Show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space
$begingroup$
Problem: Show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space.
I have tried to proceed along the lines given in the accepted answer, and subsequent comment, to this question: Example of a non complete normed vector space. Although, I am running into difficulties and wondering if the example really generalises so well when we have a general $[a,b]$. Is the Cauchy sequence I have chosen below a valid one, or have I overlooked a much simpler example to use in this case?
Attempt: In order to show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space we need to show that there is some Cauchy sequence that does not converge to it's limit in $C([a,b])$. In particular, it will suffice to show that our chosen Cauchy sequence converges under $|cdot|_2$ to a discontinuous function.
Consider the following sequence of functions, $(f_n)_{n=1}^inftysubset C([a,b])$, given by,
$$f_n(x)=begin{cases}
0,,text{when},,xin[a,frac{b-a}2-frac{2^{-n}}{b-a}] \[2ex]
1,,text{when},,xin[frac{b-a}2+frac{2^{-n}}{b-a},b] \[2ex]
frac{(2x-b+a)(b-a)-2^{-n+1}}{2^{-n+2}},,text{when},,xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]
end{cases}
$$
The definition of $f_n(x)$ when $xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]$ is the linear interpolant over that particular subinterval of $[a,b]$.
I claim that this is a Cauchy sequence. We consider:
$$|f_n-f_m|_2^2=int_a^b|f_n(x)-f_m(x)|^2dx$$
$$=int_a^{frac{b-a}2-frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2+frac{2^{-n}}{b-a}}^b|f_n(x)-f_m(x)|^2dx$$
$$=0+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+0$$
The $0$'s for the first and last integrals because we have $f_n(x)-f_m(x)=0-0$ and $f_n(x)-f_m(x)=1-1$ on each respectively. As to the middle integral, here is what I've gotten so far:
$$int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx=int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|(2x-b-a)(b-a)(2^{n-2}-2^{m-2})|^2dx$$
Based in the example I am following, supposing that $mge n$, I think that I should be able to argue that the whole thing is bound above by $frac{2^{-n}}{b-a}$ which would mean that $|f_n-f_m|_2^2lefrac{2^{-n}}{b-a}to0$ as $ntoinfty$.
I don't quite see how I'm meant to bound it above by $frac{2^{-n}}{b-a}$, as I keep tripping up over applying that $mge n$, leading me to question whether the approach I have employed here is a fruitful one after all.
real-analysis functional-analysis banach-spaces norm normed-spaces
$endgroup$
add a comment |
$begingroup$
Problem: Show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space.
I have tried to proceed along the lines given in the accepted answer, and subsequent comment, to this question: Example of a non complete normed vector space. Although, I am running into difficulties and wondering if the example really generalises so well when we have a general $[a,b]$. Is the Cauchy sequence I have chosen below a valid one, or have I overlooked a much simpler example to use in this case?
Attempt: In order to show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space we need to show that there is some Cauchy sequence that does not converge to it's limit in $C([a,b])$. In particular, it will suffice to show that our chosen Cauchy sequence converges under $|cdot|_2$ to a discontinuous function.
Consider the following sequence of functions, $(f_n)_{n=1}^inftysubset C([a,b])$, given by,
$$f_n(x)=begin{cases}
0,,text{when},,xin[a,frac{b-a}2-frac{2^{-n}}{b-a}] \[2ex]
1,,text{when},,xin[frac{b-a}2+frac{2^{-n}}{b-a},b] \[2ex]
frac{(2x-b+a)(b-a)-2^{-n+1}}{2^{-n+2}},,text{when},,xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]
end{cases}
$$
The definition of $f_n(x)$ when $xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]$ is the linear interpolant over that particular subinterval of $[a,b]$.
I claim that this is a Cauchy sequence. We consider:
$$|f_n-f_m|_2^2=int_a^b|f_n(x)-f_m(x)|^2dx$$
$$=int_a^{frac{b-a}2-frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2+frac{2^{-n}}{b-a}}^b|f_n(x)-f_m(x)|^2dx$$
$$=0+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+0$$
The $0$'s for the first and last integrals because we have $f_n(x)-f_m(x)=0-0$ and $f_n(x)-f_m(x)=1-1$ on each respectively. As to the middle integral, here is what I've gotten so far:
$$int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx=int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|(2x-b-a)(b-a)(2^{n-2}-2^{m-2})|^2dx$$
Based in the example I am following, supposing that $mge n$, I think that I should be able to argue that the whole thing is bound above by $frac{2^{-n}}{b-a}$ which would mean that $|f_n-f_m|_2^2lefrac{2^{-n}}{b-a}to0$ as $ntoinfty$.
