Mixing
Is the function uniformly continuous? .
$f_n(x)= dfrac{nx}{1+n^4x^4}$ , $x∈mathbb{R}$
I fixed $varepsilon >0$, looking for $delta >0$, so for every $x, y ∈ mho$ with $|x-y|<delta$ it is valid $|f(x)-f(y)|<epsilon$.
So: $left|dfrac{nx}{1+n^4x^4}-dfrac{ny}{1+n^4y^4}right|=left|dfrac{n(x-y)+n^5xy(y^3-x^3)}{(1+n^4x^4)(1+n^4y^4)}right|$
I don't know how can I go on to find a $delta$, is something wrong above?
real-analysis uniform-continuity
edited Nov 21 '18 at 19:20
Yadati Kiran
1,692619
asked Nov 21 '18 at 19:16
DadaDada
7010
|
show 2 more comments
$f_n(x)= dfrac{nx}{1+n^4x^4}$ , $x∈mathbb{R}$
I fixed $varepsilon >0$, looking for $delta >0$, so for every $x, y ∈ mho$ with $|x-y|<delta$ it is valid $|f(x)-f(y)|<epsilon$.
So: $left|dfrac{nx}{1+n^4x^4}-dfrac{ny}{1+n^4y^4}right|=left|dfrac{n(x-y)+n^5xy(y^3-x^3)}{(1+n^4x^4)(1+n^4y^4)}right|$
I don't know how can I go on to find a $delta$, is something wrong above?
real-analysis uniform-continuity
edited Nov 21 '18 at 19:20
Yadati Kiran
1,692619
asked Nov 21 '18 at 19:16
DadaDada
7010
Try to check that the derivative is bounded
– user8268
Nov 21 '18 at 19:19
Consider $displaystyle lim_{nrightarrowinfty}f_n(x)=lim_{nrightarrowinfty} dfrac{dfrac{x}{n^3}}{dfrac{1}{n^4}+x^4}$. Does the limit depend on $x$?
– Yadati Kiran
Nov 21 '18 at 19:22
@YadatiKiran what does this have to do with the question?
– Federico
Nov 21 '18 at 19:24
@Federico: If the $displaystyle lim_{nrightarrowinfty}f_n(x)$ is independent of $x$, then $f_n$ converges uniformly.
– Yadati Kiran
Nov 21 '18 at 19:25
But to my understanding the question is whether $f_n$ is uniformly continuous, not if $f_n$ converges uniformly to some $f$.
– Federico
Nov 21 '18 at 19:29
|
show 2 more comments
$f_n(x)= dfrac{nx}{1+n^4x^4}$ , $x∈mathbb{R}$
I fixed $varepsilon >0$, looking for $delta >0$, so for every $x, y ∈ mho$ with $|x-y|<delta$ it is valid $|f(x)-f(y)|<epsilon$.
So: $left|dfrac{nx}{1+n^4x^4}-dfrac{ny}{1+n^4y^4}right|=left|dfrac{n(x-y)+n^5xy(y^3-x^3)}{(1+n^4x^4)(1+n^4y^4)}right|$
I don't know how can I go on to find a $delta$, is something wrong above?
real-analysis uniform-continuity
edited Nov 21 '18 at 19:20
Yadati Kiran
1,692619
asked Nov 21 '18 at 19:16
DadaDada
7010
$f_n(x)= dfrac{nx}{1+n^4x^4}$ , $x∈mathbb{R}$
I fixed $varepsilon >0$, looking for $delta >0$, so for every $x, y ∈ mho$ with $|x-y|<delta$ it is valid $|f(x)-f(y)|<epsilon$.
So: $left|dfrac{nx}{1+n^4x^4}-dfrac{ny}{1+n^4y^4}right|=left|dfrac{n(x-y)+n^5xy(y^3-x^3)}{(1+n^4x^4)(1+n^4y^4)}right|$
I don't know how can I go on to find a $delta$, is something wrong above?
real-analysis uniform-continuity
real-analysis uniform-continuity
edited Nov 21 '18 at 19:20
Yadati Kiran
1,692619
asked Nov 21 '18 at 19:16
DadaDada
7010
edited Nov 21 '18 at 19:20
Yadati Kiran
1,692619
asked Nov 21 '18 at 19:16
DadaDada
7010
edited Nov 21 '18 at 19:20
Yadati Kiran
1,692619
edited Nov 21 '18 at 19:20
Yadati Kiran
1,692619
edited Nov 21 '18 at 19:20
Yadati Kiran
1,692619
1,692619
asked Nov 21 '18 at 19:16
DadaDada
7010
asked Nov 21 '18 at 19:16
DadaDada
7010
asked Nov 21 '18 at 19:16
DadaDada
7010
7010
Try to check that the derivative is bounded
– user8268
Nov 21 '18 at 19:19
Consider $displaystyle lim_{nrightarrowinfty}f_n(x)=lim_{nrightarrowinfty} dfrac{dfrac{x}{n^3}}{dfrac{1}{n^4}+x^4}$. Does the limit depend on $x$?
