Show that a bounded subset of $C^1[a, b]$ is equicontinuous












0














Let $S$ be a bounded subset of $C^1[a, b]$. If $fin S$, $exists M>0$ s.t. $|f(x)|le M$ for all $xin[a, b]$. In addition, $f$ and $f'$ are continuous on $[a, b]$.



How do I proceed?






real-analysis







share|cite|improve this question


















  • 1




    Counterexample: $f_n(x) = sin( pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)| =1$ for $x_n =1/n$. Thus, we can not have $|f_n(0)-f_n(x)| <1$ for $|x| < delta$, where $delta>0$ is independent of $n$.
    – p4sch
    Dec 4 '18 at 21:30






  • 1




    Yes, I know. The questioner doesn't said anything about the chosen norm on $C^1[a,b]$.
    – p4sch
    Dec 4 '18 at 21:34






  • 1




    Indeed. Actually the fact that questioneer mentions the existence of $M$ such that $|f(x)| leq M$ seems to suggest that he's not using the norm I mentioned.
    – MisterRiemann
    Dec 4 '18 at 21:35






  • 1




    You should mention the norm you are using. Otherwise we cannot know what boundedness means.
    – MisterRiemann
    Dec 4 '18 at 21:39








  • 1




    You say $S$ is a bounded subset of $C^1[a,b]$. Unless told to assume a nonstandard context, this should mean there exists $M>0$ such that $|f'(x)|<M$ for all $f in S$ and all $x in [a,b]$. In which case the result follows from the Mean Value Theorem.
    – Matt A Pelto
    Dec 4 '18 at 21:41


















0














Let $S$ be a bounded subset of $C^1[a, b]$. If $fin S$, $exists M>0$ s.t. $|f(x)|le M$ for all $xin[a, b]$. In addition, $f$ and $f'$ are continuous on $[a, b]$.



How do I proceed?






real-analysis







share|cite|improve this question


















  • 1




    Counterexample: $f_n(x) = sin( pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)| =1$ for $x_n =1/n$. Thus, we can not have $|f_n(0)-f_n(x)| <1$ for $|x| < delta$, where $delta>0$ is independent of $n$.
    – p4sch
    Dec 4 '18 at 21:30






  • 1




    Yes, I know. The questioner doesn't said anything about the chosen norm on $C^1[a,b]$.
    – p4sch
    Dec 4 '18 at 21:34






  • 1




    Indeed. Actually the fact that questioneer mentions the existence of $M$ such that $|f(x)| leq M$ seems to suggest that he's not using the norm I mentioned.
    – MisterRiemann
    Dec 4 '18 at 21:35






  • 1




    You should mention the norm you are using. Otherwise we cannot know what boundedness means.
    – MisterRiemann
    Dec 4 '18 at 21:39








  • 1




    You say $S$ is a bounded subset of $C^1[a,b]$. Unless told to assume a nonstandard context, this should mean there exists $M>0$ such that $|f'(x)|<M$ for all $f in S$ and all $x in [a,b]$. In which case the result follows from the Mean Value Theorem.
    – Matt A Pelto
    Dec 4 '18 at 21:41
















0












0








0







Let $S$ be a bounded subset of $C^1[a, b]$. If $fin S$, $exists M>0$ s.t. $|f(x)|le M$ for all $xin[a, b]$. In addition, $f$ and $f'$ are continuous on $[a, b]$.



How do I proceed?






real-analysis







share|cite|improve this question













Let $S$ be a bounded subset of $C^1[a, b]$. If $fin S$, $exists M>0$ s.t. $|f(x)|le M$ for all $xin[a, b]$. In addition, $f$ and $f'$ are continuous on $[a, b]$.



How do I proceed?