I don't quite see how I'm meant to bound it above by $frac{2^{-n}}{b-a}$, as I keep tripping up over applying that $mge n$, leading me to question whether the approach I have employed here is a fruitful one after all.
real-analysis functional-analysis banach-spaces norm normed-spaces
$endgroup$
$begingroup$
one of the functions dominates the other, so you just have to subtract the individual integrals. Can you integrate linear function?
$endgroup$
– Jorge Fernández
Mar 14 '18 at 22:29
1
$begingroup$
instead of $frac{b-a}{2}$ you should have $frac{b+a}{2}$
$endgroup$
– Orest Bucicovschi
Mar 15 '18 at 1:58
add a comment |
$begingroup$
Problem: Show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space.
I have tried to proceed along the lines given in the accepted answer, and subsequent comment, to this question: Example of a non complete normed vector space. Although, I am running into difficulties and wondering if the example really generalises so well when we have a general $[a,b]$. Is the Cauchy sequence I have chosen below a valid one, or have I overlooked a much simpler example to use in this case?
Attempt: In order to show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space we need to show that there is some Cauchy sequence that does not converge to it's limit in $C([a,b])$. In particular, it will suffice to show that our chosen Cauchy sequence converges under $|cdot|_2$ to a discontinuous function.
Consider the following sequence of functions, $(f_n)_{n=1}^inftysubset C([a,b])$, given by,
$$f_n(x)=begin{cases}
0,,text{when},,xin[a,frac{b-a}2-frac{2^{-n}}{b-a}] \[2ex]
1,,text{when},,xin[frac{b-a}2+frac{2^{-n}}{b-a},b] \[2ex]
frac{(2x-b+a)(b-a)-2^{-n+1}}{2^{-n+2}},,text{when},,xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]
end{cases}
$$
The definition of $f_n(x)$ when $xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]$ is the linear interpolant over that particular subinterval of $[a,b]$.
I claim that this is a Cauchy sequence. We consider:
$$|f_n-f_m|_2^2=int_a^b|f_n(x)-f_m(x)|^2dx$$
$$=int_a^{frac{b-a}2-frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2+frac{2^{-n}}{b-a}}^b|f_n(x)-f_m(x)|^2dx$$
$$=0+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+0$$
The $0$'s for the first and last integrals because we have $f_n(x)-f_m(x)=0-0$ and $f_n(x)-f_m(x)=1-1$ on each respectively. As to the middle integral, here is what I've gotten so far:
$$int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx=int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|(2x-b-a)(b-a)(2^{n-2}-2^{m-2})|^2dx$$
Based in the example I am following, supposing that $mge n$, I think that I should be able to argue that the whole thing is bound above by $frac{2^{-n}}{b-a}$ which would mean that $|f_n-f_m|_2^2lefrac{2^{-n}}{b-a}to0$ as $ntoinfty$.
I don't quite see how I'm meant to bound it above by $frac{2^{-n}}{b-a}$, as I keep tripping up over applying that $mge n$, leading me to question whether the approach I have employed here is a fruitful one after all.
real-analysis functional-analysis banach-spaces norm normed-spaces
$endgroup$
Problem: Show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space.
I have tried to proceed along the lines given in the accepted answer, and subsequent comment, to this question: Example of a non complete normed vector space. Although, I am running into difficulties and wondering if the example really generalises so well when we have a general $[a,b]$. Is the Cauchy sequence I have chosen below a valid one, or have I overlooked a much simpler example to use in this case?
Attempt: In order to show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space we need to show that there is some Cauchy sequence that does not converge to it's limit in $C([a,b])$. In particular, it will suffice to show that our chosen Cauchy sequence converges under $|cdot|_2$ to a discontinuous function.
Consider the following sequence of functions, $(f_n)_{n=1}^inftysubset C([a,b])$, given by,
$$f_n(x)=begin{cases}
0,,text{when},,xin[a,frac{b-a}2-frac{2^{-n}}{b-a}] \[2ex]
1,,text{when},,xin[frac{b-a}2+frac{2^{-n}}{b-a},b] \[2ex]
frac{(2x-b+a)(b-a)-2^{-n+1}}{2^{-n+2}},,text{when},,xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]
end{cases}
$$
The definition of $f_n(x)$ when $xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]$ is the linear interpolant over that particular subinterval of $[a,b]$.
