Let $(X,\left\|\;\cdot\;\right\|)$ be a normed space and $(x_n)_{n\in\mathbb{N}}\subseteq X$. Can we prove that there is a subsequence $\left(x_{n_k}\right)_{k\in\mathbb{N}}\subseteq(x_n)_{n\in\mathbb{N}}$ such that $$\lim_{k\to\infty}\left\|x_{n_k}\right\|=\liminf_{n\to\infty}\left\|x_n\right\|\;?$$
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Your question can be greatly reduced. The normed space and its norm can be "forgotten" because both the limit and the $\liminf$ are applied to sequences of real numbers ($||x_n||$ is a real number!), and the question simply becomes:
For a given sequence of positive real numbers $(a_n)_{n\in \mathbb N}$, does there exist a subsequence $(a_{n_k})_{k\in\mathbb N}$ such that $$\lim_{k\to\infty} a_{n_k}= \liminf_{n\to\infty} a_n$$
This question is much simpler to answer and is almost trivial, depending on your definition of $\limsup$.
5xum
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Right, in fact I favor the definition of $\liminf $ as the minimum subsequential limit. – zhw. Jun 16 '15 at 19:22
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Yes, and this holds more generally for real sequences. In fact, the limit inferior of a sequence can be defined as the least limit point of any subsequence of the sequence.
To prove that such a subsequence exists, recall that, $$\liminf\limits_{n\rightarrow \infty} a_n = \lim_{n \rightarrow \infty} \inf\limits_{k \ge n} a_k$$ Let the limit inferior be $L$ and fix an integer $m > 0$. By the above limit being $L$, we may choose some $n$ such that $$\left| \inf\limits_{k \ge n} a_k - L\right| < \frac{1}{2m}.$$ By definition of an infimum, we may choose some $n_m \ge n$ such that, $$\left| \inf\limits_{k \ge n} a_k - a_{n_m}\right| < \frac{1}{2m}.$$ Using the triangle inequality, $$\left| a_{n_m} - L\right| \le \left| \inf\limits_{k \ge n} a_k - L\right| + \left| \inf\limits_{k \ge n} a_k - a_{n_m}\right| < \frac{1}{m}.$$ By squeeze theorem, $a_{n_m} \rightarrow L$.
Theo Bendit
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It depends on how you define $\lim\inf a_n = -\infty$. What definition do you use? – Theo Bendit Apr 07 '19 at 06:25
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I define $\liminf a_n=\lim_{n\to\infty}\inf_{k\ge n}a_k$ and $\lim_{n\to\infty}b_n=-\infty$ means that for all $r\in\mathbb R$ there is a $N\in\mathbb N$ with $b_n\le r$ for all $n\ge N$. – 0xbadf00d Apr 07 '19 at 07:04
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That's the limit definition to $-\infty$. What about the definition of the limit-inferior being $-\infty$? Would you just keep it the same as the finite case, but allow $\inf_{k \ge n} a_k$ to be $-\infty$ when ${a_k : k \ge n}$ is not bounded below? – Theo Bendit Apr 07 '19 at 07:08
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As I wrote, $\liminf_{n\to\infty}a_n$ is defined in terms of the limit of the sequence $(\inf_{k\ge n}a_k)_{n\in\mathbb N}$. So the definition of the limit being $-\infty$ applies. I treat all sequences as being subsets of the extended real numbers $[-\infty,\infty]$. Maybe that's where the confusion comes from. – 0xbadf00d Apr 07 '19 at 07:31
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1OK, in that case, we know that $\inf_{k \ge n} a_k = -\infty$ by definition if ${a_k : k \ge n}$ is unbounded (finitely) from below, or contains $-\infty$. If $-\infty$ occurs infinitely often in the sequence, then we have our convergent subsequence to $-\infty$. If not, then let $x_{n_1}$ be the first term after the last occurrence of $-\infty$. Since ${a_k : k \ge n_1 + 1}$ is unbounded below, choose $n_2 > n_1$ such that $a_{n_2} < -2$. Since ${a_k : k \ge n_2 + 1}$ is bounded below, find $n_3 > n_2$ such that $a_{n_3} < -3$, etc. Find $a_{n_k}$ such that $a_{n_k} < -k$ for all $k$. – Theo Bendit Apr 07 '19 at 07:53
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Yes: this is the same as proving that for a real sequence $(a_n)_k$ there is a subsequence $(a_{n_k})_k$ such that $$ \lim_{k} a_{n_k} = \liminf_n a_n = l, $$ say.
To do this, we basically imitate one of the proofs of Bolzano–Weierstrass (the "high points" one) (or see here): pick a sequence $\varepsilon_k \to 0$. Then for each $\varepsilon_k$, there are, by the definition of the limit inferior, an infinite number of $n$ such that $ \lvert a_{n} - l \rvert < \varepsilon_k $. Choose $n_k$ as the smallest of these $n$ not already chosen, and then you find that $(a_{n_k})_k$ satisfies $ 0 \leqslant \lvert a_{n_k} - l \rvert < \varepsilon_k \to 0$, so $a_{n_k} \to l $.
