Is the following set a compact set?












2














Let $A$ be defined as



$$A:={fin C^1([0,1],mathbb{R}) : |f|_{C^1} leq 1}.$$



I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set?






general-topology functional-analysis compactness complete-spaces locally-compact-groups







share|cite|improve this question
























  • Which norm is $lVertcdotrVert_{C^1}$?
    – José Carlos Santos
    Nov 25 '18 at 9:55










  • How did You show $A$ is precompact?
    – Peter Melech
    Nov 25 '18 at 10:10
















2














Let $A$ be defined as



$$A:={fin C^1([0,1],mathbb{R}) : |f|_{C^1} leq 1}.$$



I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set?






general-topology functional-analysis compactness complete-spaces locally-compact-groups







share|cite|improve this question
























  • Which norm is $lVertcdotrVert_{C^1}$?
    – José Carlos Santos
    Nov 25 '18 at 9:55










  • How did You show $A$ is precompact?
    – Peter Melech
    Nov 25 '18 at 10:10














2












2








2







Let $A$ be defined as



$$A:={fin C^1([0,1],mathbb{R}) : |f|_{C^1} leq 1}.$$



I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set?






general-topology functional-analysis compactness complete-spaces locally-compact-groups







share|cite|improve this question















Let $A$ be defined as



$$A:={fin C^1([0,1],mathbb{R}) : |f|_{C^1} leq 1}.$$



I have shown that the set is precompact. But is $A$ a complete set? Or an other question: Is $A$ a closed set?






general-topology functional-analysis compactness complete-spaces locally-compact-groups




general-topology functional-analysis compactness complete-spaces locally-compact-groups






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 17:37









Yiorgos S. Smyrlis

62.5k1383162




62.5k1383162










asked Nov 25 '18 at 9:44









MathCracky

445212




445212












  • Which norm is $lVertcdotrVert_{C^1}$?
    – José Carlos Santos
    Nov 25 '18 at 9:55










  • How did You show $A$ is precompact?
    – Peter Melech
    Nov 25 '18 at 10:10


















  • Which norm is $lVertcdotrVert_{C^1}$?
    – José Carlos Santos
    Nov 25 '18 at 9:55










  • How did You show $A$ is precompact?
    – Peter Melech
    Nov 25 '18 at 10:10
















Which norm is $lVertcdotrVert_{C^1}$?
– José Carlos Santos
Nov 25 '18 at 9:55




Which norm is $lVertcdotrVert_{C^1}$?
– José Carlos Santos
Nov 25 '18 at 9:55












How did You show $A$ is precompact?
– Peter Melech
Nov 25 '18 at 10:10




How did You show $A$ is precompact?
– Peter Melech
Nov 25 '18 at 10:10










2 Answers
2






active

oldest

votes


















4














The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.



But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.



I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.






share|cite|improve this answer



















  • 2




    Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
    – Yanko
    Nov 25 '18 at 10:44










  • But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
    – MaoWao
    Nov 25 '18 at 17:58










  • No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
    – José Carlos Santos
    Nov 25 '18 at 18:04



















4














The answer is NO.



Consider the sequence
$$
f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
$$
Then
$$
|,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
$$
If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.



    But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.



    I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.






    share|cite|improve this answer



















    • 2




      Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
      – Yanko
      Nov 25 '18 at 10:44










    • But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
      – MaoWao
      Nov 25 '18 at 17:58










    • No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
      – José Carlos Santos
      Nov 25 '18 at 18:04
















    4














    The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.



    But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.



    I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.






    share|cite|improve this answer



















    • 2




      Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
      – Yanko
      Nov 25 '18 at 10:44










    • But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
      – MaoWao
      Nov 25 '18 at 17:58










    • No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
      – José Carlos Santos
      Nov 25 '18 at 18:04














    4












    4








    4






    The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.



    But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.



    I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.






    share|cite|improve this answer














    The set is closed since it is the inverse image of $[0,1]$ with respect to the continuous map $lVertcdotrVert_{C^1}$.



    But it cannot possibly be compact because the only normed spaces with compact closed unit balls are the finite-dimensional ones. And your space isn't.



    I don't know which norm the norm $lVertcdotrVert_{C^1}$ is, but I suppose that it is such that $C^1bigl([0,1],mathbb Rbigr)$ is complete. If that's so, then $A$ is complete (since it is closed). Therefore, it cannot possibly be precompact, since compact $iff$ precompact and complete.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 25 '18 at 10:03

























    answered Nov 25 '18 at 9:58









    José Carlos Santos

    150k22121221




    150k22121221








    • 2




      Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
      – Yanko
      Nov 25 '18 at 10:44










    • But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
      – MaoWao
      Nov 25 '18 at 17:58










    • No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
      – José Carlos Santos
      Nov 25 '18 at 18:04














    • 2




      Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
      – Yanko
      Nov 25 '18 at 10:44










    • But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
      – MaoWao
      Nov 25 '18 at 17:58










    • No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
      – José Carlos Santos
      Nov 25 '18 at 18:04








    2




    2




    Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
    – Yanko
    Nov 25 '18 at 10:44




    Good answer (+1). Just for general knowledge the norm $|cdot|_{C^1}$ is usually the sup norm + the sup norm of the derivative (i.e. $|f|_{C^1} = |f|_infty + |f'|_infty$). This is indeed a norm such that $C^1([0,1],mathbb{R})$ is complete.
    – Yanko
    Nov 25 '18 at 10:44












    But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
    – MaoWao
    Nov 25 '18 at 17:58




    But the set $A$ is precompact in $C([0,1])$ with the sup norm by Arzela-Ascoli. Maybe that's what the op meant.
    – MaoWao
    Nov 25 '18 at 17:58












    No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
    – José Carlos Santos
    Nov 25 '18 at 18:04




    No, it is not precompact. If $f_n(x)=x^n$, $(f_n)_{ninmathbb N}$ has no convergent subsequence.
    – José Carlos Santos
    Nov 25 '18 at 18:04











    4














    The answer is NO.



    Consider the sequence
    $$
    f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
    $$
    Then
    $$
    |,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
    $$
    If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
    converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.






    share|cite|improve this answer


























      4














      The answer is NO.



      Consider the sequence
      $$
      f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
      $$
      Then
      $$
      |,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
      $$
      If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
      converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.






      share|cite|improve this answer
























        4












        4








        4






        The answer is NO.



        Consider the sequence
        $$
        f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
        $$
        Then
        $$
        |,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
        $$
        If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
        converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.






        share|cite|improve this answer












        The answer is NO.



        Consider the sequence
        $$
        f_n(x)=frac{1}{n+1}sin(nx),,,ninmathbb N.
        $$
        Then
        $$
        |,f_n|_{C^1}= max |,f_n|+ max |,f_n'|=frac{1}{n+1}+frac{n}{n+1}=1.
        $$
        If ${,f_n}$ possessed a converging subsequence $,{,f_{n_k}}$, in the $C^1-$sense, with limit $,f,,$ then $,{,f_{n_k}}$ would also converge to $f$ in the uniform sense. But, ${,f_{n_k}}$
        converges uniformly to $fequiv 0$. Nevertheless, $,{,f_{n_k}}$ DOES NOT converge in the $C^1-$sense to $0$, since $,|,f_{n_k}|_{C^1}=1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 '18 at 10:55









        Yiorgos S. Smyrlis

        62.5k1383162




        62.5k1383162






























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