Is the following statement is True/false ? ..












0












$begingroup$


Is the following statement is True/false ?



Given that $f_n(x) =(-x)^n $ , $x in [0,1] $ then $f_n$ converges pointwise everywhere ?



i thinks it will be true same as $f(x) = 0 $ when x =0 ,$f(x) =1$ when



$0 <x le 1$



Am i right ??






real-analysis







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    0












    $begingroup$


    Is the following statement is True/false ?



    Given that $f_n(x) =(-x)^n $ , $x in [0,1] $ then $f_n$ converges pointwise everywhere ?



    i thinks it will be true same as $f(x) = 0 $ when x =0 ,$f(x) =1$ when



    $0 <x le 1$



    Am i right ??






    real-analysis







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Is the following statement is True/false ?



      Given that $f_n(x) =(-x)^n $ , $x in [0,1] $ then $f_n$ converges pointwise everywhere ?



      i thinks it will be true same as $f(x) = 0 $ when x =0 ,$f(x) =1$ when



      $0 <x le 1$



      Am i right ??






      real-analysis







      share|cite|improve this question









      $endgroup$




      Is the following statement is True/false ?



      Given that $f_n(x) =(-x)^n $ , $x in [0,1] $ then $f_n$ converges pointwise everywhere ?



      i thinks it will be true same as $f(x) = 0 $ when x =0 ,$f(x) =1$ when



      $0 <x le 1$



      Am i right ??






      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 Dec 12 '18 at 13:00









      jasminejasmine

      1,747417




      1,747417






















          2 Answers
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          $begingroup$

          You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$



          then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.






          share|cite|improve this answer











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          • $begingroup$
            @%xum but im confusing at negative sign $(-x)^n = x^n$
            $endgroup$
            – jasmine
            Dec 12 '18 at 13:06






          • 1




            $begingroup$
            @jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
            $endgroup$
            – 5xum
            Dec 12 '18 at 13:07



















          1












          $begingroup$

          This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$



            then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @%xum but im confusing at negative sign $(-x)^n = x^n$
              $endgroup$
              – jasmine
              Dec 12 '18 at 13:06






            • 1




              $begingroup$
              @jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
              $endgroup$
              – 5xum
              Dec 12 '18 at 13:07
















            1












            $begingroup$

            You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$



            then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              @%xum but im confusing at negative sign $(-x)^n = x^n$
              $endgroup$
              – jasmine
              Dec 12 '18 at 13:06






            • 1




              $begingroup$
              @jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
              $endgroup$
              – 5xum
              Dec 12 '18 at 13:07














            1












            1








            1





            $begingroup$

            You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$



            then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.






            share|cite|improve this answer











            $endgroup$



            You are not correct. If you define $$f(x)=begin{cases}0&xin[0,1)\1&x=1end{cases}$$



            then $f_n$ converges to $f$ pointwise on $[0,1)$, however, for $x=1$, the sequence $f_n(x)$ becomes $(-1)^n$ which is the sequence $1,-1,1,-1,dots$, a sequence that does not converge.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 12 '18 at 13:08

























            answered Dec 12 '18 at 13:04









            5xum5xum

            90.6k394161




            90.6k394161












            • $begingroup$
              @%xum but im confusing at negative sign $(-x)^n = x^n$
              $endgroup$
              – jasmine
              Dec 12 '18 at 13:06






            • 1




              $begingroup$
              @jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
              $endgroup$
              – 5xum
              Dec 12 '18 at 13:07


















            • $begingroup$
              @%xum but im confusing at negative sign $(-x)^n = x^n$
              $endgroup$
              – jasmine
              Dec 12 '18 at 13:06






            • 1




              $begingroup$
              @jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
              $endgroup$
              – 5xum
              Dec 12 '18 at 13:07
















            $begingroup$
            @%xum but im confusing at negative sign $(-x)^n = x^n$
            $endgroup$
            – jasmine
            Dec 12 '18 at 13:06




            $begingroup$
            @%xum but im confusing at negative sign $(-x)^n = x^n$
            $endgroup$
            – jasmine
            Dec 12 '18 at 13:06




            1




            1




            $begingroup$
            @jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
            $endgroup$
            – 5xum
            Dec 12 '18 at 13:07




            $begingroup$
            @jasmine Oh, I missed the negative sign! Sorry, I will edit my answer.
            $endgroup$
            – 5xum
            Dec 12 '18 at 13:07











            1












            $begingroup$

            This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.






                share|cite|improve this answer









                $endgroup$



                This is correct except for when $x=1$. When $x=1$ you get a sequence $f_n = (-1)^n$ which does not converge, and therefore does not converge pointwise at that point.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 12 '18 at 13:08









                dllegenddllegend

                163




                163






























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