Sequence of functions having a convergent subsequence












0












$begingroup$


let $V=$ space of all continuous functions on R with compact support endowed with $d(f,g)=(int_{-infty}^{infty}|f(t)-g(t)|^2dt)^{frac{1}{2}}$



Define $fin V$ .define $f_n=f(x-n)$ Then show that $f_n$ has no convergent subsequence.



I don't know how to start it.There is a similar version of this question.I am not getting it.Can anyone show an easier version of this






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's your problem with the previous version
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:48










  • $begingroup$
    I cant understand that.Please help me
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:49
















0












$begingroup$


let $V=$ space of all continuous functions on R with compact support endowed with $d(f,g)=(int_{-infty}^{infty}|f(t)-g(t)|^2dt)^{frac{1}{2}}$



Define $fin V$ .define $f_n=f(x-n)$ Then show that $f_n$ has no convergent subsequence.



I don't know how to start it.There is a similar version of this question.I am not getting it.Can anyone show an easier version of this






real-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What's your problem with the previous version
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:48










  • $begingroup$
    I cant understand that.Please help me
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:49














0












0








0


1



$begingroup$


let $V=$ space of all continuous functions on R with compact support endowed with $d(f,g)=(int_{-infty}^{infty}|f(t)-g(t)|^2dt)^{frac{1}{2}}$



Define $fin V$ .define $f_n=f(x-n)$ Then show that $f_n$ has no convergent subsequence.



I don't know how to start it.There is a similar version of this question.I am not getting it.Can anyone show an easier version of this






real-analysis







share|cite|improve this question











$endgroup$




let $V=$ space of all continuous functions on R with compact support endowed with $d(f,g)=(int_{-infty}^{infty}|f(t)-g(t)|^2dt)^{frac{1}{2}}$



Define $fin V$ .define $f_n=f(x-n)$ Then show that $f_n$ has no convergent subsequence.



I don't know how to start it.There is a similar version of this question.I am not getting it.Can anyone show an easier version of this






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 '15 at 7:01









Mhenni Benghorbal

43.2k63675




43.2k63675










asked Jan 9 '15 at 5:47









LearnmoreLearnmore

17.8k324100




17.8k324100








  • 1




    $begingroup$
    What's your problem with the previous version
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:48










  • $begingroup$
    I cant understand that.Please help me
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:49














  • 1




    $begingroup$
    What's your problem with the previous version
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:48










  • $begingroup$
    I cant understand that.Please help me
    $endgroup$
    – Learnmore
    Jan 9 '15 at 5:49








1




1




$begingroup$
What's your problem with the previous version
$endgroup$
– Learnmore
Jan 9 '15 at 5:48




$begingroup$
What's your problem with the previous version
$endgroup$
– Learnmore
Jan 9 '15 at 5:48












$begingroup$
I cant understand that.Please help me
$endgroup$
– Learnmore
Jan 9 '15 at 5:49




$begingroup$
I cant understand that.Please help me
$endgroup$
– Learnmore
Jan 9 '15 at 5:49










1 Answer
1






active

oldest

votes


















1












$begingroup$

It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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1












$begingroup$

It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35
















1












$begingroup$

It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35














1












1








1





$begingroup$

It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.






share|cite|improve this answer











$endgroup$



It seems the following.



Let $fin V$ be an arbitrary non-zero function. Since the function $f$ is continuous, $d(f,0)>0$. Since the support $text{supp } f$ of the function $f$ is compact, there exists a number $M$ such that $text{supp } fsubset[-M,M].$ Suppose that the sequence ${f_n}$ has a convergent subsequence ${f_{n_k}}$. Then the sequence ${f_{n_k}}$ is fundamental, that is for each $varepsilon>0$ there exists a number $K=K(varepsilon)>0$ such that $d(f_{n_k}, f_{n_k’})<varepsilon$ provided $k,k’>K$. Let $k>K(sqrt{2}d(f,0))$ be an arbitrary number and $k’$ be such a number that $n_k’>n_k+2M$. Since $text{supp } f_{n_k}cap text{supp } f_{n_k’}=varnothing$,



$$d(f_{n_k}, f_{n’_k})=left(int_{-infty}^{infty}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$
$$left(int_{text{supp } f_{n_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dt+int_{text{supp } f_{n’_k}}| f_{n_k}(t)-f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(int_{text{supp } f_{n_k}}| f_{n_k}(t)|^2dt+int_{text{supp } f_{n’_k}}|f_{n’_k}(t)|^2dtright)^{frac{1}{2}}=$$ $$
left(2int_{text{supp } f}| f (t)|^2dtright)^{frac{1}{2}}=sqrt{2}d(f,0),$$



a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 7:34

























answered Jan 22 '15 at 20:20









Alex RavskyAlex Ravsky

41.5k32282




41.5k32282












  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35


















  • $begingroup$
    @ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
    $endgroup$
    – jasmine
    Sep 14 '18 at 16:04








  • 1




    $begingroup$
    @jasmine Thanks for your remark. I corrected and extended my answer.
    $endgroup$
    – Alex Ravsky
    Dec 21 '18 at 7:35
















$begingroup$
@ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
$endgroup$
– jasmine
Sep 14 '18 at 16:04






$begingroup$
@ Alex im not getting how u directly write $d(f_{n_k}, f_{n_k’})=2d(f,0),$ ??
$endgroup$
– jasmine
Sep 14 '18 at 16:04






1




1




$begingroup$
@jasmine Thanks for your remark. I corrected and extended my answer.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:35




$begingroup$
@jasmine Thanks for your remark. I corrected and extended my answer.
$endgroup$
– Alex Ravsky
Dec 21 '18 at 7:35


















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