How to prove that $g(x)equiv f(x)$?

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Suppose that $f,g$ are periodic functions, and $limlimits_xto+infty [f(x)-g(x)]=0$, prove that $f(x)equiv g(x)$.

I've solved it if we add the condition that $f(x)$ is continuous.

But the continuity condition is un-needed.

==============
additional remarks =================

I know the fact that

$limlimits_xto+infty t(x)=0$ implies $forall x_nsubset Dom(t)$, if $limlimits_ntoinfty x_n=+infty$, then $limlimits_ntoinfty t(x_n)=0$,

which is called Heine Theorem in my textbook.

But I don't think you can put two different sequence here even both of them turn to infinity.

calculus

share|cite|improve this question

edited Jul 17 at 16:44

asked Jul 17 at 4:13

闫嘉琦

41219

add a comment | 

up vote
2
down vote

favorite

1

Suppose that $f,g$ are periodic functions, and $limlimits_xto+infty [f(x)-g(x)]=0$, prove that $f(x)equiv g(x)$.

I've solved it if we add the condition that $f(x)$ is continuous.

But the continuity condition is un-needed.

==============
additional remarks =================

I know the fact that

$limlimits_xto+infty t(x)=0$ implies $forall x_nsubset Dom(t)$, if $limlimits_ntoinfty x_n=+infty$, then $limlimits_ntoinfty t(x_n)=0$,

which is called Heine Theorem in my textbook.

But I don't think you can put two different sequence here even both of them turn to infinity.

calculus

share|cite|improve this question

edited Jul 17 at 16:44

asked Jul 17 at 4:13

闫嘉琦

41219

add a comment | 

up vote
2
down vote

favorite

1

up vote
2
down vote

favorite

1

1

Suppose that $f,g$ are periodic functions, and $limlimits_xto+infty [f(x)-g(x)]=0$, prove that $f(x)equiv g(x)$.

I've solved it if we add the condition that $f(x)$ is continuous.

But the continuity condition is un-needed.

==============
additional remarks =================

I know the fact that

$limlimits_xto+infty t(x)=0$ implies $forall x_nsubset Dom(t)$, if $limlimits_ntoinfty x_n=+infty$, then $limlimits_ntoinfty t(x_n)=0$,

which is called Heine Theorem in my textbook.

But I don't think you can put two different sequence here even both of them turn to infinity.

calculus

share|cite|improve this question

edited Jul 17 at 16:44

asked Jul 17 at 4:13

闫嘉琦

41219

Suppose that $f,g$ are periodic functions, and $limlimits_xto+infty [f(x)-g(x)]=0$, prove that $f(x)equiv g(x)$.

I've solved it if we add the condition that $f(x)$ is continuous.

But the continuity condition is un-needed.

==============
additional remarks =================

I know the fact that

$limlimits_xto+infty t(x)=0$ implies $forall x_nsubset Dom(t)$, if $limlimits_ntoinfty x_n=+infty$, then $limlimits_ntoinfty t(x_n)=0$,

which is called Heine Theorem in my textbook.

But I don't think you can put two different sequence here even both of them turn to infinity.

calculus

share|cite|improve this question

edited Jul 17 at 16:44

asked Jul 17 at 4:13

闫嘉琦

41219

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 16:44

edited Jul 17 at 16:44

edited Jul 17 at 16:44

asked Jul 17 at 4:13

闫嘉琦

41219

asked Jul 17 at 4:13

闫嘉琦

41219

asked Jul 17 at 4:13

闫嘉琦

41219

41219

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
6
down vote

The idea is:

Let $p_1,p_2$ the periods of $f$ and $g$ respectively (suppose that are positive, in another case rewrite the proof). Now we have that $f(x + np_1) = f(x)$ and $g(x + np_2)= g(x)$ for all $n in mathbbN$. From hypothesis, for each fixed $y in mathbbR$ we have $lim_n to +infty [f(y+ np_1) - g(y + np_2)] = 0$. By definition of limit in sequences, the last means that for every $epsilon > 0$ there exists a great $n$ such that
$$|f(y) - g(y)| = |f(y + np_1) - g(y + np_2)| < epsilon$$
Then $|f(y) - g(y)|=0$, and this implies $f(y) = g(y)$.