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 21:25









ThomasThomas

730416




730416








  • 1




    Counterexample: $f_n(x) = sin( pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)| =1$ for $x_n =1/n$. Thus, we can not have $|f_n(0)-f_n(x)| <1$ for $|x| < delta$, where $delta>0$ is independent of $n$.
    – p4sch
    Dec 4 '18 at 21:30






  • 1




    Yes, I know. The questioner doesn't said anything about the chosen norm on $C^1[a,b]$.
    – p4sch
    Dec 4 '18 at 21:34






  • 1




    Indeed. Actually the fact that questioneer mentions the existence of $M$ such that $|f(x)| leq M$ seems to suggest that he's not using the norm I mentioned.
    – MisterRiemann
    Dec 4 '18 at 21:35






  • 1




    You should mention the norm you are using. Otherwise we cannot know what boundedness means.
    – MisterRiemann
    Dec 4 '18 at 21:39








  • 1




    You say $S$ is a bounded subset of $C^1[a,b]$. Unless told to assume a nonstandard context, this should mean there exists $M>0$ such that $|f'(x)|<M$ for all $f in S$ and all $x in [a,b]$. In which case the result follows from the Mean Value Theorem.
    – Matt A Pelto
    Dec 4 '18 at 21:41
















  • 1




    Counterexample: $f_n(x) = sin( pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)| =1$ for $x_n =1/n$. Thus, we can not have $|f_n(0)-f_n(x)| <1$ for $|x| < delta$, where $delta>0$ is independent of $n$.
    – p4sch
    Dec 4 '18 at 21:30






  • 1




    Yes, I know. The questioner doesn't said anything about the chosen norm on $C^1[a,b]$.
    – p4sch
    Dec 4 '18 at 21:34






  • 1




    Indeed. Actually the fact that questioneer mentions the existence of $M$ such that $|f(x)| leq M$ seems to suggest that he's not using the norm I mentioned.
    – MisterRiemann
    Dec 4 '18 at 21:35






  • 1




    You should mention the norm you are using. Otherwise we cannot know what boundedness means.
    – MisterRiemann
    Dec 4 '18 at 21:39








  • 1




    You say $S$ is a bounded subset of $C^1[a,b]$. Unless told to assume a nonstandard context, this should mean there exists $M>0$ such that $|f'(x)|<M$ for all $f in S$ and all $x in [a,b]$. In which case the result follows from the Mean Value Theorem.
    – Matt A Pelto
    Dec 4 '18 at 21:41










1




1




Counterexample: $f_n(x) = sin( pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)| =1$ for $x_n =1/n$. Thus, we can not have $|f_n(0)-f_n(x)| <1$ for $|x| < delta$, where $delta>0$ is independent of $n$.
– p4sch
Dec 4 '18 at 21:30




Counterexample: $f_n(x) = sin( pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)| =1$ for $x_n =1/n$. Thus, we can not have $|f_n(0)-f_n(x)| <1$ for $|x| < delta$, where $delta>0$ is independent of $n$.
– p4sch
Dec 4 '18 at 21:30




1




1




Yes, I know. The questioner doesn't said anything about the chosen norm on $C^1[a,b]$.
– p4sch
Dec 4 '18 at 21:34




Yes, I know. The questioner doesn't said anything about the chosen norm on $C^1[a,b]$.
– p4sch
Dec 4 '18 at 21:34




1




1




Indeed. Actually the fact that questioneer mentions the existence of $M$ such that $|f(x)| leq M$ seems to suggest that he's not using the norm I mentioned.
– MisterRiemann
Dec 4 '18 at 21:35




Indeed. Actually the fact that questioneer mentions the existence of $M$ such that $|f(x)| leq M$ seems to suggest that he's not using the norm I mentioned.
– MisterRiemann
Dec 4 '18 at 21:35




1




1




You should mention the norm you are using. Otherwise we cannot know what boundedness means.
– MisterRiemann
Dec 4 '18 at 21:39






You should mention the norm you are using. Otherwise we cannot know what boundedness means.
– MisterRiemann
Dec 4 '18 at 21:39






1




1




You say $S$ is a bounded subset of $C^1[a,b]$. Unless told to assume a nonstandard context, this should mean there exists $M>0$ such that $|f'(x)|<M$ for all $f in S$ and all $x in [a,b]$. In which case the result follows from the Mean Value Theorem.
– Matt A Pelto
Dec 4 '18 at 21:41






You say $S$ is a bounded subset of $C^1[a,b]$. Unless told to assume a nonstandard context, this should mean there exists $M>0$ such that $|f'(x)|<M$ for all $f in S$ and all $x in [a,b]$. In which case the result follows from the Mean Value Theorem.
– Matt A Pelto
Dec 4 '18 at 21:41












1 Answer
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If you consider the norm $$|f|:= sup_{x in [a,b]} |f(x)| + sup_{x in [a,b]} |f'(x)|$$ on $C^1[a,b]$, then a set is bounded if and only if $|f| le M$ and $|f'| le M$ for all $f in S$. Thus for all $y >x$
$$|f(x)-f(y)| le int_x^y |f'(t)| , dt le M |y-x|.$$
This proves already that $M$ is equicontinuous.



