Prove that the sequence of functions $g_{n}in C[0,4]$












1












$begingroup$


I want to show that the function defined by $g_n:[0,4]to Bbb{R}$, defined bybegin{align} g_n(t)=begin{cases}0,& text{if};0leq tleq 2,\dfrac{n}{2}(t-2),& text{if};2leq tleq 2+frac{2}{n},\1 &text{if};2+frac{2}{n}leq tleq 4.end{cases} end{align}
is continuous and Cauchy.



MY TRIAL
The sequence of functions is clearly continuous by Pasting Lemma. Next, we show that the sequence of functions is Cauchy.



So, let $m,ninBbb{N},$ be given such that $mgeq n.$ Then,



$$|g_m(t)-g_n(t)|=|0-0|+left|dfrac{m}{2}(t-2)-dfrac{n}{2}(t-2)right|left|1-1right|=0<epsilon$$



I don't know if this is right. If not, kindly help please!






real-analysis functional-analysis analysis normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    functions are continuous. sequences are Cauchy
    $endgroup$
    – mathworker21
    Dec 11 '18 at 20:59










  • $begingroup$
    to prove Cauchy, just calculate the max difference between $g_n$ and $g_m$. what exactly is your difficulty?
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:00










  • $begingroup$
    @mathworker21: Edited some things. Max difference?
    $endgroup$
    – Omojola Micheal
    Dec 11 '18 at 21:06










  • $begingroup$
    wait, you're saying $|g_m(t)-g_n(t)| = 0$ for each $t$?
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:08










  • $begingroup$
    by max difference, what I mean is $max_t |g_m(t)-g_n(t)|$. you want to show this is less than $epsilon$ for $m,n$ large enough
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:08
















1












$begingroup$


I want to show that the function defined by $g_n:[0,4]to Bbb{R}$, defined bybegin{align} g_n(t)=begin{cases}0,& text{if};0leq tleq 2,\dfrac{n}{2}(t-2),& text{if};2leq tleq 2+frac{2}{n},\1 &text{if};2+frac{2}{n}leq tleq 4.end{cases} end{align}
is continuous and Cauchy.



MY TRIAL
The sequence of functions is clearly continuous by Pasting Lemma. Next, we show that the sequence of functions is Cauchy.



So, let $m,ninBbb{N},$ be given such that $mgeq n.$ Then,



$$|g_m(t)-g_n(t)|=|0-0|+left|dfrac{m}{2}(t-2)-dfrac{n}{2}(t-2)right|left|1-1right|=0<epsilon$$



I don't know if this is right. If not, kindly help please!






real-analysis functional-analysis analysis normed-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    functions are continuous. sequences are Cauchy
    $endgroup$
    – mathworker21
    Dec 11 '18 at 20:59










  • $begingroup$
    to prove Cauchy, just calculate the max difference between $g_n$ and $g_m$. what exactly is your difficulty?
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:00










  • $begingroup$
    @mathworker21: Edited some things. Max difference?
    $endgroup$
    – Omojola Micheal
    Dec 11 '18 at 21:06










  • $begingroup$
    wait, you're saying $|g_m(t)-g_n(t)| = 0$ for each $t$?
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:08










  • $begingroup$
    by max difference, what I mean is $max_t |g_m(t)-g_n(t)|$. you want to show this is less than $epsilon$ for $m,n$ large enough
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:08














1












1








1


1



$begingroup$


I want to show that the function defined by $g_n:[0,4]to Bbb{R}$, defined bybegin{align} g_n(t)=begin{cases}0,& text{if};0leq tleq 2,\dfrac{n}{2}(t-2),& text{if};2leq tleq 2+frac{2}{n},\1 &text{if};2+frac{2}{n}leq tleq 4.end{cases} end{align}
is continuous and Cauchy.



MY TRIAL
The sequence of functions is clearly continuous by Pasting Lemma. Next, we show that the sequence of functions is Cauchy.



