Does uniform continuity on a compact subset imply equicontinuity?
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 yesterday
Mike
1,506321
add a comment |
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 yesterday
Mike
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
add a comment |
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 yesterday
Mike
1,506321
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 yesterday
Mike
1,506321
asked yesterday
Mike
1,506321
edited yesterday
asked yesterday
Mike
1,506321
asked yesterday
Mike
1,506321
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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 yesterday
RRL
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
|
show 3 more comments
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1 Answer
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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 yesterday
RRL
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
|
show 3 more comments
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 yesterday
RRL
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
|
show 3 more comments
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 yesterday
RRL
49.3k42573
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 yesterday
RRL
49.3k42573
edited yesterday
answered yesterday
RRL
49.3k42573
answered yesterday
RRL
49.3k42573
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
|
show 3 more comments
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
|
show 3 more comments
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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