share|cite|improve this answer

edited Jul 17 at 4:36

answered Jul 17 at 4:28

Inzit

935412

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The $<$ doesn't hold becase generally $y+np_1=y+np_2$. And if $dfracp_1 p_2$ is not rational, we cannot find any $m,ninmathbfN_+$ such that ￥、$y+mp_1=y+np_2$.
  – é—«å˜‰ç¦
  Jul 17 at 5:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @闫嘉琦 The answer does not require $y+mp_1$ and $y+np_2$ are same. The answer just uses the definition of limit.
  – Hanul Jeon
  Jul 17 at 6:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HanulJeon Acutually I don't understand why $limlimits_ntoinfty [f(y+np_1)-g(y+np_2)]=0$ holds
  – é—«å˜‰ç¦
  Jul 17 at 6:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @闫嘉琦 Inzit use the fact that $lim_xtoinftyf(x)$ exists if and only if every sequence $(x_n)_nin Bbb N$ if for all $n$ we have $x_ninmboxDom(f)$ and $lim_ntoinfty x_n=infty$ then $lim_ntoinftyf(x_n)=lim_xtoinftyf(x)$, now define $h=f-g$ and use this fact to see how this limit holds
  – Holo
  Jul 17 at 10:21
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oops, seems I missed something reading the OP's answer.As you discussed already the OP is comparing 2 different points:$ y_1= y+np_1 not = y +np_2=y_2$.Taking limit as n goes to $infty$ one cannot use limit $ |f(y)-g(y)| $ as y goes to $infty$(same point y).If one has $n_1p_1=n_2p_2 =T$ as common period T, then comparing $|f(y+nT)-g(y+nT)| $as n goes to $infty,$ for every y, seems fine.Otherwise I would not know.
  – Peter Szilas
  Jul 17 at 18:25
  
  

  
  
  
  

 | 
show 5 more comments

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1 Answer
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oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
6
down vote

The idea is:

Let $p_1,p_2$ the periods of $f$ and $g$ respectively (suppose that are positive, in another case rewrite the proof). Now we have that $f(x + np_1) = f(x)$ and $g(x + np_2)= g(x)$ for all $n in mathbbN$. From hypothesis, for each fixed $y in mathbbR$ we have $lim_n to +infty [f(y+ np_1) - g(y + np_2)] = 0$. By definition of limit in sequences, the last means that for every $epsilon > 0$ there exists a great $n$ such that
$$|f(y) - g(y)| = |f(y + np_1) - g(y + np_2)| < epsilon$$
Then $|f(y) - g(y)|=0$, and this implies $f(y) = g(y)$.

share|cite|improve this answer

edited Jul 17 at 4:36

answered Jul 17 at 4:28

Inzit

935412

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The $<$ doesn't hold becase generally $y+np_1=y+np_2$. And if $dfracp_1 p_2$ is not rational, we cannot find any $m,ninmathbfN_+$ such that ￥、$y+mp_1=y+np_2$.
  – é—«å˜‰ç¦
  Jul 17 at 5:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @闫嘉琦 The answer does not require $y+mp_1$ and $y+np_2$ are same. The answer just uses the definition of limit.
  – Hanul Jeon
  Jul 17 at 6:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HanulJeon Acutually I don't understand why $limlimits_ntoinfty [f(y+np_1)-g(y+np_2)]=0$ holds
  – é—«å˜‰ç¦
  Jul 17 at 6:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @闫嘉琦 Inzit use the fact that $lim_xtoinftyf(x)$ exists if and only if every sequence $(x_n)_nin Bbb N$ if for all $n$ we have $x_ninmboxDom(f)$ and $lim_ntoinfty x_n=infty$ then $lim_ntoinftyf(x_n)=lim_xtoinftyf(x)$, now define $h=f-g$ and use this fact to see how this limit holds
  – Holo
  Jul 17 at 10:21
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oops, seems I missed something reading the OP's answer.As you discussed already the OP is comparing 2 different points:$ y_1= y+np_1 not = y +np_2=y_2$.Taking limit as n goes to $infty$ one cannot use limit $ |f(y)-g(y)| $ as y goes to $infty$(same point y).If one has $n_1p_1=n_2p_2 =T$ as common period T, then comparing $|f(y+nT)-g(y+nT)| $as n goes to $infty,$ for every y, seems fine.Otherwise I would not know.
  – Peter Szilas
  Jul 17 at 18:25
  