So, let $m,ninBbb{N},$ be given such that $mgeq n.$ Then,



$$|g_m(t)-g_n(t)|=|0-0|+left|dfrac{m}{2}(t-2)-dfrac{n}{2}(t-2)right|left|1-1right|=0<epsilon$$



I don't know if this is right. If not, kindly help please!






real-analysis functional-analysis analysis normed-spaces







share|cite|improve this question











$endgroup$




I want to show that the function defined by $g_n:[0,4]to Bbb{R}$, defined bybegin{align} g_n(t)=begin{cases}0,& text{if};0leq tleq 2,\dfrac{n}{2}(t-2),& text{if};2leq tleq 2+frac{2}{n},\1 &text{if};2+frac{2}{n}leq tleq 4.end{cases} end{align}
is continuous and Cauchy.



MY TRIAL
The sequence of functions is clearly continuous by Pasting Lemma. Next, we show that the sequence of functions is Cauchy.



So, let $m,ninBbb{N},$ be given such that $mgeq n.$ Then,



$$|g_m(t)-g_n(t)|=|0-0|+left|dfrac{m}{2}(t-2)-dfrac{n}{2}(t-2)right|left|1-1right|=0<epsilon$$



I don't know if this is right. If not, kindly help please!






real-analysis functional-analysis analysis normed-spaces




real-analysis functional-analysis analysis normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 22:54









user289143

913313




913313










asked Dec 11 '18 at 20:57









Omojola MichealOmojola Micheal

1,853324




1,853324












  • $begingroup$
    functions are continuous. sequences are Cauchy
    $endgroup$
    – mathworker21
    Dec 11 '18 at 20:59










  • $begingroup$
    to prove Cauchy, just calculate the max difference between $g_n$ and $g_m$. what exactly is your difficulty?
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:00










  • $begingroup$
    @mathworker21: Edited some things. Max difference?
    $endgroup$
    – Omojola Micheal
    Dec 11 '18 at 21:06










  • $begingroup$
    wait, you're saying $|g_m(t)-g_n(t)| = 0$ for each $t$?
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:08










  • $begingroup$
    by max difference, what I mean is $max_t |g_m(t)-g_n(t)|$. you want to show this is less than $epsilon$ for $m,n$ large enough
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:08


















  • $begingroup$
    functions are continuous. sequences are Cauchy
    $endgroup$
    – mathworker21
    Dec 11 '18 at 20:59










  • $begingroup$
    to prove Cauchy, just calculate the max difference between $g_n$ and $g_m$. what exactly is your difficulty?
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:00










  • $begingroup$
    @mathworker21: Edited some things. Max difference?
    $endgroup$
    – Omojola Micheal
    Dec 11 '18 at 21:06










  • $begingroup$
    wait, you're saying $|g_m(t)-g_n(t)| = 0$ for each $t$?
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:08










  • $begingroup$
    by max difference, what I mean is $max_t |g_m(t)-g_n(t)|$. you want to show this is less than $epsilon$ for $m,n$ large enough
    $endgroup$
    – mathworker21
    Dec 11 '18 at 21:08
















$begingroup$
functions are continuous. sequences are Cauchy
$endgroup$
– mathworker21
Dec 11 '18 at 20:59




$begingroup$
functions are continuous. sequences are Cauchy
$endgroup$
– mathworker21
Dec 11 '18 at 20:59












$begingroup$
to prove Cauchy, just calculate the max difference between $g_n$ and $g_m$. what exactly is your difficulty?
$endgroup$
– mathworker21
Dec 11 '18 at 21:00




$begingroup$
to prove Cauchy, just calculate the max difference between $g_n$ and $g_m$. what exactly is your difficulty?
$endgroup$
– mathworker21
Dec 11 '18 at 21:00












$begingroup$
@mathworker21: Edited some things. Max difference?
$endgroup$
– Omojola Micheal
Dec 11 '18 at 21:06