– Yadati Kiran
Nov 21 '18 at 19:22
@YadatiKiran what does this have to do with the question?
– Federico
Nov 21 '18 at 19:24
@Federico: If the $displaystyle lim_{nrightarrowinfty}f_n(x)$ is independent of $x$, then $f_n$ converges uniformly.
– Yadati Kiran
Nov 21 '18 at 19:25
But to my understanding the question is whether $f_n$ is uniformly continuous, not if $f_n$ converges uniformly to some $f$.
– Federico
Nov 21 '18 at 19:29
|
show 2 more comments
Try to check that the derivative is bounded
– user8268
Nov 21 '18 at 19:19
Consider $displaystyle lim_{nrightarrowinfty}f_n(x)=lim_{nrightarrowinfty} dfrac{dfrac{x}{n^3}}{dfrac{1}{n^4}+x^4}$. Does the limit depend on $x$?
– Yadati Kiran
Nov 21 '18 at 19:22
@YadatiKiran what does this have to do with the question?
– Federico
Nov 21 '18 at 19:24
@Federico: If the $displaystyle lim_{nrightarrowinfty}f_n(x)$ is independent of $x$, then $f_n$ converges uniformly.
– Yadati Kiran
Nov 21 '18 at 19:25
But to my understanding the question is whether $f_n$ is uniformly continuous, not if $f_n$ converges uniformly to some $f$.
– Federico
Nov 21 '18 at 19:29
Try to check that the derivative is bounded
– user8268
Nov 21 '18 at 19:19
Try to check that the derivative is bounded
– user8268
Nov 21 '18 at 19:19
Consider $displaystyle lim_{nrightarrowinfty}f_n(x)=lim_{nrightarrowinfty} dfrac{dfrac{x}{n^3}}{dfrac{1}{n^4}+x^4}$. Does the limit depend on $x$?
– Yadati Kiran
Nov 21 '18 at 19:22
Consider $displaystyle lim_{nrightarrowinfty}f_n(x)=lim_{nrightarrowinfty} dfrac{dfrac{x}{n^3}}{dfrac{1}{n^4}+x^4}$. Does the limit depend on $x$?
– Yadati Kiran
Nov 21 '18 at 19:22
@YadatiKiran what does this have to do with the question?
– Federico
Nov 21 '18 at 19:24
@YadatiKiran what does this have to do with the question?
– Federico
Nov 21 '18 at 19:24
@Federico: If the $displaystyle lim_{nrightarrowinfty}f_n(x)$ is independent of $x$, then $f_n$ converges uniformly.
– Yadati Kiran
Nov 21 '18 at 19:25
@Federico: If the $displaystyle lim_{nrightarrowinfty}f_n(x)$ is independent of $x$, then $f_n$ converges uniformly.
– Yadati Kiran
Nov 21 '18 at 19:25
But to my understanding the question is whether $f_n$ is uniformly continuous, not if $f_n$ converges uniformly to some $f$.
– Federico
Nov 21 '18 at 19:29
But to my understanding the question is whether $f_n$ is uniformly continuous, not if $f_n$ converges uniformly to some $f$.
– Federico
Nov 21 '18 at 19:29
|
show 2 more comments
2 Answers
2
active
oldest
votes
The derivative of $f_1$ is bounded: you can prove that $|f'_1|_infty=1$; so it is $1$-Lipschitz. Then $f_n(x)=f_1(nx)$ is $n$-Lipschitz.
answered Nov 21 '18 at 19:21
FedericoFederico
4,829514
add a comment |
You should find an upper bound on $|f(x)-f(y)|$ in terms of $d=y-x$. Since $y^3-x^3=(y-x) (y^2-3xy+x^2)$, we can rewrite $|f(x)-f(y)|$ as $dleft|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|$. So now if you can prove there is some constant number $M$ such that $left|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|<M$ for all $x$, $y$, you can take $delta=epsilon/M$. Then if $|d|=|y-x|<epsilon/M$, you have $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.
answered Nov 21 '18 at 19:45
AcccumulationAcccumulation
6,8162618
add a comment |
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2 Answers
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2 Answers
2
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The derivative of $f_1$ is bounded: you can prove that $|f'_1|_infty=1$; so it is $1$-Lipschitz. Then $f_n(x)=f_1(nx)$ is $n$-Lipschitz.
