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For my math class I have to prove that $\liminf_{n\rightarrow\infty}s_n\leqslant \liminf_{n\rightarrow\infty}\sigma_n$ and $\limsup_{n\rightarrow\infty}s_n\geqslant\limsup_{n\rightarrow\infty}\sigma_n$ for $\sigma_n=n^{-1}(s_1+...+s_n)$.

Now for the second part, an answer has been given already in If $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$. Unfortunately I do not really understand the whole process.

In $$\sigma_n=\frac 1n\sum_{j=1}^ks_j+\frac 1n\sum_{j=k+1}^ns_j\leqslant \frac 1n\sum_{j=1}^ks_j+\frac{n-k}n\sup_{l\geqslant k}s_l\leqslant \frac 1n\sum_{j=1}^ks_j+\sup_{l\geqslant k}s_l.$$ I do'nt understand why $\frac 1n\sum_{j=k+1}^ns_j\leqslant\frac{n-k}n\sup_{l\geqslant k}s_l$. I really don't see why this should be the case. If you could give me any help with this it would be much appreciated!

Also, any tips on the first part of my exercise are welcome!

Community
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Soof_fie
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  • For $k + 1 \leqslant j \leqslant n$, you have $s_j \leqslant \sup\limits_{l \geqslant k} s_l$, by definition of $\sup$. In the inequality you want to prove, replace $\sup$ with $\inf$, the inequalities go in the other direction then. – Daniel Fischer Feb 13 '16 at 20:31
  • But I don't have to prove that $$s_j\leqslant\sup_{l\geqslant k}s_l$$ but that $$\sum_{j=k+1}^ns_j\leqslant\sup_{l\geqslant k}s_l$$ – Soof_fie Feb 13 '16 at 20:42
  • No, you have to show that $$\frac{1}{n}\sum_{j = k+1}^n s_j \leqslant \frac{n-k}{n} \sup_{l \geqslant k} s_l.$$ – Daniel Fischer Feb 13 '16 at 20:50
  • Fair point. But how does proving that $s_j\leqslant\sup_{l\geqslant k}s_l$ suffice for proving that? – Soof_fie Feb 13 '16 at 20:52
  • $$s_j \leqslant S \implies \sum_{j = k+1}^n s_j \leqslant \sum_{j = k+1}^n S.$$ – Daniel Fischer Feb 13 '16 at 20:54
  • But $S$ is just one value though, only the supremum. So if I think for instance of a simple sequence where $s_j=(1,2,3)$, then $S=3$, but the sum of the elements is 6. So how can you say that? – Soof_fie Feb 13 '16 at 21:01

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Just in case it helps, here is the solution to your question written out explicitly. You say you don't understand why $\frac 1n\sum_{j=k+1}^ns_j\leqslant\frac{n-k}n\sup_{l\geqslant k}s_l$.

In the sum there are $n-k$ terms. Each of those terms is $s_j$, where $j$ goes from $k+1$ to $n$. Now, if you look at $\sup_{l\geqslant k} s_l$, it must, by the definition of $\sup$, be greater than or equal to $s_{k+1}$, $s_{k+2}, \ldots, s_n$, because all of the latter have an index $l$ which is $\geqslant k$.

A sum of $n-k$ terms, each of which is less than or equal to $\sup_{l\geqslant k} s_l$, must be $\leqslant (n-k)\sup_{l\geqslant k} s_l$. Multiply both sides by $1/n$ and you get the desired result.

Edited to answer further question. For fixed $k$ we have $\sup_{l\ge k} s_l$ constant. Furthermore, $\lim_{n\to\infty} {1\over n}\sum_{j=1}^k s_j= 0$, again because $k$ is fixed. As a result $$\limsup_n \left({1\over n}\sum_{j=1}^k s_j + \sup_{l\ge k} s_l\right) = \lim_{n\to\infty} {1\over n}\sum_{j=1}^k s_j + \sup_{l\ge k} s_l = \sup_{l\ge k} s_l$$

Thus, we have $\limsup\sigma_n\le \sup_{l\ge k} s_l$ for all $k$, and thus $\limsup \sigma_n\le \inf_k \sup_{l\ge k} s_l$. We'd like to show that this implies $\limsup\sigma_n\le\limsup s_n$.

Write $U_k=\sup_{l\ge k} s_l$, a non-increasing sequence so that $U_{k+1}\le U_k$ for all $k$. Let $U=\inf_k U_k$; we have $U=\lim_k U_k$ on account of the sequence $U_k$ being non-increasing. Let $V=\limsup s_n$. We want to show $V\ge U$.

Given $\epsilon>0$, for each $k$ pick $j(k)$ such that $s_{j(k)} > U_k-\epsilon$, with $j(k)\ge k$. We can do this because of how $U_k$ is defined as a supremum. Then the sequence $s_{j(k)}$, being bounded, must have a convergent subsequence. Taking limits as $k\to\infty$ on both sides of $s_{j(k)} > U_k-\epsilon$, we see that the limit of that subsequence is $\ge U-\epsilon$. Recall that $V$ is the supremum of all limits of subsequences of $s_n$. Thus $V\ge U-\epsilon$, and since $\epsilon$ is arbitrary, $V\ge U$.

This proof may be suboptimal. There's probably a neat theorem somewhere that makes it much shorter.

ForgotALot
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  • Thank you so much, this explains a lot! Now I only still have a problem with the last step. So they would take the $\limsup$ of both sides. So you'd get: $$\limsup_{n\rightarrow\infty}\sigma_n\leqslant\limsup_{n\rightarrow\infty} (\frac{1}{n} \sum_{j=1}^{k}s_j+\sup_{l\geqslant k}s_l)$$

    How would we go from here to $\limsup_{n\rightarrow\infty}\sigma_n\leqslant\limsup_{n\rightarrow\infty}s_n$?

    I would like to pretend that we can take $\limsup(A+B)=\limsup A+\limsup B$, but I don't think you can.

     – Soof_fie Feb 13 '16 at 21:46
  • We have $\limsup f_n = \lim f_n$ whenever the latter exists, so we can make progress. Please see additional material in answer (but it may be quite suboptimal). – ForgotALot Feb 14 '16 at 05:44