Does uniform continuity on a compact subset imply equicontinuity?
$begingroup$
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
asked Jan 5 at 7:54
Omojola MichealOmojola Micheal
1,971324
$endgroup$
add a comment |
$begingroup$
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
asked Jan 5 at 7:54
Omojola MichealOmojola Micheal
1,971324
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 5 at 7:54
Omojola MichealOmojola Micheal
1,971324
$endgroup$
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
asked Jan 5 at 7:54
Omojola MichealOmojola Micheal
1,971324
asked Jan 5 at 7:54
Omojola MichealOmojola Micheal
1,971324
edited Jan 5 at 8:09
Omojola Micheal
asked Jan 5 at 7:54
Omojola MichealOmojola Micheal
1,971324
asked Jan 5 at 7:54
Omojola MichealOmojola Micheal
1,971324
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 5 at 8:20
RRLRRL
52.7k42573
$endgroup$
$begingroup$
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
|
show 3 more comments
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1 Answer
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$begingroup$
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.
answered Jan 5 at 8:20
RRLRRL
52.7k42573
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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.
answered Jan 5 at 8:20
RRLRRL
52.7k42573
$endgroup$
$begingroup$
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
|
show 3 more comments
$begingroup$
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.
answered Jan 5 at 8:20
RRLRRL
52.7k42573
$endgroup$
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.
answered Jan 5 at 8:20
RRLRRL
52.7k42573
edited Jan 5 at 8:35
answered Jan 5 at 8:20
RRLRRL
52.7k42573
answered Jan 5 at 8:20
RRLRRL
52.7k42573
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
|
show 3 more comments
$begingroup$
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
|
show 3 more comments
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$begingroup$
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