Let $\{ y_{k}\}_{k}$ a positive sequence and $\alpha$ a scalar continuous function s.t. $\alpha(0) =0$ and $\alpha(s) > s$ for every $s >0$.
I would like to prove the following equality:
$\limsup\limits_{k \rightarrow \infty} \alpha(y_k) = \alpha(\limsup\limits_{k \rightarrow \infty} y_k) $
Are there any other assumptions that I should take under consideration?
Thank you in advance!
I think you shoule add the condition : $\alpha(s)$ is monotone increasing .
W.L.G, we may assume the sequence $\{ y_{k}\}$ is bounded. First, we prove $$\limsup\limits_{k \to \infty} \alpha(y_k)\geq\alpha(\limsup\limits_{k \to \infty} y_k).$$ For the sequence $\{ y_{k}\}$, there exists subsequence $\{ y_{k_{j}}\}$ such that $$\lim_{j\to\infty}y_{k_{j}}=\limsup\limits_{k \to \infty} y_k:=y,$$ due to the continuity of $\alpha(s)$, we have $$\lim_{j\to\infty}\alpha(y_{k_{j}})=\alpha(y),$$ obviously, $$\limsup\limits_{k \to \infty} \alpha(y_k)\geq\lim_{j\to\infty}\alpha(y_{k_{j}})=\alpha(y)=\alpha(\limsup\limits_{k \to \infty} y_k).$$ (because $\{\alpha(y_{k_{j}})\}$ is subsequence of $\{\alpha(y_{k})\}$.) Then we show $$\limsup\limits_{k \to \infty} \alpha(y_k)\leq\alpha(\limsup\limits_{k \to \infty} y_k).$$ For the sequence $\{\alpha(y_{k})\}$, there is a subsequence $\{\alpha(y_{k_{j}})\}$ such that $$\lim_{j\to \infty}\alpha(y_{k_{j}})=\limsup\limits_{k \to \infty} \alpha(y_k)=Y.$$ Then there is a convergent subsequence $\{y_{k_{j_i}\}}$of $\{y_{k_{j}}\}$ such that $$\lim_{i\to \infty} y_{k_{j_i}}=y,$$
by the continuity of $\alpha(s)$, we get $$\lim_{i\to \infty}\alpha(y_{k_{j_i}})=\alpha(y)=Y.$$ So $$\limsup\limits_{k \to \infty} \alpha(y_k)= Y={\color{red}{\alpha(y)\leq\alpha(\limsup\limits_{k \to \infty} y_k)}}.$$ The red part uses the monotonicity!