answered Nov 21 '18 at 19:21
FedericoFederico
4,829514
add a comment |
The derivative of $f_1$ is bounded: you can prove that $|f'_1|_infty=1$; so it is $1$-Lipschitz. Then $f_n(x)=f_1(nx)$ is $n$-Lipschitz.
answered Nov 21 '18 at 19:21
FedericoFederico
4,829514
add a comment |
The derivative of $f_1$ is bounded: you can prove that $|f'_1|_infty=1$; so it is $1$-Lipschitz. Then $f_n(x)=f_1(nx)$ is $n$-Lipschitz.
answered Nov 21 '18 at 19:21
FedericoFederico
4,829514
The derivative of $f_1$ is bounded: you can prove that $|f'_1|_infty=1$; so it is $1$-Lipschitz. Then $f_n(x)=f_1(nx)$ is $n$-Lipschitz.
answered Nov 21 '18 at 19:21
FedericoFederico
4,829514
answered Nov 21 '18 at 19:21
FedericoFederico
4,829514
answered Nov 21 '18 at 19:21
FedericoFederico
4,829514
answered Nov 21 '18 at 19:21
FedericoFederico
4,829514
4,829514
add a comment |
add a comment |
You should find an upper bound on $|f(x)-f(y)|$ in terms of $d=y-x$. Since $y^3-x^3=(y-x) (y^2-3xy+x^2)$, we can rewrite $|f(x)-f(y)|$ as $dleft|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|$. So now if you can prove there is some constant number $M$ such that $left|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|<M$ for all $x$, $y$, you can take $delta=epsilon/M$. Then if $|d|=|y-x|<epsilon/M$, you have $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.
answered Nov 21 '18 at 19:45
AcccumulationAcccumulation
6,8162618
add a comment |
You should find an upper bound on $|f(x)-f(y)|$ in terms of $d=y-x$. Since $y^3-x^3=(y-x) (y^2-3xy+x^2)$, we can rewrite $|f(x)-f(y)|$ as $dleft|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|$. So now if you can prove there is some constant number $M$ such that $left|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|<M$ for all $x$, $y$, you can take $delta=epsilon/M$. Then if $|d|=|y-x|<epsilon/M$, you have $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.
answered Nov 21 '18 at 19:45
AcccumulationAcccumulation
6,8162618
add a comment |
You should find an upper bound on $|f(x)-f(y)|$ in terms of $d=y-x$. Since $y^3-x^3=(y-x) (y^2-3xy+x^2)$, we can rewrite $|f(x)-f(y)|$ as $dleft|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|$. So now if you can prove there is some constant number $M$ such that $left|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|<M$ for all $x$, $y$, you can take $delta=epsilon/M$. Then if $|d|=|y-x|<epsilon/M$, you have $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.
answered Nov 21 '18 at 19:45
AcccumulationAcccumulation
6,8162618
You should find an upper bound on $|f(x)-f(y)|$ in terms of $d=y-x$. Since $y^3-x^3=(y-x) (y^2-3xy+x^2)$, we can rewrite $|f(x)-f(y)|$ as $dleft|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|$. So now if you can prove there is some constant number $M$ such that $left|dfrac{n+n^5xy(y^2-3xy+x^2)}{(1+n^4x^4)(1+n^4y^4)}right|<M$ for all $x$, $y$, you can take $delta=epsilon/M$. Then if $|d|=|y-x|<epsilon/M$, you have $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.
answered Nov 21 '18 at 19:45
AcccumulationAcccumulation
6,8162618
answered Nov 21 '18 at 19:45
AcccumulationAcccumulation
6,8162618
answered Nov 21 '18 at 19:45
AcccumulationAcccumulation
6,8162618
answered Nov 21 '18 at 19:45
AcccumulationAcccumulation
6,8162618
6,8162618
add a comment |
add a comment |
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Try to check that the derivative is bounded
– user8268
Nov 21 '18 at 19:19
Consider $displaystyle lim_{nrightarrowinfty}f_n(x)=lim_{nrightarrowinfty} dfrac{dfrac{x}{n^3}}{dfrac{1}{n^4}+x^4}$. Does the limit depend on $x$?
– Yadati Kiran
Nov 21 '18 at 19:22
@YadatiKiran what does this have to do with the question?
– Federico
Nov 21 '18 at 19:24
@Federico: If the $displaystyle lim_{nrightarrowinfty}f_n(x)$ is independent of $x$, then $f_n$ converges uniformly.
– Yadati Kiran
Nov 21 '18 at 19:25
But to my understanding the question is whether $f_n$ is uniformly continuous, not if $f_n$ converges uniformly to some $f$.
– Federico
Nov 21 '18 at 19:29