# Prove that f(underset(x -> +oo)(lim sup x_n)) = underset(x -> +oo)(lim sup ) f(_xn)

Prove that $f\left(\underset{x\to +\mathrm{\infty }}{lim sup}{x}_{n}\right)$= $\underset{x\to +\mathrm{\infty }}{lim sup}f\left({x}_{n}\right)$
The problem is divided into two questions:
${x}_{n}$ is a real valued sequence.
Prove that $f\left(\underset{n\to +\mathrm{\infty }}{lim sup}{x}_{n}\right)$ = $\underset{n\to +\mathrm{\infty }}{lim sup}f\left({x}_{n}\right)$ given that f is continuous and increasing.
What can we say when f is decreasing.
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For any sequence $\left({y}_{n}\right)$ let ${y}_{n}^{+}=\underset{k\ge n}{sup}{y}_{k}$. Let $L=lim sup{x}_{n}$ be a real number. Since $L=lim{x}_{n}^{+}$ and f is continuous, we have
$f\left(L\right)=limf\left({x}_{n}^{+}\right).$
Now we show that $f\left({x}_{n}^{+}\right)=f\left({x}_{n}{\right)}^{+}$ for each n. Since ${x}_{n}^{+}\ge {x}_{k}$ for each $k\ge n$ and f is increasing, we have $f\left({x}_{n}^{+}\right)\ge f\left({x}_{n}{\right)}^{+}$. For any $\epsilon >0$ there exists $k\ge n$ such that ${x}_{n}^{+}-\epsilon <{x}_{k}$. Again using the fact that f is increasing we get $f\left({x}_{n}^{+}-\epsilon \right)\le f\left({x}_{k}\right)\le f\left({x}_{n}{\right)}^{+}$. Letting $\epsilon \to 0$ and using the continuity of f we get $f\left({x}_{n}^{+}\right)\le f\left({x}_{n}{\right)}^{+}$
If f is decreasing then (−f) is increasing and thus $\left(-f\right)\left(lim sup{x}_{n}\right)=lim sup\phantom{\rule{mediummathspace}{0ex}}\left(-f\right)\left({x}_{n}\right)$(xn). This reduces to $f\left(lim sup{x}_{n}\right)=lim inff\left({x}_{n}\right)$ after simplification.