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) $

edited Aug 16 at 9:49

asked Aug 16 at 9:43

Jhon Doe

397211

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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) $

edited Aug 16 at 9:49

asked Aug 16 at 9:43

Jhon Doe

397211

add a commentÂ |Â

up vote
0
down vote

favorite

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) $

edited Aug 16 at 9:49

asked Aug 16 at 9:43

Jhon Doe

397211

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) $

edited Aug 16 at 9:49

asked Aug 16 at 9:43

Jhon Doe

397211

edited Aug 16 at 9:49

asked Aug 16 at 9:43

Jhon Doe

397211

asked Aug 16 at 9:43

Jhon Doe

397211

asked Aug 16 at 9:43

Jhon Doe

397211

add a commentÂ |Â

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