$\liminf b_n \le \liminf a_n$

Question

Say we have a sequence of reals $a_n, b_n$ both from $n = 1$ to infinity. $$ b_n = 1/n\cdot(a_1+...+a_n) $$ How could we prove that $\liminf_{n\to\infty}a_n\leq\liminf b_n$.

Answer 1

This is true with $\le$ instead; let $a=\liminf_{n\to\infty}a_n$, and wlog we can assume $a > -\infty$ as there is nothing to prove otherwise; similarly if $a_n \to \infty$ then clearly $b_n \to \infty$ so we can assume $a$ finite and $a_n \ge K$ for some real $K$; then for every $\epsilon >0$ there is $n(\epsilon)$ st $a_n \ge a-\epsilon, n \ge n(\epsilon)$.

Then $a_1+...+a_{n(\epsilon)} \ge n(\epsilon)K$, hence $b_n \ge (n-n(\epsilon))(a-\epsilon)/n +n(\epsilon)K/n, n \ge n(\epsilon)$ so $\liminf b_n \ge a-\epsilon$.

Letting $\epsilon \to 0$ proves our claim

Answer 2

If $(a_{n})$ is bounded, then $(\inf\{a_{n}: n\geq N\}), N \in \mathbb{N}$ is bounded and monotone and thus convergent. Call its limit inferior $L_{\inf a}$. If $(a_{n})$ is bounded, $(b_{n})$ is bounded too, call its limit inferior $L_{\inf b}$. Going forward we prove statement assuming $(a_{n})$ is bounded.

From definition of infimum, we can deduce $\forall \epsilon>0, \space \exists N \in \mathbb{N} \mid \forall n \geq N, |\inf\{a_k:k\geq n\}-L_{\inf a}| < \epsilon \implies \forall \epsilon>0, \exists N \in \mathbb{N} \mid \forall n \geq N, a_n > L_{\inf a}-\epsilon. $

Then, $\forall n\geq N$, $b_{n}=\frac{1}{n}\sum_{k=1}^{n}a_{k}=\frac{1}{n}(\sum_{k=1}^{N-1}a_{k}+\sum_{k=N}^{n}a_{k})>\frac{1}{n}(\sum_{k=1}^{N-1}a_{k}+\sum_{k=N}^{n}(L_{\inf a}-\epsilon)) \implies $ for all convergent subsequences $(b_{n_{i}}), \lim_{i\rightarrow\infty}b_{n_{i}} > \lim_{i\rightarrow\infty} \frac{1}{n_{i}}(\sum_{k=1}^{N-1}a_{k}+\sum_{k=N}^{n_{i}}(L_{\inf a}-\epsilon))=\lim_{i\rightarrow\infty}\frac{(n_{i}-N+1)(L_{\inf a}-\epsilon)}{n_{i}}=L_{\inf a}-\epsilon.$

This must be true for all $\epsilon > 0 \implies$ for all convergent subsequences $(b_{n_{i}})$ of $b_{n}$, $\lim_{i\rightarrow\infty}b_{n_{i}} \geq \lim\inf_{n\rightarrow\infty}a_{n}$. If $\lim\inf_{n\rightarrow\infty}b_{n}<\lim\inf_{n\rightarrow\infty}a_{n} \implies \exists (b_{n_{i}})$ convergent $\mid \lim_{i\rightarrow\infty}b_{n_{i}}<\lim\inf_{n\rightarrow\infty}a_{n}$, contrary to the proof. Thus $\lim\inf_{n\rightarrow\infty}b_{n}\geq \lim\inf_{n\rightarrow\infty}a_{n}$ when $(a_{n})$ is bounded.

If $(a_{n})$ is unbounded:

• if $(a_{n})$ has a subsequence approaching $-\infty$, then $\lim\inf_{n\rightarrow\infty}a_{n} \rightarrow -\infty \leq \lim\inf_{n\rightarrow\infty}b_{n}$.
• if $(a_{n})$ has a subsequence approaching $\infty$: if every subsequence of $(a_{n})$ is unbounded from above, same is true for $(b_{n})$, and their limit inferiors will be approaching $\infty$. If $(a_{n})$ has a subsequence $(a_{k_{i}})$ bounded from above, by the proof above $L_{\inf a}$ of this subsequence exists and, since the complement of subsequence $(a_{k_{i}})$ relative to $(a_{n})$ (call it $(a_{m_{i}})$) is unbounded from above, $\forall K>0, \exists M \in \mathbb{N} \mid a_{m_{i}}>K, \forall i\geq M$. Then, by the second paragraph, $\forall \epsilon>0$, if $K=L_{\inf a}-\epsilon$, $\exists P=\max(M,N) \mid \forall n\geq P, a_{n}>L_{\inf a}-\epsilon$. Then $\forall n\geq P$ we can bound $(b_{n})$ below and the rest of the argument for convergent subsequences $(b_{n_{i}})$ (which exist because if $(a_{k_{i}})$ is bounded, then $(b_{k_{i}})$ is bounded) holds and thus the whole statement.