Does uniform continuity on a compact subset imply equicontinuity?












1














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
























  • What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
    – RRL
    yesterday










  • @RRL: Thanks for that! But how do I get an example?
    – Mike
    yesterday
















1














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
























  • What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
    – RRL
    yesterday










  • @RRL: Thanks for that! But how do I get an example?
    – Mike
    yesterday














1












1








1


1





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















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 yesterday

























asked yesterday









Mike

1,506321




1,506321












  • What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
    – RRL
    yesterday










  • @RRL: Thanks for that! But how do I get an example?
    – Mike
    yesterday


















  • What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
    – RRL
    yesterday










  • @RRL: Thanks for that! But how do I get an example?
    – Mike
    yesterday
















What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
– RRL
yesterday




What is the min of an infinite set of numbers? What is $min{1/n : n in mathbb{N}}$? Your proof is flawed.
– RRL
yesterday












@RRL: Thanks for that! But how do I get an example?
– Mike
yesterday




@RRL: Thanks for that! But how do I get an example?
– Mike
yesterday










1 Answer
1






active

oldest

votes


















1














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























  • So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    – Mike
    yesterday










  • If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    – RRL
    yesterday












  • 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 $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    – Mike
    yesterday








  • 1




    "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!
    – Mike
    yesterday






  • 1




    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    – Mike
    yesterday











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1 Answer
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1 Answer
1






active

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active

oldest

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active

oldest

votes









1














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























  • So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    – Mike
    yesterday










  • If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    – RRL
    yesterday












  • 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 $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    – Mike
    yesterday








  • 1




    "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!
    – Mike
    yesterday






  • 1




    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    – Mike
    yesterday
















1














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























  • So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    – Mike
    yesterday










  • If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    – RRL
    yesterday












  • 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 $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    – Mike
    yesterday








  • 1




    "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!
    – Mike
    yesterday






  • 1




    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    – Mike
    yesterday














1












1








1






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














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 yesterday

























answered yesterday









RRL

49.3k42573




49.3k42573












  • So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    – Mike
    yesterday










  • If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    – RRL
    yesterday












  • 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 $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    – Mike
    yesterday








  • 1




    "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!
    – Mike
    yesterday






  • 1




    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    – Mike
    yesterday


















  • So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
    – Mike
    yesterday










  • If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
    – RRL
    yesterday












  • 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 $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
    – Mike
    yesterday








  • 1




    "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!
    – Mike
    yesterday






  • 1




    Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
    – Mike
    yesterday
















So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
– Mike
yesterday




So, a sequence $(f_n)$ of continuous functions on a compact is cannot be regarded uniformly continuous, right?
– Mike
yesterday












If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
– RRL
yesterday






If each function $f_n$ is continuous on compact $K$ then it is uniformly continuous, but the family ${f_n}$ may not be equicontinuous.
– RRL
yesterday














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 $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
– Mike
yesterday






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 $$|f_n(x)-f_n(y)| < epsilon,forall;nin Bbb{N}?$$
– Mike
yesterday






1




1




"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!
– Mike
yesterday




"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!
– Mike
yesterday




1




1




Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
– Mike
yesterday




Since the limit is not continuous, then it is not! But for future purposes, you can also add up the proof!
– Mike
yesterday


















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