Dec 9 '18 at 9:16
Gaby Alfonso
1,1791318
asked Oct 1 '18 at 3:39
Joe Man AnalysisJoe Man Analysis
56419
edited Dec 9 '18 at 9:16
Gaby Alfonso
1,1791318
edited Dec 9 '18 at 9:16
Gaby Alfonso
1,1791318
edited Dec 9 '18 at 9:16
Gaby Alfonso
1,1791318
1,1791318
asked Oct 1 '18 at 3:39
Joe Man AnalysisJoe Man Analysis
56419
asked Oct 1 '18 at 3:39
Joe Man AnalysisJoe Man Analysis
56419
asked Oct 1 '18 at 3:39
Joe Man AnalysisJoe Man Analysis
56419
56419
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
add a comment |
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06
add a comment |
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$begingroup$
In your statement of your question, when you say "for some $x in [0, 1]$", do you mean "for all $x in [0, 1]$"?
$endgroup$
– Theo Bendit
Oct 1 '18 at 3:46
$begingroup$
It is for some $xin [0,1]$.
$endgroup$
– Joe Man Analysis
Oct 1 '18 at 3:52
1
$begingroup$
In your last line you have $lim_{nto infty}A(n)=0$ where $A(n)=sum_{k=1}^{2^n}frac {1}{2^{n+1}}$ but this is $false$ because $A(n)=1-2^{-(n+1)}$. Furthermore what you are interested in is whether $int_0^1 (sum_{k=1}^nI_k)^2dx $ goes to $0.$
$endgroup$
– DanielWainfleet
Oct 1 '18 at 5:57
1
$begingroup$
BTW you can use text {....} between dollar-signs or between double-dollar signs, around material (like the word "if") that should be un-Latexed... E.g .text { if }. You can also use , and ; and quad to insert various sizes of extra space. E.g. in a displayed line I like to write ; text { if }; for the word "if"
$endgroup$
– DanielWainfleet
Oct 1 '18 at 6:06