Does uniform continuity on a compact subset imply equicontinuity?












1












$begingroup$


I gave a proof here some time ago but was wondering about the following:



Suppose that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies
$$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}.$$
Taking $delta=min{delta_n:;nin Bbb{N}}$ (which might not be true), then we have the definition given here, that
"for every $epsilon > 0$, there exists $delta>0$ such that $forall,x,yin K, ;|x − y| < delta,$ implies
$$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}.$$
which implies that ${f_n}$ is equicontinuous.



QUESTION: I'm I right? If not, can you please provide a counter-example?






real-analysis functional-analysis analysis equicontinuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
    $endgroup$
    – RRL
    Jan 5 at 8:04










  • $begingroup$
    @RRL: Thanks for that! But how do I get an example?
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:07
















1












$begingroup$


I gave a proof here some time ago but was wondering about the following:



Suppose that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies
$$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}.$$
Taking $delta=min{delta_n:;nin Bbb{N}}$ (which might not be true), then we have the definition given here, that
"for every $epsilon > 0$, there exists $delta>0$ such that $forall,x,yin K, ;|x − y| < delta,$ implies
$$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}.$$
which implies that ${f_n}$ is equicontinuous.



QUESTION: I'm I right? If not, can you please provide a counter-example?






real-analysis functional-analysis analysis equicontinuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
    $endgroup$
    – RRL
    Jan 5 at 8:04










  • $begingroup$
    @RRL: Thanks for that! But how do I get an example?
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:07














1












1








1


1



$begingroup$


I gave a proof here some time ago but was wondering about the following:



Suppose that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies
$$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}.$$
Taking $delta=min{delta_n:;nin Bbb{N}}$ (which might not be true), then we have the definition given here, that
"for every $epsilon > 0$, there exists $delta>0$ such that $forall,x,yin K, ;|x − y| < delta,$ implies
$$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}.$$
which implies that ${f_n}$ is equicontinuous.



QUESTION: I'm I right? If not, can you please provide a counter-example?






real-analysis functional-analysis analysis equicontinuity







share|cite|improve this question











$endgroup$




I gave a proof here some time ago but was wondering about the following:



Suppose that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies
$$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}.$$
Taking $delta=min{delta_n:;nin Bbb{N}}$ (which might not be true), then we have the definition given here, that
"for every $epsilon > 0$, there exists $delta>0$ such that $forall,x,yin K, ;|x − y| < delta,$ implies
$$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}.$$
which implies that ${f_n}$ is equicontinuous.



QUESTION: I'm I right? If not, can you please provide a counter-example?






real-analysis functional-analysis analysis equicontinuity




real-analysis functional-analysis analysis equicontinuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 8:09







Omojola Micheal

















asked Jan 5 at 7:54









Omojola MichealOmojola Micheal

1,971324




1,971324












  • $begingroup$
    What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
    $endgroup$
    – RRL
    Jan 5 at 8:04










  • $begingroup$
    @RRL: Thanks for that! But how do I get an example?
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:07


















  • $begingroup$
    What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
    $endgroup$
    – RRL
    Jan 5 at 8:04










  • $begingroup$
    @RRL: Thanks for that! But how do I get an example?
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:07
















$begingroup$
What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
$endgroup$
– RRL
Jan 5 at 8:04




$begingroup$
What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
$endgroup$
– RRL
Jan 5 at 8:04












$begingroup$
@RRL: Thanks for that! But how do I get an example?
$endgroup$
– Omojola Micheal
Jan 5 at 8:07




$begingroup$
@RRL: Thanks for that! But how do I get an example?
$endgroup$
– Omojola Micheal
Jan 5 at 8:07










1 Answer
1






active

oldest

votes


















1












$begingroup$

To construct a sequence $(f_n)$ of continuous functions on a compact set that is not equicontinuous, select one that is pointwise but not uniformly convergent and with a discontinuous limit function.



An example is $f_n(x) = x^n$ with $K = [0,1]$. To see that these are not equicontinuous, suppose that there exists $delta > 0$ such that if $|x - 1| < delta$ then $|f_n(x) - f_n(1)| =|x^n - 1| < 1/4$ for every $n$. However, since $x^n to 0$ when $1- delta < x < 1$, with $x$ fixed we can find $n$ such that $|x^n - 1| > 1/4$ a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:25










  • $begingroup$
    If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    $endgroup$
    – RRL
    Jan 5 at 8:27












  • $begingroup$
    So, is it right to say that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:27








  • 1




    $begingroup$
    "If each function fn is continuous on compact K then it is uniformly continuous, but the family {fn} may not be equicontinuous." Okay, I get you now!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:28






  • 1




    $begingroup$
    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:30











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

To construct a sequence $(f_n)$ of continuous functions on a compact set that is not equicontinuous, select one that is pointwise but not uniformly convergent and with a discontinuous limit function.