$C^1[a,b]$ with this norm is a Banach space. Note that $|f|_2 := sup_{x in [a,b]} |f(x)|$ is another norm and now $C^1[a,b]$ is not a Banach space. In this case, the statement is false. Take as in my comment $f_n(x)=sin(pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)|=1$ for $x_n=1/n$. Thus, we can not have $|f_n(0)−f_n(x)|<1$ for $|x| < delta$, where $delta >0$ is independent of $n$.



However, this property doesn't depend on the completness: We can take
$$|f|_3:= sup_{x in [a,b]} |f(x)| + sqrt{int_a^b |f'(x)|^2 , dx}.$$
And here we also have that a bounded set is equicontinuous. (We can argue as above in combination with the Hölder inequality.) But the space is not complete with this norm! (Just take sequence $g_n$ of continuous functions with $g_n rightarrow g$ in $L^2$, where $g$ is not continuous and set $f_n(t):= int_a^t g_n(h) , dh$. Note that $|f_n-f_m|_infty le sqrt{(b-a)} |g_n - g_m|_{L^2} $.)






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    Good answer (+1)
    – MisterRiemann
    Dec 4 '18 at 22:04











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If you consider the norm $$|f|:= sup_{x in [a,b]} |f(x)| + sup_{x in [a,b]} |f'(x)|$$ on $C^1[a,b]$, then a set is bounded if and only if $|f| le M$ and $|f'| le M$ for all $f in S$. Thus for all $y >x$
$$|f(x)-f(y)| le int_x^y |f'(t)| , dt le M |y-x|.$$
This proves already that $M$ is equicontinuous.



$C^1[a,b]$ with this norm is a Banach space. Note that $|f|_2 := sup_{x in [a,b]} |f(x)|$ is another norm and now $C^1[a,b]$ is not a Banach space. In this case, the statement is false. Take as in my comment $f_n(x)=sin(pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)|=1$ for $x_n=1/n$. Thus, we can not have $|f_n(0)−f_n(x)|<1$ for $|x| < delta$, where $delta >0$ is independent of $n$.



However, this property doesn't depend on the completness: We can take
$$|f|_3:= sup_{x in [a,b]} |f(x)| + sqrt{int_a^b |f'(x)|^2 , dx}.$$
And here we also have that a bounded set is equicontinuous. (We can argue as above in combination with the Hölder inequality.) But the space is not complete with this norm! (Just take sequence $g_n$ of continuous functions with $g_n rightarrow g$ in $L^2$, where $g$ is not continuous and set $f_n(t):= int_a^t g_n(h) , dh$. Note that $|f_n-f_m|_infty le sqrt{(b-a)} |g_n - g_m|_{L^2} $.)






share|cite|improve this answer

















  • 1




    Good answer (+1)
    – MisterRiemann
    Dec 4 '18 at 22:04
















3














If you consider the norm $$|f|:= sup_{x in [a,b]} |f(x)| + sup_{x in [a,b]} |f'(x)|$$ on $C^1[a,b]$, then a set is bounded if and only if $|f| le M$ and $|f'| le M$ for all $f in S$. Thus for all $y >x$
$$|f(x)-f(y)| le int_x^y |f'(t)| , dt le M |y-x|.$$
This proves already that $M$ is equicontinuous.



$C^1[a,b]$ with this norm is a Banach space. Note that $|f|_2 := sup_{x in [a,b]} |f(x)|$ is another norm and now $C^1[a,b]$ is not a Banach space. In this case, the statement is false. Take as in my comment $f_n(x)=sin(pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)|=1$ for $x_n=1/n$. Thus, we can not have $|f_n(0)−f_n(x)|<1$ for $|x| < delta$, where $delta >0$ is independent of $n$.



