A question on max norm












1














This is question 30 in chapter 7 from royden



"Let ${f_n}$ be a sequence in C[a, b] and $sum a_k$ a convergent series of positive numbers such
that $|| f_{k+1}- f_k||_{max} leq a_k, forall k$
Prove that
$|f_{k+n}(x)- f_k(x) | leq || f_{k+n}- f_k||_{max} leq sum_{j=n}^{infty}a_j , forall k,n $



Conclude that there is a function f$in$ C[a, b] such that ${fn} rightarrow f$ uniformly on [a, b] "



I know that, it is easy to solve this question if the series above starts from j=k , by adding and subtracting $f_{k+1},..., f_{k+n-1 } $ to the leftside and useing the triangular inequality.



But in this problem the seris starts from j=n.
I tried hard to solve it., however I have not any good idea.
Can you help me??
Any suggestion will be appreciated.






real-analysis linear-algebra functional-analysis measure-theory







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  • This seems like a typo to me. What you say you can prove ought to be the indended meaning.
    – quid♦
    yesterday








  • 1




    Well, the statement is not true and that's why you couldn't solve it. It looks like a typo. To see this, suppose it's true and take $n$ to $infty$.Then we get $lim_n f_n = f_k$ for all $k$ which is absurd.
    – Song
    yesterday








  • 1




    To elaborate slightly on my first comment, as a "test" consider constant functions $f_k$ defined via $f_{k+1}(x) = f_k(x)+a_k$.
    – quid♦
    yesterday










  • Thank you so much
    – Mano.kiko
    yesterday
















1














This is question 30 in chapter 7 from royden



"Let ${f_n}$ be a sequence in C[a, b] and $sum a_k$ a convergent series of positive numbers such
that $|| f_{k+1}- f_k||_{max} leq a_k, forall k$
Prove that
$|f_{k+n}(x)- f_k(x) | leq || f_{k+n}- f_k||_{max} leq sum_{j=n}^{infty}a_j , forall k,n $



Conclude that there is a function f$in$ C[a, b] such that ${fn} rightarrow f$ uniformly on [a, b] "



I know that, it is easy to solve this question if the series above starts from j=k , by adding and subtracting $f_{k+1},..., f_{k+n-1 } $ to the leftside and useing the triangular inequality.



But in this problem the seris starts from j=n.
I tried hard to solve it., however I have not any good idea.
Can you help me??
Any suggestion will be appreciated.






real-analysis linear-algebra functional-analysis measure-theory







share|cite|improve this question









New contributor




Mano.kiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • This seems like a typo to me. What you say you can prove ought to be the indended meaning.
    – quid♦
    yesterday








  • 1




    Well, the statement is not true and that's why you couldn't solve it. It looks like a typo. To see this, suppose it's true and take $n$ to $infty$.Then we get $lim_n f_n = f_k$ for all $k$ which is absurd.
    – Song
    yesterday








  • 1




    To elaborate slightly on my first comment, as a "test" consider constant functions $f_k$ defined via $f_{k+1}(x) = f_k(x)+a_k$.
    – quid♦
    yesterday










  • Thank you so much
    – Mano.kiko
    yesterday














1












1








1







This is question 30 in chapter 7 from royden



"Let ${f_n}$ be a sequence in C[a, b] and $sum a_k$ a convergent series of positive numbers such
that $|| f_{k+1}- f_k||_{max} leq a_k, forall k$
Prove that
$|f_{k+n}(x)- f_k(x) | leq || f_{k+n}- f_k||_{max} leq sum_{j=n}^{infty}a_j , forall k,n $



Conclude that there is a function f$in$ C[a, b] such that ${fn} rightarrow f$ uniformly on [a, b] "



I know that, it is easy to solve this question if the series above starts from j=k , by adding and subtracting $f_{k+1},..., f_{k+n-1 } $ to the leftside and useing the triangular inequality.



But in this problem the seris starts from j=n.
I tried hard to solve it., however I have not any good idea.
Can you help me??
Any suggestion will be appreciated.






real-analysis linear-algebra functional-analysis measure-theory







share|cite|improve this question









New contributor




Mano.kiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











This is question 30 in chapter 7 from royden



"Let ${f_n}$ be a sequence in C[a, b] and $sum a_k$ a convergent series of positive numbers such
that $|| f_{k+1}- f_k||_{max} leq a_k, forall k$
Prove that
$|f_{k+n}(x)- f_k(x) | leq || f_{k+n}- f_k||_{max} leq sum_{j=n}^{infty}a_j , forall k,n $



Conclude that there is a function f$in$ C[a, b] such that ${fn} rightarrow f$ uniformly on [a, b] "



I know that, it is easy to solve this question if the series above starts from j=k , by adding and subtracting $f_{k+1},..., f_{k+n-1 } $ to the leftside and useing the triangular inequality.



But in this problem the seris starts from j=n.
I tried hard to solve it., however I have not any good idea.
Can you help me??
Any suggestion will be appreciated.






real-analysis linear-algebra functional-analysis measure-theory




real-analysis linear-algebra functional-analysis measure-theory






share|cite|improve this question









New contributor




Mano.kiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Mano.kiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited yesterday









Davide Giraudo

125k16150260




125k16150260






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asked yesterday









Mano.kiko

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Mano.kiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Mano.kiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Mano.kiko is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • This seems like a typo to me. What you say you can prove ought to be the indended meaning.
    – quid♦
    yesterday








  • 1




    Well, the statement is not true and that's why you couldn't solve it. It looks like a typo. To see this, suppose it's true and take $n$ to $infty$.Then we get $lim_n f_n = f_k$ for all $k$ which is absurd.
    – Song
    yesterday








  • 1




    To elaborate slightly on my first comment, as a "test" consider constant functions $f_k$ defined via $f_{k+1}(x) = f_k(x)+a_k$.
    – quid♦
    yesterday










  • Thank you so much
    – Mano.kiko
    yesterday


















  • This seems like a typo to me. What you say you can prove ought to be the indended meaning.
    – quid♦
    yesterday








  • 1




    Well, the statement is not true and that's why you couldn't solve it. It looks like a typo. To see this, suppose it's true and take $n$ to $infty$.Then we get $lim_n f_n = f_k$ for all $k$ which is absurd.
    – Song
    yesterday








  • 1




    To elaborate slightly on my first comment, as a "test" consider constant functions $f_k$ defined via $f_{k+1}(x) = f_k(x)+a_k$.
    – quid♦
    yesterday










  • Thank you so much
    – Mano.kiko
    yesterday
















This seems like a typo to me. What you say you can prove ought to be the indended meaning.
– quid♦
yesterday






This seems like a typo to me. What you say you can prove ought to be the indended meaning.
– quid♦
yesterday






1




1




Well, the statement is not true and that's why you couldn't solve it. It looks like a typo. To see this, suppose it's true and take $n$ to $infty$.Then we get $lim_n f_n = f_k$ for all $k$ which is absurd.
– Song
yesterday






Well, the statement is not true and that's why you couldn't solve it. It looks like a typo. To see this, suppose it's true and take $n$ to $infty$.Then we get $lim_n f_n = f_k$ for all $k$ which is absurd.
– Song
yesterday






1




1




To elaborate slightly on my first comment, as a "test" consider constant functions $f_k$ defined via $f_{k+1}(x) = f_k(x)+a_k$.
– quid♦
yesterday




To elaborate slightly on my first comment, as a "test" consider constant functions $f_k$ defined via $f_{k+1}(x) = f_k(x)+a_k$.
– quid♦
yesterday












Thank you so much
– Mano.kiko
yesterday




Thank you so much
– Mano.kiko
yesterday










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