Show that $\liminf(\sqrt{x_n})=\sqrt{\liminf x_n}$

Question

Let $(x_n)$ be a bounded sequence of real numbers such that $x_n \ge0$ for each $n \in \mathbb N$. Show that $\liminf(\sqrt{x_n})=\sqrt{\liminf x_n}$.

I assume we need to consider the subsequence of ($\sqrt{x_n}$), but I'm really confused with the notion on "boundedness". Can I assume the question implies an upper bound on $(x_n)$?

Even then wouldn't the equation be implied? Given how roots generally affect limits.

Answer 1

I'll be using the following property: Given a sequence $x_n$, $\lim \inf x_n$ is the smallest limit a convergent subsequence of $x_n$ can have.

Let $x_{n_k}$ be a subsequence of $x_n$ that converges to $\lim \inf x_n$.

Then $\sqrt{x_{n_k}}$ converges to $\sqrt{\lim \inf x_n}$. But $\sqrt{x_{n_k}}$ is a subsequence of $\sqrt{x_{n}}$. Hence $\lim \inf \sqrt{x_{n}}\leq \sqrt{\lim \inf x_n}$.

The reverse inequality can be proved in a similar fashion.