Is the function uniformly continuous? .












0















$f_n(x)= dfrac{nx}{1+n^4x^4}$ , $x∈mathbb{R}$




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







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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
















0















$f_n(x)= dfrac{nx}{1+n^4x^4}$ , $x∈mathbb{R}$




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







share|cite|improve this question
























  • 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














0












0








0








$f_n(x)= dfrac{nx}{1+n^4x^4}$ , $x∈mathbb{R}$




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







share|cite|improve this question
















$f_n(x)= dfrac{nx}{1+n^4x^4}$ , $x∈mathbb{R}$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 21 '18 at 19:20









Yadati Kiran

1,692619




1,692619










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


















  • 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










2 Answers
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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.






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    0














    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 $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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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.






      share|cite|improve this answer


























        1














        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.






        share|cite|improve this answer
























          1












          1








          1






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 21 '18 at 19:21









          FedericoFederico

          4,829514




          4,829514























              0














              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 $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.






              share|cite|improve this answer


























                0














                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 $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  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 $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.






                  share|cite|improve this answer












                  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 $|f(y)-f(x)|<dM=frac{epsilon M}M=epsilon$.







                  share|cite|improve this answer












                  share|cite|improve this answer



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                  answered Nov 21 '18 at 19:45









                  AcccumulationAcccumulation

                  6,8162618




                  6,8162618






























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