$begingroup$
@mathworker21: Edited some things. Max difference?
$endgroup$
– Omojola Micheal
Dec 11 '18 at 21:06












$begingroup$
wait, you're saying $|g_m(t)-g_n(t)| = 0$ for each $t$?
$endgroup$
– mathworker21
Dec 11 '18 at 21:08




$begingroup$
wait, you're saying $|g_m(t)-g_n(t)| = 0$ for each $t$?
$endgroup$
– mathworker21
Dec 11 '18 at 21:08












$begingroup$
by max difference, what I mean is $max_t |g_m(t)-g_n(t)|$. you want to show this is less than $epsilon$ for $m,n$ large enough
$endgroup$
– mathworker21
Dec 11 '18 at 21:08




$begingroup$
by max difference, what I mean is $max_t |g_m(t)-g_n(t)|$. you want to show this is less than $epsilon$ for $m,n$ large enough
$endgroup$
– mathworker21
Dec 11 '18 at 21:08










2 Answers
2






active

oldest

votes


















1












$begingroup$

What meaning exactly do you assign to the sentence “the sequence $g_n$ is Cauchy”?



This sequence of functions converges pointwise to the function $h$ such that $h(x)=0$ if $0 leq x leq 2$ and $h(x)=1$ otherwise.



Now $h$ is not continuous whereas every $g_n$ is. So there cannot be max-norm convergence.



Edit: Just to provide some even more explicit argument because it does not seem to be accepted that the space of continuous functions from $[0,4]$ to $mathbb{R}$ is complete for the max-norm.



$g_nleft(2+frac{2}{2n}right)=frac{n}{2}frac{2}{2n}=frac{1}{2}$, while $g_{2n}left(2+frac{2}{2n}right)=1$.
Thus the max-norm of $g_n-g_{2n}$ is not lower than $frac{1}{2}$, hence the sequence is not Cauchy.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    C([0,4]) is complete for the max-norm.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:40






  • 1




    $begingroup$
    Besides, I do not have neough reputation to comment your post, but I fail to see how you can have $(N-n)/N < 1/N$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:49












  • $begingroup$
    If I am not mistaken, you are thus proving that the max-norm of $|g_n-g_m|$ is lower than $(N-n)/n$, which would be negative?
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:25










  • $begingroup$
    John B: I do not have another question. I am not convinced by user289143’s proof and I cannot comment it.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:30






  • 1




    $begingroup$
    User289143: then you cannot have $|N-n|<1$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:31



















0












$begingroup$

A sequence is Cauchy if for every $epsilon$ there exists $N in mathbb{N}$ such that $forall n, m > N$ we have $|g_n-g_m|< epsilon$.
Now let $m > n$. How does the function $(g_m-g_n)(t)$ look like?
begin{align}
(g_m-g_n)(t)= begin{cases}
0 & if 0 leq t leq 2
\(frac{m}{2}-frac{n}{2})(t-2) & if 2 < t leq 2+frac{2}{m}
\1-frac{n}{2}(t-2) & if 2+frac{2}{m} < t leq 2+frac{2}{n}
\ 0 & if 2+frac{2}{n} < t leq 4
end{cases}
end{align}
Now $||g_m-g_n||_1=int_2^{2+frac{2}{m}}|(frac{m}{2}-frac{n}{2})(t-2)|dt+int_{2+frac{2}{m}}^{2+frac{2}{n}}|1-frac{n}{2}(t-2)|dt$

If we compute explicitly the integrals we get $||g_m-g_n||_1=frac{m-n}{m^2}+frac{(m-n)^2}{nm^2}=frac{m-n}{mn}<frac{m}{mn}=frac{1}{n}$

Let $N in mathbb{N}$ such that $N > frac{1}{epsilon}$ and we have $||g_m-g_m||_1<frac{1}{n}<frac{1}{N}<epsilon$ and this proves the sequence is Cauchy






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Could you please explain why |N-n| < 1?
    $endgroup$
    – Mindlack
    Dec 12 '18 at 0:01










  • $begingroup$
    Edited the answer with the $||cdot||_1$ norm, to prove the space is not complete
    $endgroup$
    – user289143
    Dec 12 '18 at 13:22











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

What meaning exactly do you assign to the sentence “the sequence $g_n$ is Cauchy”?



