Proving that $\limsup_{n\to\infty}\frac{1}{n}\sum_{m=1}^n s_m\leq \limsup_{n\to\infty}s_n.$

I am reviewing for my first year analysis exam and am stuck on a problem.

Let $\sigma_n=\frac{1}{n}\sum_{m=1}^n s_m$. I am trying to show that,

if $(s_n)$ is a bounded sequence of real numbers, $$\limsup_{n\to\infty}\sigma_n\leq \limsup_{n\to\infty}s_n.$$

I understand what's supposed to happen here, but I'm having trouble showing it. This is what I have so far:

I know from a previous exercise that $s_n\rightarrow s$ implies that $\sigma_n\rightarrow s$. I'll call this fact $\star$.

If $(\sigma_{n_k})_{k\in K}$ is a convergent subsequence of $(\sigma_n)$, let $(s_{n_i})_{i\in I\subseteq K}$ be a convergent subsequence of $(s_{n_k})_{k\in K}$. We know this must exist because $(s_{n_k})_{k\in K}$ is bounded (because $(s_n)$ is bounded). Let $I_k=\{i\in I:i \leq k\}$, $I_k^\prime =\{i\in K\setminus I:i \leq k\}$, and $m_k=|I_k|$. Then $$\sigma_{n_k}=\frac{m_k}{n_k}\left(\frac{1}{m_k}\sum_{i\in I_k}s_{n_i}\right)+\frac{n_k-m_k}{n_k}\left(\frac{1}{n_k-m_k}\sum_{i\in I_k^\prime}s_{n_i}\right)$$ Because $s_{n_i}$ is convergent by assumption, $\frac{1}{m_k}\sum_{i\in I_k}s_{n_i}$ is convergent by $\star$.

But, now I don't know where to go. The coefficients are a little ugly and I don't know what the other sum might converge to. What should I do next? Or am I taking the wrong approach?

You can do things formally, you idea is not completely off. Pick $M>0$ large and consider $n>M$. Then $$\frac 1n\sum_{k=1}^n a_k=\frac 1 n\sum_{k=1}^M a_k+\frac 1 n\sum_{k=M+1}^n a_k\leqslant \frac 1n\sum_{k=1}^M a_k+\frac {n-M} n\sup_{k\geqslant M+1}a_k$$

Taking $\limsup$ gives $$\limsup_{n\to\infty}\frac 1n\sum_{k=1}^n a_k\leqslant \sup_{k\geqslant M+1}a_k$$ and taking $M\to\infty$ gives $$\limsup_{n\to\infty}\frac 1 n\sum_{k=1}^n a_k\leqslant \lim_{M\to\infty}\sup_{k\geqslant M+1}a_k=\limsup_{n\to\infty}a_n$$

My intuition is that this the wrong approach, and I would rather look at the proof of fact $\star$ for inspiration than the statement of fact $\star$.

What happens if you take a subsequence of $\sigma_n$ that converges to $\limsup\sigma_n$ and try to find indices for an $s_n$-subsequence that converges to something $\ge\limsup\sigma_n$?

let $L = \lim \sup s$

if $L = \infty$ there is nothing to prove. if $L$ is finite there is no loss of generality in assuming $L=0$

given $\epsilon \gt 0$ then $\exists N. n \gt N \Rightarrow s_n \lt \frac12 \epsilon$ $$\sigma_n \lt \frac1{n}\left(\sum_{k=1}^N s_k +\frac12(n-N)\epsilon \right)$$ set $M$ so that $\frac1{M} \sum_{k=1}^Ns_k \lt \frac12 \epsilon$ then for $n \gt M$ we have $$\sigma_n \lt \epsilon$$ hence $$\lim \sup \sigma \lt \epsilon$$

I don't think your solution is going down a good path; it makes no reference to the fact that this is the $\limsup$ we are interested in, not merely any old limit. There's really no reason why two convergent subsequences of $\sigma_n$ and $s_n$ where the indices of the latter are a subset of the indices of the former have to concur in any way - $\sigma_n$ is a running average, more or less resistant to any variations in $s_n$, so we could easily choose subsequences for which the limit $\sigma_n$ is greater than that of $s_n$ (e.g. in the sequence $s_n=0$ if $n$ is square and $1$ otherwise; the subsequence of square numbers converges, for $s_n$, to $0$, but for $\sigma_n$, to $1$)

Stepping back a bit, this is a very intuitive property; loosely, the property reads:

There must be infinitely many values of a sequence greater or equal to its average.

In particular, notice that you can prove that, if, for some $x$, there are only finitely many indexes at which $s_n>x$, then it is clear that those terms make a vanishingly small contribution to $\sigma_n$ as $n$ grows - in particular, one could bound $\sigma_n$ as $$\sigma_n = \frac{1}{n}\sum_{i=1}^{n}s_n\leq \frac{k}n+\frac{1}n\sum_{i}^n x=\frac{k}n+x$$ where $k$ is the sum of the differnece $s_n-x$ take over all terms of $s_n$ greater than $x$ (which is finite, since there are only finitely many such terms). The inequality holds since, for every other term, $s_n\leq x$, so replacing $s_n$ with $x$ in every case and accounting for the possible difference yields an upper ound. However, from $\sigma_n\leq \frac{k}n + x$, we obviously have that $\limsup \sigma_n \leq x$, since the $\frac{k}n$ term vanishes.

Using the definition of $\limsup$, this suffices to show the desired inequality.

• The average of a sequence is less than the highest value that appears infinitely often in the sequence. So I am a bit more familiar with probability proofs using limsups than those in real analysis (for reasons of taking classes out of order). But, I don't see how this applies. For example if I have a monotonic injective sequence that converges to a limit, nothing appears infinitely often,right? So I'm not sure I understand how this helps. – stuckanalysis Oct 15 '14 at 1:12
• I edited that part; I was being a bit sloppy there, thinking "highest value" could be taken to include "a value, and everything greater" rather than just the value. What I have now should be far clearer. – Milo Brandt Oct 15 '14 at 1:27