However, this property doesn't depend on the completness: We can take
$$|f|_3:= sup_{x in [a,b]} |f(x)| + sqrt{int_a^b |f'(x)|^2 , dx}.$$
And here we also have that a bounded set is equicontinuous. (We can argue as above in combination with the Hölder inequality.) But the space is not complete with this norm! (Just take sequence $g_n$ of continuous functions with $g_n rightarrow g$ in $L^2$, where $g$ is not continuous and set $f_n(t):= int_a^t g_n(h) , dh$. Note that $|f_n-f_m|_infty le sqrt{(b-a)} |g_n - g_m|_{L^2} $.)






share|cite|improve this answer

















  • 1




    Good answer (+1)
    – MisterRiemann
    Dec 4 '18 at 22:04














3












3








3






If you consider the norm $$|f|:= sup_{x in [a,b]} |f(x)| + sup_{x in [a,b]} |f'(x)|$$ on $C^1[a,b]$, then a set is bounded if and only if $|f| le M$ and $|f'| le M$ for all $f in S$. Thus for all $y >x$
$$|f(x)-f(y)| le int_x^y |f'(t)| , dt le M |y-x|.$$
This proves already that $M$ is equicontinuous.



$C^1[a,b]$ with this norm is a Banach space. Note that $|f|_2 := sup_{x in [a,b]} |f(x)|$ is another norm and now $C^1[a,b]$ is not a Banach space. In this case, the statement is false. Take as in my comment $f_n(x)=sin(pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)|=1$ for $x_n=1/n$. Thus, we can not have $|f_n(0)−f_n(x)|<1$ for $|x| < delta$, where $delta >0$ is independent of $n$.



However, this property doesn't depend on the completness: We can take
$$|f|_3:= sup_{x in [a,b]} |f(x)| + sqrt{int_a^b |f'(x)|^2 , dx}.$$
And here we also have that a bounded set is equicontinuous. (We can argue as above in combination with the Hölder inequality.) But the space is not complete with this norm! (Just take sequence $g_n$ of continuous functions with $g_n rightarrow g$ in $L^2$, where $g$ is not continuous and set $f_n(t):= int_a^t g_n(h) , dh$. Note that $|f_n-f_m|_infty le sqrt{(b-a)} |g_n - g_m|_{L^2} $.)






share|cite|improve this answer












If you consider the norm $$|f|:= sup_{x in [a,b]} |f(x)| + sup_{x in [a,b]} |f'(x)|$$ on $C^1[a,b]$, then a set is bounded if and only if $|f| le M$ and $|f'| le M$ for all $f in S$. Thus for all $y >x$
$$|f(x)-f(y)| le int_x^y |f'(t)| , dt le M |y-x|.$$
This proves already that $M$ is equicontinuous.



$C^1[a,b]$ with this norm is a Banach space. Note that $|f|_2 := sup_{x in [a,b]} |f(x)|$ is another norm and now $C^1[a,b]$ is not a Banach space. In this case, the statement is false. Take as in my comment $f_n(x)=sin(pi n x)$ on $[0,1]$. We have $|f_n| le 1$, but $|f_n(x_n)|=1$ for $x_n=1/n$. Thus, we can not have $|f_n(0)−f_n(x)|<1$ for $|x| < delta$, where $delta >0$ is independent of $n$.



However, this property doesn't depend on the completness: We can take
$$|f|_3:= sup_{x in [a,b]} |f(x)| + sqrt{int_a^b |f'(x)|^2 , dx}.$$
And here we also have that a bounded set is equicontinuous. (We can argue as above in combination with the Hölder inequality.) But the space is not complete with this norm! (Just take sequence $g_n$ of continuous functions with $g_n rightarrow g$ in $L^2$, where $g$ is not continuous and set $f_n(t):= int_a^t g_n(h) , dh$. Note that $|f_n-f_m|_infty le sqrt{(b-a)} |g_n - g_m|_{L^2} $.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 21:58









p4schp4sch

4,770217




4,770217








  • 1




    Good answer (+1)
    – MisterRiemann
    Dec 4 '18 at 22:04














  • 1




    Good answer (+1)
    – MisterRiemann
    Dec 4 '18 at 22:04








1




1




Good answer (+1)
– MisterRiemann
Dec 4 '18 at 22:04




Good answer (+1)
– MisterRiemann
Dec 4 '18 at 22:04


















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