  

  
  
  
  

 | 
show 5 more comments

up vote
6
down vote

The idea is:

Let $p_1,p_2$ the periods of $f$ and $g$ respectively (suppose that are positive, in another case rewrite the proof). Now we have that $f(x + np_1) = f(x)$ and $g(x + np_2)= g(x)$ for all $n in mathbbN$. From hypothesis, for each fixed $y in mathbbR$ we have $lim_n to +infty [f(y+ np_1) - g(y + np_2)] = 0$. By definition of limit in sequences, the last means that for every $epsilon > 0$ there exists a great $n$ such that
$$|f(y) - g(y)| = |f(y + np_1) - g(y + np_2)| < epsilon$$
Then $|f(y) - g(y)|=0$, and this implies $f(y) = g(y)$.

share|cite|improve this answer

edited Jul 17 at 4:36

answered Jul 17 at 4:28

Inzit

935412

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The $<$ doesn't hold becase generally $y+np_1=y+np_2$. And if $dfracp_1 p_2$ is not rational, we cannot find any $m,ninmathbfN_+$ such that ￥、$y+mp_1=y+np_2$.
  – é—«å˜‰ç¦
  Jul 17 at 5:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @闫嘉琦 The answer does not require $y+mp_1$ and $y+np_2$ are same. The answer just uses the definition of limit.
  – Hanul Jeon
  Jul 17 at 6:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HanulJeon Acutually I don't understand why $limlimits_ntoinfty [f(y+np_1)-g(y+np_2)]=0$ holds
  – é—«å˜‰ç¦
  Jul 17 at 6:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @闫嘉琦 Inzit use the fact that $lim_xtoinftyf(x)$ exists if and only if every sequence $(x_n)_nin Bbb N$ if for all $n$ we have $x_ninmboxDom(f)$ and $lim_ntoinfty x_n=infty$ then $lim_ntoinftyf(x_n)=lim_xtoinftyf(x)$, now define $h=f-g$ and use this fact to see how this limit holds
  – Holo
  Jul 17 at 10:21
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oops, seems I missed something reading the OP's answer.As you discussed already the OP is comparing 2 different points:$ y_1= y+np_1 not = y +np_2=y_2$.Taking limit as n goes to $infty$ one cannot use limit $ |f(y)-g(y)| $ as y goes to $infty$(same point y).If one has $n_1p_1=n_2p_2 =T$ as common period T, then comparing $|f(y+nT)-g(y+nT)| $as n goes to $infty,$ for every y, seems fine.Otherwise I would not know.
  – Peter Szilas
  Jul 17 at 18:25
  
  

  
  
  
  

 | 
show 5 more comments

up vote
6
down vote

up vote
6
down vote

The idea is:

Let $p_1,p_2$ the periods of $f$ and $g$ respectively (suppose that are positive, in another case rewrite the proof). Now we have that $f(x + np_1) = f(x)$ and $g(x + np_2)= g(x)$ for all $n in mathbbN$. From hypothesis, for each fixed $y in mathbbR$ we have $lim_n to +infty [f(y+ np_1) - g(y + np_2)] = 0$. By definition of limit in sequences, the last means that for every $epsilon > 0$ there exists a great $n$ such that
$$|f(y) - g(y)| = |f(y + np_1) - g(y + np_2)| < epsilon$$
Then $|f(y) - g(y)|=0$, and this implies $f(y) = g(y)$.

