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$lim_n{f(frac{x}{n})}=0$, for all $xin (0,1)$ implies $lim_{xto0}f(x)=0$ (edit, $f$ is not continuous)...
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This question already has an answer here:
If $f(x/n)to0$ when $ntoinfty$, for every $x$ and $f$ is continuous, then $f(x)to0$ when $xto0$, or not?
2 answers
Let be $f:(0,1)rightarrow mathbb{R}$. For all $xin (0,1)$ I have that $lim_n{f(frac{x}{n})}=0$.
It's true that $lim_{xto0}f(x)=0$?
And if $f$ is continuous?
For the second point I just use the definition of continuity and limits:
Let be $x_0in(0,1)$ and let be the succession ${f(frac{x_0}{n})}$, and I know that for all $varepsilon_1>0$ $exists N>0$ such that if $n>N$ then $|f(frac{x_0}{n})|<varepsilon_1$.
$f$ is continuos in $frac{x_0}{n}$, so for all $varepsilon_2>0$ $exists delta_2>0$ such that if $|x-frac{x_0}{n}|<delta_2$ then $|f(x)-f(frac{x_0}{n})|<varepsilon_2$.
So I have that $|x|le|x-frac{x_0}{n}|+|frac{x_0}{n}|<delta_2+|frac{x_0}{n}|=delta^*$, because $x_0$ is fixed.
On the other hand $|f(x)|le|f(x)-f(frac{x_0}{n})|+|f(frac{x_0}{n})|<varepsilon_1+varepsilon_2$.
But I don't know how solve the second. I try to prove that is false using many non-continuous function. but I failed.
(Sorry for my bad English, I'm Italian)
real-analysis limits continuity
asked Jan 27 at 1:08
SimmetricoSimmetrico
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marked as duplicate by Mike Earnest, RRL real-analysis
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Jan 27 at 7:05
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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This question already has an answer here:
If $f(x/n)to0$ when $ntoinfty$, for every $x$ and $f$ is continuous, then $f(x)to0$ when $xto0$, or not?
2 answers
Let be $f:(0,1)rightarrow mathbb{R}$. For all $xin (0,1)$ I have that $lim_n{f(frac{x}{n})}=0$.
It's true that $lim_{xto0}f(x)=0$?
And if $f$ is continuous?
For the second point I just use the definition of continuity and limits:
Let be $x_0in(0,1)$ and let be the succession ${f(frac{x_0}{n})}$, and I know that for all $varepsilon_1>0$ $exists N>0$ such that if $n>N$ then $|f(frac{x_0}{n})|<varepsilon_1$.
$f$ is continuos in $frac{x_0}{n}$, so for all $varepsilon_2>0$ $exists delta_2>0$ such that if $|x-frac{x_0}{n}|<delta_2$ then $|f(x)-f(frac{x_0}{n})|<varepsilon_2$.
So I have that $|x|le|x-frac{x_0}{n}|+|frac{x_0}{n}|<delta_2+|frac{x_0}{n}|=delta^*$, because $x_0$ is fixed.
On the other hand $|f(x)|le|f(x)-f(frac{x_0}{n})|+|f(frac{x_0}{n})|<varepsilon_1+varepsilon_2$.
But I don't know how solve the second. I try to prove that is false using many non-continuous function. but I failed.
(Sorry for my bad English, I'm Italian)
real-analysis limits continuity
asked Jan 27 at 1:08
SimmetricoSimmetrico
93
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marked as duplicate by Mike Earnest, RRL real-analysis
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Jan 27 at 7:05
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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This question already has an answer here:
If $f(x/n)to0$ when $ntoinfty$, for every $x$ and $f$ is continuous, then $f(x)to0$ when $xto0$, or not?
2 answers
Let be $f:(0,1)rightarrow mathbb{R}$. For all $xin (0,1)$ I have that $lim_n{f(frac{x}{n})}=0$.
It's true that $lim_{xto0}f(x)=0$?
And if $f$ is continuous?
For the second point I just use the definition of continuity and limits:
Let be $x_0in(0,1)$ and let be the succession ${f(frac{x_0}{n})}$, and I know that for all $varepsilon_1>0$ $exists N>0$ such that if $n>N$ then $|f(frac{x_0}{n})|<varepsilon_1$.
$f$ is continuos in $frac{x_0}{n}$, so for all $varepsilon_2>0$ $exists delta_2>0$ such that if $|x-frac{x_0}{n}|<delta_2$ then $|f(x)-f(frac{x_0}{n})|<varepsilon_2$.
So I have that $|x|le|x-frac{x_0}{n}|+|frac{x_0}{n}|<delta_2+|frac{x_0}{n}|=delta^*$, because $x_0$ is fixed.
On the other hand $|f(x)|le|f(x)-f(frac{x_0}{n})|+|f(frac{x_0}{n})|<varepsilon_1+varepsilon_2$.
But I don't know how solve the second. I try to prove that is false using many non-continuous function. but I failed.
