If $|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$, then exists a subsequence which converges uniformly [duplicate]
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Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.
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Let $f_n:[0,1]rightarrow mathbbR$ be continuous functions such that:
$|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$ for every $nin N$. Prove that exists a subsequence of $(f_n)$ which converges uniformly.
I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence $(f_n)$ is uniformly bounded and that it is equicontinuous.
What I have noted/done so far:
Since $|f'_n(x)|leq frac1sqrt x$, if $x,yin [0,1],xneq y$, then
$$|f_n(y)-f_n(x)|=|int_x^yf'_n(t)dt |leqint_x^y|f'_n(t)|dtleqint_x^yfrac1sqrt tdtleqint_0^1frac1sqrt tdt = K$$
but I can't use this to solve the problem. Any help? Thanks in advance
real-analysis arzela-ascoli
marked as duplicate by Xander Henderson, Sil, Taroccoesbrocco, Theoretical Economist, Lord Shark the Unknown Aug 12 at 11:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
2
down vote
favorite
This question already has an answer here:
Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.
1 answer
Let $f_n:[0,1]rightarrow mathbbR$ be continuous functions such that:
$|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$ for every $nin N$. Prove that exists a subsequence of $(f_n)$ which converges uniformly.
I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence $(f_n)$ is uniformly bounded and that it is equicontinuous.
What I have noted/done so far:
Since $|f'_n(x)|leq frac1sqrt x$, if $x,yin [0,1],xneq y$, then
$$|f_n(y)-f_n(x)|=|int_x^yf'_n(t)dt |leqint_x^y|f'_n(t)|dtleqint_x^yfrac1sqrt tdtleqint_0^1frac1sqrt tdt = K$$
but I can't use this to solve the problem. Any help? Thanks in advance
real-analysis arzela-ascoli
marked as duplicate by Xander Henderson, Sil, Taroccoesbrocco, Theoretical Economist, Lord Shark the Unknown Aug 12 at 11:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question already has an answer here:
Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.
1 answer
Let $f_n:[0,1]rightarrow mathbbR$ be continuous functions such that:
$|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$ for every $nin N$. Prove that exists a subsequence of $(f_n)$ which converges uniformly.
I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence $(f_n)$ is uniformly bounded and that it is equicontinuous.
What I have noted/done so far:
Since $|f'_n(x)|leq frac1sqrt x$, if $x,yin [0,1],xneq y$, then
$$|f_n(y)-f_n(x)|=|int_x^yf'_n(t)dt |leqint_x^y|f'_n(t)|dtleqint_x^yfrac1sqrt tdtleqint_0^1frac1sqrt tdt = K$$
but I can't use this to solve the problem. Any help? Thanks in advance
real-analysis arzela-ascoli
This question already has an answer here:
Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.
1 answer
Let $f_n:[0,1]rightarrow mathbbR$ be continuous functions such that:
$|f'_n(x)|leq frac1sqrt x$ and $int_0^1f_n(x)dx =0$ for every $nin N$. Prove that exists a subsequence of $(f_n)$ which converges uniformly.
I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence $(f_n)$ is uniformly bounded and that it is equicontinuous.
What I have noted/done so far:
Since $|f'_n(x)|leq frac1sqrt x$, if $x,yin [0,1],xneq y$, then
$$|f_n(y)-f_n(x)|=|int_x^yf'_n(t)dt |leqint_x^y|f'_n(t)|dtleqint_x^yfrac1sqrt tdtleqint_0^1frac1sqrt tdt = K$$
but I can't use this to solve the problem. Any help? Thanks in advance
This question already has an answer here:
Prove $f_n$ has a uniformly convergent subsequence on $[0,1]$.
1 answer
real-analysis arzela-ascoli
edited Aug 12 at 7:02
Sil
5,14121443
5,14121443
asked Jun 20 at 11:26
math.h
1,034516
1,034516
marked as duplicate by Xander Henderson, Sil, Taroccoesbrocco, Theoretical Economist, Lord Shark the Unknown Aug 12 at 11:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Xander Henderson, Sil, Taroccoesbrocco, Theoretical Economist, Lord Shark the Unknown Aug 12 at 11:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
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up vote
3
down vote
accepted
Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
$$
|f_n(y) - f_n(x)| leq left|
int_x^y frac1sqrts, ds
right|
= 2 left| sqrty - sqrtxright|
leq 2 sqrt,
$$
so that the sequence $(f_n)$ is equicontinuous.
On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
$$
|f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
qquad forall xin [0,1].
$$
perfectly noted, thank you
â math.h
Jun 20 at 11:46
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
$$
|f_n(y) - f_n(x)| leq left|
int_x^y frac1sqrts, ds
right|
= 2 left| sqrty - sqrtxright|
leq 2 sqrt,
$$
so that the sequence $(f_n)$ is equicontinuous.
On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
$$
|f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
qquad forall xin [0,1].
$$
perfectly noted, thank you
â math.h
Jun 20 at 11:46
add a comment |Â
up vote
3
down vote
accepted
Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
$$
|f_n(y) - f_n(x)| leq left|
int_x^y frac1sqrts, ds
right|
= 2 left| sqrty - sqrtxright|
leq 2 sqrt,
$$
so that the sequence $(f_n)$ is equicontinuous.
On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
$$
|f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
qquad forall xin [0,1].
$$
perfectly noted, thank you
â math.h
Jun 20 at 11:46
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
$$
|f_n(y) - f_n(x)| leq left|
int_x^y frac1sqrts, ds
right|
= 2 left| sqrty - sqrtxright|
leq 2 sqrt,
$$
so that the sequence $(f_n)$ is equicontinuous.
On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
$$
|f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
qquad forall xin [0,1].
$$
Observe that, for all $ninmathbbN$ and every $x,yin [0,1]$,
$$
|f_n(y) - f_n(x)| leq left|
int_x^y frac1sqrts, ds
right|
= 2 left| sqrty - sqrtxright|
leq 2 sqrt,
$$
so that the sequence $(f_n)$ is equicontinuous.
On the other hand, the condition $int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_nin [0,1]$ such that $f_n(x_n) = 0$, so that
$$
|f_n(x)| = |f_n(x) - f_n(x_n)| leq left| int_x_n^x frac1sqrts, dsright| leq 2,
qquad forall xin [0,1].
$$
answered Jun 20 at 11:39
Rigel
9,61511319
9,61511319
perfectly noted, thank you
â math.h
Jun 20 at 11:46
add a comment |Â
perfectly noted, thank you
â math.h
Jun 20 at 11:46
perfectly noted, thank you
â math.h
Jun 20 at 11:46
perfectly noted, thank you
â math.h
Jun 20 at 11:46
add a comment |Â