I claim that this is a Cauchy sequence. We consider:
$$|f_n-f_m|_2^2=int_a^b|f_n(x)-f_m(x)|^2dx$$
$$=int_a^{frac{b-a}2-frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2+frac{2^{-n}}{b-a}}^b|f_n(x)-f_m(x)|^2dx$$
$$=0+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+0$$
The $0$'s for the first and last integrals because we have $f_n(x)-f_m(x)=0-0$ and $f_n(x)-f_m(x)=1-1$ on each respectively. As to the middle integral, here is what I've gotten so far:
$$int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx=int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|(2x-b-a)(b-a)(2^{n-2}-2^{m-2})|^2dx$$
Based in the example I am following, supposing that $mge n$, I think that I should be able to argue that the whole thing is bound above by $frac{2^{-n}}{b-a}$ which would mean that $|f_n-f_m|_2^2lefrac{2^{-n}}{b-a}to0$ as $ntoinfty$.
I don't quite see how I'm meant to bound it above by $frac{2^{-n}}{b-a}$, as I keep tripping up over applying that $mge n$, leading me to question whether the approach I have employed here is a fruitful one after all.
real-analysis functional-analysis banach-spaces norm normed-spaces
real-analysis functional-analysis banach-spaces norm normed-spaces
edited Dec 4 '18 at 10:37
Masacroso
13.1k41746
13.1k41746
asked Mar 14 '18 at 22:20
Jeremy Jeffrey JamesJeremy Jeffrey James
966615
966615
$begingroup$
one of the functions dominates the other, so you just have to subtract the individual integrals. Can you integrate linear function?
$endgroup$
– Jorge Fernández
Mar 14 '18 at 22:29
1
$begingroup$
instead of $frac{b-a}{2}$ you should have $frac{b+a}{2}$
$endgroup$
– Orest Bucicovschi
Mar 15 '18 at 1:58
add a comment |
$begingroup$
one of the functions dominates the other, so you just have to subtract the individual integrals. Can you integrate linear function?
$endgroup$
– Jorge Fernández
Mar 14 '18 at 22:29
1
$begingroup$
instead of $frac{b-a}{2}$ you should have $frac{b+a}{2}$
$endgroup$
– Orest Bucicovschi
Mar 15 '18 at 1:58
$begingroup$
one of the functions dominates the other, so you just have to subtract the individual integrals. Can you integrate linear function?
$endgroup$
– Jorge Fernández
Mar 14 '18 at 22:29
$begingroup$
one of the functions dominates the other, so you just have to subtract the individual integrals. Can you integrate linear function?
$endgroup$
– Jorge Fernández
Mar 14 '18 at 22:29
1
1
$begingroup$
instead of $frac{b-a}{2}$ you should have $frac{b+a}{2}$
$endgroup$
– Orest Bucicovschi
Mar 15 '18 at 1:58
$begingroup$
instead of $frac{b-a}{2}$ you should have $frac{b+a}{2}$
$endgroup$
– Orest Bucicovschi
Mar 15 '18 at 1:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Use the mean value theorem:
$$int_a^b f(x),dx = (b-a)f(c), text{ for some } cin langle a , brangle$$
In particular, since your $f_n(x) in [0,1]$, we have:
begin{align}
int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2,dx &= left(frac{b-a}2+frac{2^{-n}}{b-a} - frac{b-a}2+frac{2^{-n}}{b-a} right)underbrace{|f_n(c) - f_m(c)|^2}_{le 2^2 = 4}\
&le 8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0
end{align}
Therefore, your sequence is Cauchy.
$endgroup$
$begingroup$
I would never have thought to use the MVT, thanks. I guess it's the exact same procedure for showing that $f_nto f$ as $ntoinfty$ where $f$ is the step function on $[a,b]$? In fact, one gets, exactly as you did here, a bound above of $8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0$ - is this correct?
$endgroup$
– Jeremy Jeffrey James
Mar 15 '18 at 13:22
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
Use the mean value theorem:
$$int_a^b f(x),dx = (b-a)f(c), text{ for some } cin langle a , brangle$$
In particular, since your $f_n(x) in [0,1]$, we have:
begin{align}
int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2,dx &= left(frac{b-a}2+frac{2^{-n}}{b-a} - frac{b-a}2+frac{2^{-n}}{b-a} right)underbrace{|f_n(c) - f_m(c)|^2}_{le 2^2 = 4}\
&le 8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0
end{align}
Therefore, your sequence is Cauchy.
