How can I show rigorously that $$ \limsup \sqrt{x_n} = \sqrt{\limsup x_n} $$ where $ (x_n) $ is a bounded sequence of real numbers such that $ x_n \geq 0 $ for each $ n \in \mathbb{N} $.
Prove that the limit superior of square root is the square root of limit superior
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real-analysis
limits
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0The function $f(x) = \sqrt{x}$ is continuous and non decreasing. – 2017-02-22
3 Answers
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Recall that for sequences with limit supremums, there exists a sub sequence that attains that limit.
Hence there exists some $(x_n)_k $ such that $\lim x_{n_k} = \lim\sup x_n=s$. Hence $\sqrt{s}=\sqrt{\lim x_{n_k}}= \lim \sqrt{x_{n_k}}\leq \lim\sup \sqrt{x_{n}}$.
We can show the other direction of the inequality similarly, so we have equality
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Hint:
If $L=\limsup x_n$, $$|\sqrt{x_n}-\sqrt L|=\frac{|x_n-L|}{\sqrt{x_n}+\sqrt L}<\frac1{\sqrt L}|x_n-L|$$
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If $f$ is non decreasing and continuous, then $\sup f(S) = f(\sup S)$.
To see this, note that if $s \in S$ then $s \le \sup S$ and so $f(s) \le f(\sup S)$ and so $\sup f(S) \le f(\sup S)$. Now let $s_k \in S$ such that $s_k \to \sup S$, then $f(s_k) \to f(\sup S)$ and hence $\sup f(S) \ge f(\sup S)$.
Then $\limsup_n f(x_n) = \lim_n \sup_{k \ge n} f(x_k) = \lim_n f(\sup_{k \ge n} x_k) = f( \limsup_n x_n)$.
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0Perhaps the $ \sup f(S) \le \sup f(S) $ is a typo? – 2017-02-23
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0@KendryJohnson: Thanks for catching the typo. – 2017-02-23