share|cite|improve this answer

edited Jul 17 at 4:36

answered Jul 17 at 4:28

Inzit

935412

The idea is:

Let $p_1,p_2$ the periods of $f$ and $g$ respectively (suppose that are positive, in another case rewrite the proof). Now we have that $f(x + np_1) = f(x)$ and $g(x + np_2)= g(x)$ for all $n in mathbbN$. From hypothesis, for each fixed $y in mathbbR$ we have $lim_n to +infty [f(y+ np_1) - g(y + np_2)] = 0$. By definition of limit in sequences, the last means that for every $epsilon > 0$ there exists a great $n$ such that
$$|f(y) - g(y)| = |f(y + np_1) - g(y + np_2)| < epsilon$$
Then $|f(y) - g(y)|=0$, and this implies $f(y) = g(y)$.

share|cite|improve this answer

edited Jul 17 at 4:36

answered Jul 17 at 4:28

Inzit

935412

share|cite|improve this answer

share|cite|improve this answer

edited Jul 17 at 4:36

edited Jul 17 at 4:36

edited Jul 17 at 4:36

answered Jul 17 at 4:28

Inzit

935412

answered Jul 17 at 4:28

Inzit

935412

answered Jul 17 at 4:28

Inzit

935412

935412

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The $<$ doesn't hold becase generally $y+np_1=y+np_2$. And if $dfracp_1 p_2$ is not rational, we cannot find any $m,ninmathbfN_+$ such that ￥、$y+mp_1=y+np_2$.
  – é—«å˜‰ç¦
  Jul 17 at 5:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @闫嘉琦 The answer does not require $y+mp_1$ and $y+np_2$ are same. The answer just uses the definition of limit.
  – Hanul Jeon
  Jul 17 at 6:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HanulJeon Acutually I don't understand why $limlimits_ntoinfty [f(y+np_1)-g(y+np_2)]=0$ holds
  – é—«å˜‰ç¦
  Jul 17 at 6:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @闫嘉琦 Inzit use the fact that $lim_xtoinftyf(x)$ exists if and only if every sequence $(x_n)_nin Bbb N$ if for all $n$ we have $x_ninmboxDom(f)$ and $lim_ntoinfty x_n=infty$ then $lim_ntoinftyf(x_n)=lim_xtoinftyf(x)$, now define $h=f-g$ and use this fact to see how this limit holds
  – Holo
  Jul 17 at 10:21
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oops, seems I missed something reading the OP's answer.As you discussed already the OP is comparing 2 different points:$ y_1= y+np_1 not = y +np_2=y_2$.Taking limit as n goes to $infty$ one cannot use limit $ |f(y)-g(y)| $ as y goes to $infty$(same point y).If one has $n_1p_1=n_2p_2 =T$ as common period T, then comparing $|f(y+nT)-g(y+nT)| $as n goes to $infty,$ for every y, seems fine.Otherwise I would not know.
  – Peter Szilas
  Jul 17 at 18:25
  
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The $<$ doesn't hold becase generally $y+np_1=y+np_2$. And if $dfracp_1 p_2$ is not rational, we cannot find any $m,ninmathbfN_+$ such that ￥、$y+mp_1=y+np_2$.
  – é—«å˜‰ç¦
  Jul 17 at 5:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @闫嘉琦 The answer does not require $y+mp_1$ and $y+np_2$ are same. The answer just uses the definition of limit.
  – Hanul Jeon
  Jul 17 at 6:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @HanulJeon Acutually I don't understand why $limlimits_ntoinfty [f(y+np_1)-g(y+np_2)]=0$ holds
  – é—«å˜‰ç¦
  Jul 17 at 6:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @闫嘉琦 Inzit use the fact that $lim_xtoinftyf(x)$ exists if and only if every sequence $(x_n)_nin Bbb N$ if for all $n$ we have $x_ninmboxDom(f)$ and $lim_ntoinfty x_n=infty$ then $lim_ntoinftyf(x_n)=lim_xtoinftyf(x)$, now define $h=f-g$ and use this fact to see how this limit holds
  – Holo
  Jul 17 at 10:21
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Oops, seems I missed something reading the OP's answer.As you discussed already the OP is comparing 2 different points:$ y_1= y+np_1 not = y +np_2=y_2$.Taking limit as n goes to $infty$ one cannot use limit $ |f(y)-g(y)| $ as y goes to $infty$(same point y).If one has $n_1p_1=n_2p_2 =T$ as common period T, then comparing $|f(y+nT)-g(y+nT)| $as n goes to $infty,$ for every y, seems fine.Otherwise I would not know.
  – Peter Szilas
  Jul 17 at 18:25
  
