Define $\phi={a_1+...+a_n\over n}$ and $(a_n)$ be a sequence of real number. Is it true that $\limsup \phi\le\limsup a_n?$
Intuitively i am wondering why the inequality holds as the sup of a sequence could appear all at the begining and form a decreasing sequence so the average seems larger than the tails of the sequence but this is a question I found in some books, so that's not likely to be false.
The answer is yes.
Let's assume that $\limsup a_n =b\neq+\infty$ (The case $b=+\infty$ is easy). Then for $\epsilon>0$ exists $n_0\in\mathbb N$ such that $a_n<b+\epsilon$ for all
$n>n_0$. For any $n>n_0, \ \dfrac{a_1+a_2+\cdots+a_n}{n}<\dfrac{a_1+a_2+\ldots+a_{n_0}}{n}+\dfrac{n-n_0}{n}(b+\epsilon)$.
Now prove that exists $N\in\mathbb N$ such that $\dfrac{a_1+a_2+\cdots+a_n}{n}<b+3\epsilon, \ \forall \ n>N$.
(Hint: If $b+\epsilon>0, \ \dfrac{n-n_0}{n}(b+\epsilon)<b+\epsilon$.
If $b+\epsilon<0$ find $n_1\in\mathbb N$ such that $\dfrac{n-n_0}{n}(b+\epsilon)<b+r\epsilon, \ \forall n\geq n_1$ where $r$ is small enough s.t. $b+r\epsilon<0$ and $1<r<2$.
Use that $\dfrac{a_1+a_2+\ldots+a_{n_0}}{n}\xrightarrow[n\to\infty]{}0$.)