(Sorry for my bad English, I'm Italian)
real-analysis limits continuity
asked Jan 27 at 1:08
SimmetricoSimmetrico
93
$endgroup$
This question already has an answer here:
If $f(x/n)to0$ when $ntoinfty$, for every $x$ and $f$ is continuous, then $f(x)to0$ when $xto0$, or not?
2 answers
Let be $f:(0,1)rightarrow mathbb{R}$. For all $xin (0,1)$ I have that $lim_n{f(frac{x}{n})}=0$.
It's true that $lim_{xto0}f(x)=0$?
And if $f$ is continuous?
For the second point I just use the definition of continuity and limits:
Let be $x_0in(0,1)$ and let be the succession ${f(frac{x_0}{n})}$, and I know that for all $varepsilon_1>0$ $exists N>0$ such that if $n>N$ then $|f(frac{x_0}{n})|<varepsilon_1$.
$f$ is continuos in $frac{x_0}{n}$, so for all $varepsilon_2>0$ $exists delta_2>0$ such that if $|x-frac{x_0}{n}|<delta_2$ then $|f(x)-f(frac{x_0}{n})|<varepsilon_2$.
So I have that $|x|le|x-frac{x_0}{n}|+|frac{x_0}{n}|<delta_2+|frac{x_0}{n}|=delta^*$, because $x_0$ is fixed.
On the other hand $|f(x)|le|f(x)-f(frac{x_0}{n})|+|f(frac{x_0}{n})|<varepsilon_1+varepsilon_2$.
But I don't know how solve the second. I try to prove that is false using many non-continuous function. but I failed.
(Sorry for my bad English, I'm Italian)
This question already has an answer here:
If $f(x/n)to0$ when $ntoinfty$, for every $x$ and $f$ is continuous, then $f(x)to0$ when $xto0$, or not?
2 answers
real-analysis limits continuity
real-analysis limits continuity
asked Jan 27 at 1:08
SimmetricoSimmetrico
93
asked Jan 27 at 1:08
SimmetricoSimmetrico
93
edited Jan 27 at 12:43
Simmetrico
asked Jan 27 at 1:08
SimmetricoSimmetrico
93
asked Jan 27 at 1:08
SimmetricoSimmetrico
93
asked Jan 27 at 1:08
SimmetricoSimmetrico
93
93
marked as duplicate by Mike Earnest, RRL real-analysis
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Jan 27 at 7:05
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Jan 27 at 7:05
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2 Answers
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This is wrong. As a counterexample, take a function $f$ which is zero except at points $1/sqrt{p}$ for $p$ prime, where the function takes the value 1 instead.
More generally, define $f(x)=1$ for $xin A,$ and $f(x)=0$ for $xnotin A,$ where $A$ is any set which does not contain $0,$ has $0$ as an accumulation point, and has the property that $xin A$ and $ninmathbb{N}$ with $n>1$ implies $x/n notin A.$
answered Jan 27 at 1:11
Reiner MartinReiner Martin
3,509414
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I will provide a proof for the continuous case. This result requires Baire Category Theorem. If $epsilon >0$ then $(0,1)=cup_n A_n$ where $A_k={x:|f(frac x n )| leq epsilon forall k geq n}$. Since $(0,1)$ is of second category it follows that there is some interval $(a,b)$ and some $n_0$ such that $|f(frac x n )| <epsilon$ whenever $a<x<b$ and $ n geq n_0$ $cdots (1)$. Now the length of the interval $(frac a x, frac b x)$ is greater than $1$ whenever $0<x<b-a$. Hence there is an integer $n$ in this interval. This integer is also greater than $n_0$ if $frac a x >n_0$ or $x <frac a {n_0}$. Replacing $x$ by $nx$ in (1) we get $|f(x)| <epsilon$ whenever $x$ is sufficiently small.
answered Jan 27 at 5:49
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
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2 Answers
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2 Answers
2
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$begingroup$
This is wrong. As a counterexample, take a function $f$ which is zero except at points $1/sqrt{p}$ for $p$ prime, where the function takes the value 1 instead.
More generally, define $f(x)=1$ for $xin A,$ and $f(x)=0$ for $xnotin A,$ where $A$ is any set which does not contain $0,$ has $0$ as an accumulation point, and has the property that $xin A$ and $ninmathbb{N}$ with $n>1$ implies $x/n notin A.$
answered Jan 27 at 1:11
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
This is wrong. As a counterexample, take a function $f$ which is zero except at points $1/sqrt{p}$ for $p$ prime, where the function takes the value 1 instead.
More generally, define $f(x)=1$ for $xin A,$ and $f(x)=0$ for $xnotin A,$ where $A$ is any set which does not contain $0,$ has $0$ as an accumulation point, and has the property that $xin A$ and $ninmathbb{N}$ with $n>1$ implies $x/n notin A.$
answered Jan 27 at 1:11
Reiner MartinReiner Martin
3,509414
$endgroup$
add a comment |
$begingroup$
This is wrong. As a counterexample, take a function $f$ which is zero except at points $1/sqrt{p}$ for $p$ prime, where the function takes the value 1 instead.
