the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly Choose the correct option












1














let $f :mathbb{R} rightarrow mathbb{R}$ be non zero function such that $|f(x)| le frac{1}{1+2x^2}$ for all $x in mathbb{R}$ .Define real value function $f_n$ on $mathbb{R}$ for all $n in mathbb{N}$ by $f_n(x) = f(x +n)$.



Then ,the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly



Choose the correct option



$a)$ on $[0,1]$



$b)$ on $[-1,0]$



$c)$ on $[-1,1]$



$d)$ None of these



I thinks option $a) ,b)$ and $c)$ will correct because derivative of $f$ is bounded



Is its True??






real-analysis







share|cite|improve this question






















  • See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
    – Yadati Kiran
    Nov 21 '18 at 1:04


















1














let $f :mathbb{R} rightarrow mathbb{R}$ be non zero function such that $|f(x)| le frac{1}{1+2x^2}$ for all $x in mathbb{R}$ .Define real value function $f_n$ on $mathbb{R}$ for all $n in mathbb{N}$ by $f_n(x) = f(x +n)$.



Then ,the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly



Choose the correct option



$a)$ on $[0,1]$



$b)$ on $[-1,0]$



$c)$ on $[-1,1]$



$d)$ None of these



I thinks option $a) ,b)$ and $c)$ will correct because derivative of $f$ is bounded



Is its True??






real-analysis







share|cite|improve this question






















  • See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
    – Yadati Kiran
    Nov 21 '18 at 1:04
















1












1








1







let $f :mathbb{R} rightarrow mathbb{R}$ be non zero function such that $|f(x)| le frac{1}{1+2x^2}$ for all $x in mathbb{R}$ .Define real value function $f_n$ on $mathbb{R}$ for all $n in mathbb{N}$ by $f_n(x) = f(x +n)$.



Then ,the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly



Choose the correct option



$a)$ on $[0,1]$



$b)$ on $[-1,0]$



$c)$ on $[-1,1]$



$d)$ None of these



I thinks option $a) ,b)$ and $c)$ will correct because derivative of $f$ is bounded



Is its True??






real-analysis







share|cite|improve this question













let $f :mathbb{R} rightarrow mathbb{R}$ be non zero function such that $|f(x)| le frac{1}{1+2x^2}$ for all $x in mathbb{R}$ .Define real value function $f_n$ on $mathbb{R}$ for all $n in mathbb{N}$ by $f_n(x) = f(x +n)$.



Then ,the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly



Choose the correct option



$a)$ on $[0,1]$



$b)$ on $[-1,0]$



$c)$ on $[-1,1]$



$d)$ None of these



I thinks option $a) ,b)$ and $c)$ will correct because derivative of $f$ is bounded



Is its True??






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 21 '18 at 0:51









Messi fifa

51611




51611












  • See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
    – Yadati Kiran
    Nov 21 '18 at 1:04




















  • See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
    – Yadati Kiran
    Nov 21 '18 at 1:04


















See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
– Yadati Kiran
Nov 21 '18 at 1:04






See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
– Yadati Kiran
Nov 21 '18 at 1:04












1 Answer
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oldest

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HINT:



We are not given that $f$ is differentiable.



So, note that for $xin [-1,1]$



$$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$



Appeal to the Weierstrass M-Test. Can you finish?






share|cite|improve this answer





















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    HINT:



    We are not given that $f$ is differentiable.



    So, note that for $xin [-1,1]$



    $$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$



    Appeal to the Weierstrass M-Test. Can you finish?






    share|cite|improve this answer


























      1














      HINT:



      We are not given that $f$ is differentiable.



      So, note that for $xin [-1,1]$



      $$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$



      Appeal to the Weierstrass M-Test. Can you finish?






      share|cite|improve this answer
























        1












        1








        1






        HINT:



        We are not given that $f$ is differentiable.



        So, note that for $xin [-1,1]$



        $$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$



        Appeal to the Weierstrass M-Test. Can you finish?






        share|cite|improve this answer












        HINT:



        We are not given that $f$ is differentiable.



        So, note that for $xin [-1,1]$



        $$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$



        Appeal to the Weierstrass M-Test. Can you finish?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 '18 at 2:15









        Mark Viola

        130k1274170




        130k1274170






























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