Given a non negative sequance $(a_n)$, is it true that $$ \varlimsup_{n\to\infty} (a_n^2)=(\varlimsup_{n\to\infty} a_n)^2\quad? $$ Thanks!
Is $\varlimsup_{n\to\infty} (a_n^2)=(\varlimsup_{n\to\infty} a_n)^2\quad?$
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calculus
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0Yes, take the subsequence that converges to the limsup and use continuity. – 2017-01-03
1 Answers
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First, try to prove that $L$ is a cluster point of $(a_n)$ $\iff$ $L^2$ is a cluster point of $(a_n^2)$.
Now let $A$ be the set of cluster points of $(a_n)$ and $A^2$ be that of $(a_n^2)$. If $\sup A = \infty$, then for $M > 0$, there exists $a\in A$ such that $a > \sqrt M$, so $A^2 \ni b:=a^2 > M$, and therefore $\sup A^2 = \infty$. Otherwise, we have for $a^2 \in A^2$, $a \in A$, so $a \le \sup A$, so $a^2 \le (\sup A)^2$, $\therefore \sup A^2 \le (\sup A)^2$ (in particular, $< \infty$). Also, for $a\in A$, $a^2 \in A^2$, so $a^2 \le \sup A^2$, hence $(\sup A)^2 \le \sup A^2$. $\therefore \sup A^2 = (\sup A)^2$.