This sequence of functions converges pointwise to the function $h$ such that $h(x)=0$ if $0 leq x leq 2$ and $h(x)=1$ otherwise.



Now $h$ is not continuous whereas every $g_n$ is. So there cannot be max-norm convergence.



Edit: Just to provide some even more explicit argument because it does not seem to be accepted that the space of continuous functions from $[0,4]$ to $mathbb{R}$ is complete for the max-norm.



$g_nleft(2+frac{2}{2n}right)=frac{n}{2}frac{2}{2n}=frac{1}{2}$, while $g_{2n}left(2+frac{2}{2n}right)=1$.
Thus the max-norm of $g_n-g_{2n}$ is not lower than $frac{1}{2}$, hence the sequence is not Cauchy.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    C([0,4]) is complete for the max-norm.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:40






  • 1




    $begingroup$
    Besides, I do not have neough reputation to comment your post, but I fail to see how you can have $(N-n)/N < 1/N$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:49












  • $begingroup$
    If I am not mistaken, you are thus proving that the max-norm of $|g_n-g_m|$ is lower than $(N-n)/n$, which would be negative?
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:25










  • $begingroup$
    John B: I do not have another question. I am not convinced by user289143’s proof and I cannot comment it.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:30






  • 1




    $begingroup$
    User289143: then you cannot have $|N-n|<1$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:31
















1












$begingroup$

What meaning exactly do you assign to the sentence “the sequence $g_n$ is Cauchy”?



This sequence of functions converges pointwise to the function $h$ such that $h(x)=0$ if $0 leq x leq 2$ and $h(x)=1$ otherwise.



Now $h$ is not continuous whereas every $g_n$ is. So there cannot be max-norm convergence.



Edit: Just to provide some even more explicit argument because it does not seem to be accepted that the space of continuous functions from $[0,4]$ to $mathbb{R}$ is complete for the max-norm.



$g_nleft(2+frac{2}{2n}right)=frac{n}{2}frac{2}{2n}=frac{1}{2}$, while $g_{2n}left(2+frac{2}{2n}right)=1$.
Thus the max-norm of $g_n-g_{2n}$ is not lower than $frac{1}{2}$, hence the sequence is not Cauchy.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    C([0,4]) is complete for the max-norm.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:40






  • 1




    $begingroup$
    Besides, I do not have neough reputation to comment your post, but I fail to see how you can have $(N-n)/N < 1/N$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:49












  • $begingroup$
    If I am not mistaken, you are thus proving that the max-norm of $|g_n-g_m|$ is lower than $(N-n)/n$, which would be negative?
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:25










  • $begingroup$
    John B: I do not have another question. I am not convinced by user289143’s proof and I cannot comment it.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:30






  • 1




    $begingroup$
    User289143: then you cannot have $|N-n|<1$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:31














1












1








1





$begingroup$

What meaning exactly do you assign to the sentence “the sequence $g_n$ is Cauchy”?



This sequence of functions converges pointwise to the function $h$ such that $h(x)=0$ if $0 leq x leq 2$ and $h(x)=1$ otherwise.



Now $h$ is not continuous whereas every $g_n$ is. So there cannot be max-norm convergence.



Edit: Just to provide some even more explicit argument because it does not seem to be accepted that the space of continuous functions from $[0,4]$ to $mathbb{R}$ is complete for the max-norm.



$g_nleft(2+frac{2}{2n}right)=frac{n}{2}frac{2}{2n}=frac{1}{2}$, while $g_{2n}left(2+frac{2}{2n}right)=1$.
Thus the max-norm of $g_n-g_{2n}$ is not lower than $frac{1}{2}$, hence the sequence is not Cauchy.






share|cite|improve this answer











$endgroup$



What meaning exactly do you assign to the sentence “the sequence $g_n$ is Cauchy”?



