Theorem by Cesàro: how to prove?

Question

The following was introduced to me in class as «an old theorem, almost forgotten, by an old mathematician, Cesàro, 1960».

Theorem

Suppose $\{a_n\}$ is a real sequence. Set:

$$A_n=\frac{1}{n+1}\sum_{k=1}^na_k.$$

Then:

$$\liminf_na_n\leq\liminf_NA_N\leq\limsup_NA_N\leq\limsup_na_n.$$

I tried googling for it, but was unable to find it. I cannot seem to be able to prove it myself. How would I go about this?

Not sure what you mean by 1960, but: "Ernesto Cesàro (March 12, 1859 – September 12, 1906)" — Thomas Andrews, Apr 13 '17 at 17:32
As written, it doesn't seem to be true - I think you might mean $$\liminf a_n\leq \liminf \frac{A_n}{n}\leq \limsup \frac{A_n}{n}\leq \limsup a_n$$ or maybe you meant $$A_n=\frac{1}{n}\sum_{k=1}^{n} a_k$$ — Thomas Andrews, Apr 13 '17 at 17:33
@ThomasAndrews: comment 1: that's probably my teacher getting the year wrong :); comment 2: not quite, but I had an error, edited to fix. — MickG, Apr 13 '17 at 17:37
Hardly forgotten. Cesàro means are well known in real analysis, especially summability theory. — Robert Israel, Apr 13 '17 at 17:37
There's really no difference between having a factor $\frac{1}{n}$ and a factor of $\frac{1}{n+1}$ in the limit. — Thomas Andrews, Apr 13 '17 at 17:38
Yeah, I wonder if your professor meant that Cesàro himself was "almost forgotten," not the theorem. — Thomas Andrews, Apr 13 '17 at 17:38
@ThomasAndrews again, that is just what my teacher said to introduce this statement. As for the factors, yeah I figured that, the ratio tends to 1 when $n\to\infty$. — MickG, Apr 13 '17 at 17:39
Possibly. But the way the sentence is phrased seems to refer "almost forgotten" to the theorem. I don't know anyways. — MickG, Apr 13 '17 at 17:40
Are you sure it's $\frac1{n+1}$? It looks funny to me. It won't alter the asymptotic but makes many arguments more complicated than necessary. — user251257, Apr 13 '17 at 17:43

score 1 · Accepted Answer · answered Apr 13 '17 at 17:55

We'll prove that $$\liminf a_n\leq \liminf A_n$$

If $\liminf a_n=-\infty$ this is obviously true.

Otherwise, let $\alpha<\liminf a_n$. We will show that $\alpha\leq\liminf A_n$.

By the definition of $\liminf$, this means that there exists a $K$ such that for all $n>K$, $\alpha<a_n$.

Now, if $N>K$ then:

$$A_N = \frac{1}{N+1}\left((K+1)A_K+\sum_{n=K+1}^{N} a_k\right)>\frac{1}{N+1}\left((K+1)A_K+(N-K)\alpha\right)$$

Now, $\lim_{N\to\infty} \frac{(K+1)A_K}{N+1} =0$. And $\lim_{N\to\infty} \frac{N-K}{N+1}=1$. So $\liminf A_n\geq \alpha$.

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