An example is $f_n(x) = x^n$ with $K = [0,1]$. To see that these are not equicontinuous, suppose that there exists $delta > 0$ such that if $|x - 1| < delta$ then $|f_n(x) - f_n(1)| =|x^n - 1| < 1/4$ for every $n$. However, since $x^n to 0$ when $1- delta < x < 1$, with $x$ fixed we can find $n$ such that $|x^n - 1| > 1/4$ a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:25










  • $begingroup$
    If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    $endgroup$
    – RRL
    Jan 5 at 8:27












  • $begingroup$
    So, is it right to say that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:27








  • 1




    $begingroup$
    "If each function fn is continuous on compact K then it is uniformly continuous, but the family {fn} may not be equicontinuous." Okay, I get you now!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:28






  • 1




    $begingroup$
    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:30
















1












$begingroup$

To construct a sequence $(f_n)$ of continuous functions on a compact set that is not equicontinuous, select one that is pointwise but not uniformly convergent and with a discontinuous limit function.



An example is $f_n(x) = x^n$ with $K = [0,1]$. To see that these are not equicontinuous, suppose that there exists $delta > 0$ such that if $|x - 1| < delta$ then $|f_n(x) - f_n(1)| =|x^n - 1| < 1/4$ for every $n$. However, since $x^n to 0$ when $1- delta < x < 1$, with $x$ fixed we can find $n$ such that $|x^n - 1| > 1/4$ a contradiction.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:25










  • $begingroup$
    If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    $endgroup$
    – RRL
    Jan 5 at 8:27












  • $begingroup$
    So, is it right to say that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:27








  • 1




    $begingroup$
    "If each function fn is continuous on compact K then it is uniformly continuous, but the family {fn} may not be equicontinuous." Okay, I get you now!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:28






  • 1




    $begingroup$
    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:30














1












1








1





$begingroup$

To construct a sequence $(f_n)$ of continuous functions on a compact set that is not equicontinuous, select one that is pointwise but not uniformly convergent and with a discontinuous limit function.



An example is $f_n(x) = x^n$ with $K = [0,1]$. To see that these are not equicontinuous, suppose that there exists $delta > 0$ such that if $|x - 1| < delta$ then $|f_n(x) - f_n(1)| =|x^n - 1| < 1/4$ for every $n$. However, since $x^n to 0$ when $1- delta < x < 1$, with $x$ fixed we can find $n$ such that $|x^n - 1| > 1/4$ a contradiction.






share|cite|improve this answer











$endgroup$



To construct a sequence $(f_n)$ of continuous functions on a compact set that is not equicontinuous, select one that is pointwise but not uniformly convergent and with a discontinuous limit function.



An example is $f_n(x) = x^n$ with $K = [0,1]$. To see that these are not equicontinuous, suppose that there exists $delta > 0$ such that if $|x - 1| < delta$ then $|f_n(x) - f_n(1)| =|x^n - 1| < 1/4$ for every $n$. However, since $x^n to 0$ when $1- delta < x < 1$, with $x$ fixed we can find $n$ such that $|x^n - 1| > 1/4$ a contradiction.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 5 at 8:35

























answered Jan 5 at 8:20









RRLRRL

52.7k42573




52.7k42573












  • $begingroup$
    So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:25










  • $begingroup$
    If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    $endgroup$
    – RRL
    Jan 5 at 8:27












  • $begingroup$
    So, is it right to say that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:27








  • 1




    $begingroup$
    "If each function fn is continuous on compact K then it is uniformly continuous, but the family {fn} may not be equicontinuous." Okay, I get you now!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:28






  • 1




    $begingroup$
    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:30


















  • $begingroup$
    So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:25










  • $begingroup$
    If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    $endgroup$
    – RRL
    Jan 5 at 8:27












  • $begingroup$
    So, is it right to say that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:27








  • 1




    $begingroup$
    "If each function fn is continuous on compact K then it is uniformly continuous, but the family {fn} may not be equicontinuous." Okay, I get you now!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:28






  • 1




    $begingroup$
    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    $endgroup$
    – Omojola Micheal
    Jan 5 at 8:30
















$begingroup$
So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
$endgroup$
– Omojola Micheal
Jan 5 at 8:25




$begingroup$
So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
$endgroup$
– Omojola Micheal
Jan 5 at 8:25












$begingroup$
If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
$endgroup$
– RRL
Jan 5 at 8:27






$begingroup$
If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
$endgroup$
– RRL
Jan 5 at 8:27














$begingroup$
So, is it right to say that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
$endgroup$
– Omojola Micheal
Jan 5 at 8:27






$begingroup$
So, is it right to say that $f_n:Ksubseteq Bbb{R}toBbb{R}$ is continuous for each $nin Bbb{N}$, then ${f_n}$ is uniformly continuous. So, for every $epsilon > 0$ and $nin Bbb{N}$, there exists $delta_n>0$ such that $forall,x,yin K, ;|x − y| < delta_n,$ implies $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
$endgroup$
– Omojola Micheal
Jan 5 at 8:27






1




1




$begingroup$
"If each function fn is continuous on compact K then it is uniformly continuous, but the family {fn} may not be equicontinuous." Okay, I get you now!
$endgroup$
– Omojola Micheal
Jan 5 at 8:28




$begingroup$
"If each function fn is continuous on compact K then it is uniformly continuous, but the family {fn} may not be equicontinuous." Okay, I get you now!
$endgroup$
– Omojola Micheal
Jan 5 at 8:28




1




1




$begingroup$
Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
$endgroup$
– Omojola Micheal
Jan 5 at 8:30




$begingroup$
Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
$endgroup$
– Omojola Micheal
Jan 5 at 8:30


















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.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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