This sequence of functions converges pointwise to the function $h$ such that $h(x)=0$ if $0 leq x leq 2$ and $h(x)=1$ otherwise.



Now $h$ is not continuous whereas every $g_n$ is. So there cannot be max-norm convergence.



Edit: Just to provide some even more explicit argument because it does not seem to be accepted that the space of continuous functions from $[0,4]$ to $mathbb{R}$ is complete for the max-norm.



$g_nleft(2+frac{2}{2n}right)=frac{n}{2}frac{2}{2n}=frac{1}{2}$, while $g_{2n}left(2+frac{2}{2n}right)=1$.
Thus the max-norm of $g_n-g_{2n}$ is not lower than $frac{1}{2}$, hence the sequence is not Cauchy.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 11:57

























answered Dec 11 '18 at 22:06









MindlackMindlack

3,87018




3,87018








  • 1




    $begingroup$
    C([0,4]) is complete for the max-norm.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:40






  • 1




    $begingroup$
    Besides, I do not have neough reputation to comment your post, but I fail to see how you can have $(N-n)/N < 1/N$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:49












  • $begingroup$
    If I am not mistaken, you are thus proving that the max-norm of $|g_n-g_m|$ is lower than $(N-n)/n$, which would be negative?
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:25










  • $begingroup$
    John B: I do not have another question. I am not convinced by user289143’s proof and I cannot comment it.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:30






  • 1




    $begingroup$
    User289143: then you cannot have $|N-n|<1$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:31














  • 1




    $begingroup$
    C([0,4]) is complete for the max-norm.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:40






  • 1




    $begingroup$
    Besides, I do not have neough reputation to comment your post, but I fail to see how you can have $(N-n)/N < 1/N$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 22:49












  • $begingroup$
    If I am not mistaken, you are thus proving that the max-norm of $|g_n-g_m|$ is lower than $(N-n)/n$, which would be negative?
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:25










  • $begingroup$
    John B: I do not have another question. I am not convinced by user289143’s proof and I cannot comment it.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:30






  • 1




    $begingroup$
    User289143: then you cannot have $|N-n|<1$.
    $endgroup$
    – Mindlack
    Dec 11 '18 at 23:31








1




1




$begingroup$
C([0,4]) is complete for the max-norm.
$endgroup$
– Mindlack
Dec 11 '18 at 22:40




$begingroup$
C([0,4]) is complete for the max-norm.
$endgroup$
– Mindlack
Dec 11 '18 at 22:40




1




1




$begingroup$
Besides, I do not have neough reputation to comment your post, but I fail to see how you can have $(N-n)/N < 1/N$.
$endgroup$
– Mindlack
Dec 11 '18 at 22:49






$begingroup$
Besides, I do not have neough reputation to comment your post, but I fail to see how you can have $(N-n)/N < 1/N$.
$endgroup$
– Mindlack
Dec 11 '18 at 22:49














$begingroup$
If I am not mistaken, you are thus proving that the max-norm of $|g_n-g_m|$ is lower than $(N-n)/n$, which would be negative?
$endgroup$
– Mindlack
Dec 11 '18 at 23:25




$begingroup$
If I am not mistaken, you are thus proving that the max-norm of $|g_n-g_m|$ is lower than $(N-n)/n$, which would be negative?
$endgroup$
– Mindlack
Dec 11 '18 at 23:25












$begingroup$
John B: I do not have another question. I am not convinced by user289143’s proof and I cannot comment it.
$endgroup$
– Mindlack
Dec 11 '18 at 23:30




$begingroup$
John B: I do not have another question. I am not convinced by user289143’s proof and I cannot comment it.
$endgroup$
– Mindlack
Dec 11 '18 at 23:30




1




1




$begingroup$
User289143: then you cannot have $|N-n|<1$.
$endgroup$
– Mindlack
Dec 11 '18 at 23:31