More generally, define $f(x)=1$ for $xin A,$ and $f(x)=0$ for $xnotin A,$ where $A$ is any set which does not contain $0,$ has $0$ as an accumulation point, and has the property that $xin A$ and $ninmathbb{N}$ with $n>1$ implies $x/n notin A.$
answered Jan 27 at 1:11
Reiner MartinReiner Martin
3,509414
$endgroup$
This is wrong. As a counterexample, take a function $f$ which is zero except at points $1/sqrt{p}$ for $p$ prime, where the function takes the value 1 instead.
More generally, define $f(x)=1$ for $xin A,$ and $f(x)=0$ for $xnotin A,$ where $A$ is any set which does not contain $0,$ has $0$ as an accumulation point, and has the property that $xin A$ and $ninmathbb{N}$ with $n>1$ implies $x/n notin A.$
answered Jan 27 at 1:11
Reiner MartinReiner Martin
3,509414
answered Jan 27 at 1:11
Reiner MartinReiner Martin
3,509414
answered Jan 27 at 1:11
Reiner MartinReiner Martin
3,509414
answered Jan 27 at 1:11
Reiner MartinReiner Martin
3,509414
3,509414
add a comment |
add a comment |
$begingroup$
I will provide a proof for the continuous case. This result requires Baire Category Theorem. If $epsilon >0$ then $(0,1)=cup_n A_n$ where $A_k={x:|f(frac x n )| leq epsilon forall k geq n}$. Since $(0,1)$ is of second category it follows that there is some interval $(a,b)$ and some $n_0$ such that $|f(frac x n )| <epsilon$ whenever $a<x<b$ and $ n geq n_0$ $cdots (1)$. Now the length of the interval $(frac a x, frac b x)$ is greater than $1$ whenever $0<x<b-a$. Hence there is an integer $n$ in this interval. This integer is also greater than $n_0$ if $frac a x >n_0$ or $x <frac a {n_0}$. Replacing $x$ by $nx$ in (1) we get $|f(x)| <epsilon$ whenever $x$ is sufficiently small.
answered Jan 27 at 5:49
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
$endgroup$
add a comment |
$begingroup$
I will provide a proof for the continuous case. This result requires Baire Category Theorem. If $epsilon >0$ then $(0,1)=cup_n A_n$ where $A_k={x:|f(frac x n )| leq epsilon forall k geq n}$. Since $(0,1)$ is of second category it follows that there is some interval $(a,b)$ and some $n_0$ such that $|f(frac x n )| <epsilon$ whenever $a<x<b$ and $ n geq n_0$ $cdots (1)$. Now the length of the interval $(frac a x, frac b x)$ is greater than $1$ whenever $0<x<b-a$. Hence there is an integer $n$ in this interval. This integer is also greater than $n_0$ if $frac a x >n_0$ or $x <frac a {n_0}$. Replacing $x$ by $nx$ in (1) we get $|f(x)| <epsilon$ whenever $x$ is sufficiently small.
answered Jan 27 at 5:49
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
$endgroup$
add a comment |
$begingroup$
I will provide a proof for the continuous case. This result requires Baire Category Theorem. If $epsilon >0$ then $(0,1)=cup_n A_n$ where $A_k={x:|f(frac x n )| leq epsilon forall k geq n}$. Since $(0,1)$ is of second category it follows that there is some interval $(a,b)$ and some $n_0$ such that $|f(frac x n )| <epsilon$ whenever $a<x<b$ and $ n geq n_0$ $cdots (1)$. Now the length of the interval $(frac a x, frac b x)$ is greater than $1$ whenever $0<x<b-a$. Hence there is an integer $n$ in this interval. This integer is also greater than $n_0$ if $frac a x >n_0$ or $x <frac a {n_0}$. Replacing $x$ by $nx$ in (1) we get $|f(x)| <epsilon$ whenever $x$ is sufficiently small.
answered Jan 27 at 5:49
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
$endgroup$
I will provide a proof for the continuous case. This result requires Baire Category Theorem. If $epsilon >0$ then $(0,1)=cup_n A_n$ where $A_k={x:|f(frac x n )| leq epsilon forall k geq n}$. Since $(0,1)$ is of second category it follows that there is some interval $(a,b)$ and some $n_0$ such that $|f(frac x n )| <epsilon$ whenever $a<x<b$ and $ n geq n_0$ $cdots (1)$. Now the length of the interval $(frac a x, frac b x)$ is greater than $1$ whenever $0<x<b-a$. Hence there is an integer $n$ in this interval. This integer is also greater than $n_0$ if $frac a x >n_0$ or $x <frac a {n_0}$. Replacing $x$ by $nx$ in (1) we get $|f(x)| <epsilon$ whenever $x$ is sufficiently small.
answered Jan 27 at 5:49
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
edited Jan 27 at 5:54
answered Jan 27 at 5:49
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
answered Jan 27 at 5:49
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
answered Jan 27 at 5:49
Kavi Rama MurthyKavi Rama Murthy
69.3k53169
69.3k53169
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