$endgroup$
$begingroup$
I would never have thought to use the MVT, thanks. I guess it's the exact same procedure for showing that $f_nto f$ as $ntoinfty$ where $f$ is the step function on $[a,b]$? In fact, one gets, exactly as you did here, a bound above of $8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0$ - is this correct?
$endgroup$
– Jeremy Jeffrey James
Mar 15 '18 at 13:22
add a comment |
$begingroup$
Use the mean value theorem:
$$int_a^b f(x),dx = (b-a)f(c), text{ for some } cin langle a , brangle$$
In particular, since your $f_n(x) in [0,1]$, we have:
begin{align}
int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2,dx &= left(frac{b-a}2+frac{2^{-n}}{b-a} - frac{b-a}2+frac{2^{-n}}{b-a} right)underbrace{|f_n(c) - f_m(c)|^2}_{le 2^2 = 4}\
&le 8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0
end{align}
Therefore, your sequence is Cauchy.
$endgroup$
$begingroup$
I would never have thought to use the MVT, thanks. I guess it's the exact same procedure for showing that $f_nto f$ as $ntoinfty$ where $f$ is the step function on $[a,b]$? In fact, one gets, exactly as you did here, a bound above of $8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0$ - is this correct?
$endgroup$
– Jeremy Jeffrey James
Mar 15 '18 at 13:22
add a comment |
$begingroup$
Use the mean value theorem:
$$int_a^b f(x),dx = (b-a)f(c), text{ for some } cin langle a , brangle$$
In particular, since your $f_n(x) in [0,1]$, we have:
begin{align}
int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2,dx &= left(frac{b-a}2+frac{2^{-n}}{b-a} - frac{b-a}2+frac{2^{-n}}{b-a} right)underbrace{|f_n(c) - f_m(c)|^2}_{le 2^2 = 4}\
&le 8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0
end{align}
Therefore, your sequence is Cauchy.
$endgroup$
Use the mean value theorem:
$$int_a^b f(x),dx = (b-a)f(c), text{ for some } cin langle a , brangle$$
In particular, since your $f_n(x) in [0,1]$, we have:
begin{align}
int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2,dx &= left(frac{b-a}2+frac{2^{-n}}{b-a} - frac{b-a}2+frac{2^{-n}}{b-a} right)underbrace{|f_n(c) - f_m(c)|^2}_{le 2^2 = 4}\
&le 8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0
end{align}
Therefore, your sequence is Cauchy.
answered Mar 14 '18 at 22:35
mechanodroidmechanodroid
27.3k62446
27.3k62446
$begingroup$
I would never have thought to use the MVT, thanks. I guess it's the exact same procedure for showing that $f_nto f$ as $ntoinfty$ where $f$ is the step function on $[a,b]$? In fact, one gets, exactly as you did here, a bound above of $8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0$ - is this correct?
$endgroup$
– Jeremy Jeffrey James
Mar 15 '18 at 13:22
add a comment |
$begingroup$
I would never have thought to use the MVT, thanks. I guess it's the exact same procedure for showing that $f_nto f$ as $ntoinfty$ where $f$ is the step function on $[a,b]$? In fact, one gets, exactly as you did here, a bound above of $8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0$ - is this correct?
$endgroup$
– Jeremy Jeffrey James
Mar 15 '18 at 13:22
$begingroup$
I would never have thought to use the MVT, thanks. I guess it's the exact same procedure for showing that $f_nto f$ as $ntoinfty$ where $f$ is the step function on $[a,b]$? In fact, one gets, exactly as you did here, a bound above of $8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0$ - is this correct?
$endgroup$
– Jeremy Jeffrey James
Mar 15 '18 at 13:22
$begingroup$
I would never have thought to use the MVT, thanks. I guess it's the exact same procedure for showing that $f_nto f$ as $ntoinfty$ where $f$ is the step function on $[a,b]$? In fact, one gets, exactly as you did here, a bound above of $8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0$ - is this correct?
$endgroup$
– Jeremy Jeffrey James
Mar 15 '18 at 13:22
add a comment |
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$begingroup$
one of the functions dominates the other, so you just have to subtract the individual integrals. Can you integrate linear function?
$endgroup$
– Jorge Fernández
Mar 14 '18 at 22:29
1
$begingroup$
instead of $frac{b-a}{2}$ you should have $frac{b+a}{2}$
$endgroup$
– Orest Bucicovschi
Mar 15 '18 at 1:58