$begingroup$
User289143: then you cannot have $|N-n|<1$.
$endgroup$
– Mindlack
Dec 11 '18 at 23:31











0












$begingroup$

A sequence is Cauchy if for every $epsilon$ there exists $N in mathbb{N}$ such that $forall n, m > N$ we have $|g_n-g_m|< epsilon$.
Now let $m > n$. How does the function $(g_m-g_n)(t)$ look like?
begin{align}
(g_m-g_n)(t)= begin{cases}
0 & if 0 leq t leq 2
\(frac{m}{2}-frac{n}{2})(t-2) & if 2 < t leq 2+frac{2}{m}
\1-frac{n}{2}(t-2) & if 2+frac{2}{m} < t leq 2+frac{2}{n}
\ 0 & if 2+frac{2}{n} < t leq 4
end{cases}
end{align}
Now $||g_m-g_n||_1=int_2^{2+frac{2}{m}}|(frac{m}{2}-frac{n}{2})(t-2)|dt+int_{2+frac{2}{m}}^{2+frac{2}{n}}|1-frac{n}{2}(t-2)|dt$

If we compute explicitly the integrals we get $||g_m-g_n||_1=frac{m-n}{m^2}+frac{(m-n)^2}{nm^2}=frac{m-n}{mn}<frac{m}{mn}=frac{1}{n}$

Let $N in mathbb{N}$ such that $N > frac{1}{epsilon}$ and we have $||g_m-g_m||_1<frac{1}{n}<frac{1}{N}<epsilon$ and this proves the sequence is Cauchy






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Could you please explain why |N-n| < 1?
    $endgroup$
    – Mindlack
    Dec 12 '18 at 0:01










  • $begingroup$
    Edited the answer with the $||cdot||_1$ norm, to prove the space is not complete
    $endgroup$
    – user289143
    Dec 12 '18 at 13:22
















0












$begingroup$

A sequence is Cauchy if for every $epsilon$ there exists $N in mathbb{N}$ such that $forall n, m > N$ we have $|g_n-g_m|< epsilon$.
Now let $m > n$. How does the function $(g_m-g_n)(t)$ look like?
begin{align}
(g_m-g_n)(t)= begin{cases}
0 & if 0 leq t leq 2
\(frac{m}{2}-frac{n}{2})(t-2) & if 2 < t leq 2+frac{2}{m}
\1-frac{n}{2}(t-2) & if 2+frac{2}{m} < t leq 2+frac{2}{n}
\ 0 & if 2+frac{2}{n} < t leq 4
end{cases}
end{align}
Now $||g_m-g_n||_1=int_2^{2+frac{2}{m}}|(frac{m}{2}-frac{n}{2})(t-2)|dt+int_{2+frac{2}{m}}^{2+frac{2}{n}}|1-frac{n}{2}(t-2)|dt$

If we compute explicitly the integrals we get $||g_m-g_n||_1=frac{m-n}{m^2}+frac{(m-n)^2}{nm^2}=frac{m-n}{mn}<frac{m}{mn}=frac{1}{n}$

Let $N in mathbb{N}$ such that $N > frac{1}{epsilon}$ and we have $||g_m-g_m||_1<frac{1}{n}<frac{1}{N}<epsilon$ and this proves the sequence is Cauchy






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    Could you please explain why |N-n| < 1?
    $endgroup$
    – Mindlack
    Dec 12 '18 at 0:01










  • $begingroup$
    Edited the answer with the $||cdot||_1$ norm, to prove the space is not complete
    $endgroup$
    – user289143
    Dec 12 '18 at 13:22














