Clarification on prove regarding Continuous, increasing functions and $limsup$

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For every continuous increasing function $u:mathbb R_+tomathbb R_+$ and every nonnegative sequence $(x_n)$, $$limsuplimits_nu(x_n)=uleft(limsuplimits_nx_nright).$$




let $x_n_i rightarrow limsuplimits_nx_n$



Hence $limsuplimits_nu(x_n)geqlim u(x_n_i)=u(limsuplimits_nx_n) $, since $u$ is continuous.



Let $u(x_n_j)rightarrow limsuplimits_nu(x_n) $



$forall epsilon>0$ $exists J$ $forall jgeq J$ $ x_n_j< limsuplimits_nx_n + epsilon $



Since $u$ is increasing,



$u(x_n_j)leq u(limsuplimits_nx_n + epsilon)$



Hence, $limsuplimits_nu(x_n)= lim u(x_n_j)leq u(limsuplimits_nx_n + epsilon) $ for all $epsilon$



Since $u$ is continuous let $epsilon rightarrow 0$



Hence $limsuplimits_nu(x_n)leq u(limsuplimits_nx_n) $




calculus continuity limsup-and-liminf




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    For every continuous increasing function $u:mathbb R_+tomathbb R_+$ and every nonnegative sequence $(x_n)$, $$limsuplimits_nu(x_n)=uleft(limsuplimits_nx_nright).$$




    let $x_n_i rightarrow limsuplimits_nx_n$



    Hence $limsuplimits_nu(x_n)geqlim u(x_n_i)=u(limsuplimits_nx_n) $, since $u$ is continuous.



    Let $u(x_n_j)rightarrow limsuplimits_nu(x_n) $



    $forall epsilon>0$ $exists J$ $forall jgeq J$ $ x_n_j< limsuplimits_nx_n + epsilon $



    Since $u$ is increasing,



    $u(x_n_j)leq u(limsuplimits_nx_n + epsilon)$



    Hence, $limsuplimits_nu(x_n)= lim u(x_n_j)leq u(limsuplimits_nx_n + epsilon) $ for all $epsilon$



    Since $u$ is continuous let $epsilon rightarrow 0$



    Hence $limsuplimits_nu(x_n)leq u(limsuplimits_nx_n) $




    calculus continuity limsup-and-liminf




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      For every continuous increasing function $u:mathbb R_+tomathbb R_+$ and every nonnegative sequence $(x_n)$, $$limsuplimits_nu(x_n)=uleft(limsuplimits_nx_nright).$$




      let $x_n_i rightarrow limsuplimits_nx_n$



      Hence $limsuplimits_nu(x_n)geqlim u(x_n_i)=u(limsuplimits_nx_n) $, since $u$ is continuous.



      Let $u(x_n_j)rightarrow limsuplimits_nu(x_n) $



      $forall epsilon>0$ $exists J$ $forall jgeq J$ $ x_n_j< limsuplimits_nx_n + epsilon $



      Since $u$ is increasing,



      $u(x_n_j)leq u(limsuplimits_nx_n + epsilon)$



      Hence, $limsuplimits_nu(x_n)= lim u(x_n_j)leq u(limsuplimits_nx_n + epsilon) $ for all $epsilon$



      Since $u$ is continuous let $epsilon rightarrow 0$



      Hence $limsuplimits_nu(x_n)leq u(limsuplimits_nx_n) $




      calculus continuity limsup-and-liminf




      share|cite|improve this question















      For every continuous increasing function $u:mathbb R_+tomathbb R_+$ and every nonnegative sequence $(x_n)$, $$limsuplimits_nu(x_n)=uleft(limsuplimits_nx_nright).$$




      let $x_n_i rightarrow limsuplimits_nx_n$



      Hence $limsuplimits_nu(x_n)geqlim u(x_n_i)=u(limsuplimits_nx_n) $, since $u$ is continuous.



      Let $u(x_n_j)rightarrow limsuplimits_nu(x_n) $



      $forall epsilon>0$ $exists J$ $forall jgeq J$ $ x_n_j< limsuplimits_nx_n + epsilon $



      Since $u$ is increasing,



      $u(x_n_j)leq u(limsuplimits_nx_n + epsilon)$



      Hence, $limsuplimits_nu(x_n)= lim u(x_n_j)leq u(limsuplimits_nx_n + epsilon) $ for all $epsilon$



      Since $u$ is continuous let $epsilon rightarrow 0$



      Hence $limsuplimits_nu(x_n)leq u(limsuplimits_nx_n) $





      calculus continuity limsup-and-liminf





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      edited Aug 16 at 9:49

























      asked Aug 16 at 9:43









      Jhon Doe

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