  

  
  
  
  

The $<$ doesn't hold becase generally $y+np_1=y+np_2$. And if $dfracp_1 p_2$ is not rational, we cannot find any $m,ninmathbfN_+$ such that ￥、$y+mp_1=y+np_2$.
– é—«å˜‰ç¦
Jul 17 at 5:07

The $<$ doesn't hold becase generally $y+np_1=y+np_2$. And if $dfracp_1 p_2$ is not rational, we cannot find any $m,ninmathbfN_+$ such that ￥、$y+mp_1=y+np_2$.
– é—«å˜‰ç¦
Jul 17 at 5:07

@闫嘉琦 The answer does not require $y+mp_1$ and $y+np_2$ are same. The answer just uses the definition of limit.
– Hanul Jeon
Jul 17 at 6:00

@闫嘉琦 The answer does not require $y+mp_1$ and $y+np_2$ are same. The answer just uses the definition of limit.
– Hanul Jeon
Jul 17 at 6:00

@HanulJeon Acutually I don't understand why $limlimits_ntoinfty [f(y+np_1)-g(y+np_2)]=0$ holds
– é—«å˜‰ç¦
Jul 17 at 6:22

@HanulJeon Acutually I don't understand why $limlimits_ntoinfty [f(y+np_1)-g(y+np_2)]=0$ holds
– é—«å˜‰ç¦
Jul 17 at 6:22

1

1

@闫嘉琦 Inzit use the fact that $lim_xtoinftyf(x)$ exists if and only if every sequence $(x_n)_nin Bbb N$ if for all $n$ we have $x_ninmboxDom(f)$ and $lim_ntoinfty x_n=infty$ then $lim_ntoinftyf(x_n)=lim_xtoinftyf(x)$, now define $h=f-g$ and use this fact to see how this limit holds
– Holo
Jul 17 at 10:21

@闫嘉琦 Inzit use the fact that $lim_xtoinftyf(x)$ exists if and only if every sequence $(x_n)_nin Bbb N$ if for all $n$ we have $x_ninmboxDom(f)$ and $lim_ntoinfty x_n=infty$ then $lim_ntoinftyf(x_n)=lim_xtoinftyf(x)$, now define $h=f-g$ and use this fact to see how this limit holds
– Holo
Jul 17 at 10:21

2

2

Oops, seems I missed something reading the OP's answer.As you discussed already the OP is comparing 2 different points:$ y_1= y+np_1 not = y +np_2=y_2$.Taking limit as n goes to $infty$ one cannot use limit $ |f(y)-g(y)| $ as y goes to $infty$(same point y).If one has $n_1p_1=n_2p_2 =T$ as common period T, then comparing $|f(y+nT)-g(y+nT)| $as n goes to $infty,$ for every y, seems fine.Otherwise I would not know.
– Peter Szilas
Jul 17 at 18:25

Oops, seems I missed something reading the OP's answer.As you discussed already the OP is comparing 2 different points:$ y_1= y+np_1 not = y +np_2=y_2$.Taking limit as n goes to $infty$ one cannot use limit $ |f(y)-g(y)| $ as y goes to $infty$(same point y).If one has $n_1p_1=n_2p_2 =T$ as common period T, then comparing $|f(y+nT)-g(y+nT)| $as n goes to $infty,$ for every y, seems fine.Otherwise I would not know.
– Peter Szilas
Jul 17 at 18:25

 | 
show 5 more comments

 

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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

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StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

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Sign up using Google

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Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email