0












0








0





$begingroup$

A sequence is Cauchy if for every $epsilon$ there exists $N in mathbb{N}$ such that $forall n, m > N$ we have $|g_n-g_m|< epsilon$.
Now let $m > n$. How does the function $(g_m-g_n)(t)$ look like?
begin{align}
(g_m-g_n)(t)= begin{cases}
0 & if 0 leq t leq 2
\(frac{m}{2}-frac{n}{2})(t-2) & if 2 < t leq 2+frac{2}{m}
\1-frac{n}{2}(t-2) & if 2+frac{2}{m} < t leq 2+frac{2}{n}
\ 0 & if 2+frac{2}{n} < t leq 4
end{cases}
end{align}
Now $||g_m-g_n||_1=int_2^{2+frac{2}{m}}|(frac{m}{2}-frac{n}{2})(t-2)|dt+int_{2+frac{2}{m}}^{2+frac{2}{n}}|1-frac{n}{2}(t-2)|dt$

If we compute explicitly the integrals we get $||g_m-g_n||_1=frac{m-n}{m^2}+frac{(m-n)^2}{nm^2}=frac{m-n}{mn}<frac{m}{mn}=frac{1}{n}$

Let $N in mathbb{N}$ such that $N > frac{1}{epsilon}$ and we have $||g_m-g_m||_1<frac{1}{n}<frac{1}{N}<epsilon$ and this proves the sequence is Cauchy






share|cite|improve this answer











$endgroup$



A sequence is Cauchy if for every $epsilon$ there exists $N in mathbb{N}$ such that $forall n, m > N$ we have $|g_n-g_m|< epsilon$.
Now let $m > n$. How does the function $(g_m-g_n)(t)$ look like?
begin{align}
(g_m-g_n)(t)= begin{cases}
0 & if 0 leq t leq 2
\(frac{m}{2}-frac{n}{2})(t-2) & if 2 < t leq 2+frac{2}{m}
\1-frac{n}{2}(t-2) & if 2+frac{2}{m} < t leq 2+frac{2}{n}
\ 0 & if 2+frac{2}{n} < t leq 4
end{cases}
end{align}
Now $||g_m-g_n||_1=int_2^{2+frac{2}{m}}|(frac{m}{2}-frac{n}{2})(t-2)|dt+int_{2+frac{2}{m}}^{2+frac{2}{n}}|1-frac{n}{2}(t-2)|dt$

If we compute explicitly the integrals we get $||g_m-g_n||_1=frac{m-n}{m^2}+frac{(m-n)^2}{nm^2}=frac{m-n}{mn}<frac{m}{mn}=frac{1}{n}$

Let $N in mathbb{N}$ such that $N > frac{1}{epsilon}$ and we have $||g_m-g_m||_1<frac{1}{n}<frac{1}{N}<epsilon$ and this proves the sequence is Cauchy







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 13:21

























answered Dec 11 '18 at 21:48









user289143user289143

913313




913313








  • 2




    $begingroup$
    Could you please explain why |N-n| < 1?
    $endgroup$
    – Mindlack
    Dec 12 '18 at 0:01










  • $begingroup$
    Edited the answer with the $||cdot||_1$ norm, to prove the space is not complete
    $endgroup$
    – user289143
    Dec 12 '18 at 13:22














  • 2




    $begingroup$
    Could you please explain why |N-n| < 1?
    $endgroup$
    – Mindlack
    Dec 12 '18 at 0:01










  • $begingroup$
    Edited the answer with the $||cdot||_1$ norm, to prove the space is not complete
    $endgroup$
    – user289143
    Dec 12 '18 at 13:22








2




2




$begingroup$
Could you please explain why |N-n| < 1?
$endgroup$
– Mindlack
Dec 12 '18 at 0:01




$begingroup$
Could you please explain why |N-n| < 1?
$endgroup$
– Mindlack
Dec 12 '18 at 0:01












$begingroup$
Edited the answer with the $||cdot||_1$ norm, to prove the space is not complete
$endgroup$
– user289143
Dec 12 '18 at 13:22




$begingroup$
Edited the answer with the $||cdot||_1$ norm, to prove the space is not complete
$endgroup$
– user289143
Dec 12 '18 at 13:22


















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