Counter example of bounds on limit implies bounds on the sequence

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Let $F_n(x)$ be a sequence of functions from $[0, infty]$ to [0,1], suppose $F_n(x)$ is monotone in x, and $F_n(x)$ converges uniformly to $F(x) = e^-x$. Find a counter example such that the following statement is not true:



$exists NinmathbbN, text s.t. forall n>N, forall x, F_n(x)<e^-fracx2$.



This is in analog to convergence of sequence:



$a_nto a, |a|<infty$, then $forall b>a, exists NinmathbbN, texts.t. forall n>N, a_n<b$.



Remark: an non counter example is $F_n(x) = e^-fracx1+frac1n$




real-analysis




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    Let $F_n(x)$ be a sequence of functions from $[0, infty]$ to [0,1], suppose $F_n(x)$ is monotone in x, and $F_n(x)$ converges uniformly to $F(x) = e^-x$. Find a counter example such that the following statement is not true:



    $exists NinmathbbN, text s.t. forall n>N, forall x, F_n(x)<e^-fracx2$.



    This is in analog to convergence of sequence:



    $a_nto a, |a|<infty$, then $forall b>a, exists NinmathbbN, texts.t. forall n>N, a_n<b$.



    Remark: an non counter example is $F_n(x) = e^-fracx1+frac1n$




    real-analysis




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      Let $F_n(x)$ be a sequence of functions from $[0, infty]$ to [0,1], suppose $F_n(x)$ is monotone in x, and $F_n(x)$ converges uniformly to $F(x) = e^-x$. Find a counter example such that the following statement is not true:



      $exists NinmathbbN, text s.t. forall n>N, forall x, F_n(x)<e^-fracx2$.



      This is in analog to convergence of sequence:



      $a_nto a, |a|<infty$, then $forall b>a, exists NinmathbbN, texts.t. forall n>N, a_n<b$.



      Remark: an non counter example is $F_n(x) = e^-fracx1+frac1n$




      real-analysis




      share|cite|improve this question














      Let $F_n(x)$ be a sequence of functions from $[0, infty]$ to [0,1], suppose $F_n(x)$ is monotone in x, and $F_n(x)$ converges uniformly to $F(x) = e^-x$. Find a counter example such that the following statement is not true:



      $exists NinmathbbN, text s.t. forall n>N, forall x, F_n(x)<e^-fracx2$.



      This is in analog to convergence of sequence:



      $a_nto a, |a|<infty$, then $forall b>a, exists NinmathbbN, texts.t. forall n>N, a_n<b$.



      Remark: an non counter example is $F_n(x) = e^-fracx1+frac1n$





      real-analysis





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      edited Aug 16 at 10:22

























      asked Aug 16 at 10:14









      kevin

      986




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          2 Answers
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          accepted










          Take $F_n(x)=frac e^-x+frac 1 n 1+frac 1 n $. For large $x$ we don't have $F_n(x) <e^-x/2$.






          share|cite|improve this answer




















          • +1 Sorry, did not see your nice example while typing.
            – Severin Schraven
            Aug 16 at 10:28

















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          Take
          $$ F_n(x) = begincases 1,& 0leq x leq 1/n \
          e^-x,& xgeq 2/n, \
          endcases $$
          and interpolate between the end points.






          share|cite|improve this answer




















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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

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            active

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            up vote
            2
            down vote



            accepted










            Take $F_n(x)=frac e^-x+frac 1 n 1+frac 1 n $. For large $x$ we don't have $F_n(x) <e^-x/2$.






            share|cite|improve this answer




















            • +1 Sorry, did not see your nice example while typing.
              – Severin Schraven
              Aug 16 at 10:28














            up vote
            2
            down vote



            accepted










            Take $F_n(x)=frac e^-x+frac 1 n 1+frac 1 n $. For large $x$ we don't have $F_n(x) <e^-x/2$.






            share|cite|improve this answer




















            • +1 Sorry, did not see your nice example while typing.
              – Severin Schraven
              Aug 16 at 10:28












            up vote
            2
            down vote



            accepted







            up vote
            2
            down vote



            accepted






            Take $F_n(x)=frac e^-x+frac 1 n 1+frac 1 n $. For large $x$ we don't have $F_n(x) <e^-x/2$.






            share|cite|improve this answer












            Take $F_n(x)=frac e^-x+frac 1 n 1+frac 1 n $. For large $x$ we don't have $F_n(x) <e^-x/2$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 16 at 10:24









            Kavi Rama Murthy

            22.6k2933




            22.6k2933











            • +1 Sorry, did not see your nice example while typing.
              – Severin Schraven
              Aug 16 at 10:28
















            • +1 Sorry, did not see your nice example while typing.
              – Severin Schraven
              Aug 16 at 10:28















            +1 Sorry, did not see your nice example while typing.
            – Severin Schraven
            Aug 16 at 10:28




            +1 Sorry, did not see your nice example while typing.
            – Severin Schraven
            Aug 16 at 10:28










            up vote
            1
            down vote













            Take
            $$ F_n(x) = begincases 1,& 0leq x leq 1/n \
            e^-x,& xgeq 2/n, \
            endcases $$
            and interpolate between the end points.






            share|cite|improve this answer
























              up vote
              1
              down vote













              Take
              $$ F_n(x) = begincases 1,& 0leq x leq 1/n \
              e^-x,& xgeq 2/n, \
              endcases $$
              and interpolate between the end points.






              share|cite|improve this answer






















                up vote
                1
                down vote










                up vote
                1
                down vote









                Take
                $$ F_n(x) = begincases 1,& 0leq x leq 1/n \
                e^-x,& xgeq 2/n, \
                endcases $$
                and interpolate between the end points.






                share|cite|improve this answer












                Take
                $$ F_n(x) = begincases 1,& 0leq x leq 1/n \
                e^-x,& xgeq 2/n, \
                endcases $$
                and interpolate between the end points.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 16 at 10:26









                Severin Schraven

                4,9061831




                4,9061831






















                     

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