Does $\frac{1}{n}\sum_{i=1}^n|x_i|\to L<\infty$ imply $\frac{1}{n}\max_{1\leq i\leq n}|x_i|=0$? To simplify notation, let us assume that $\{x_n\}_{n\geq 1}$ is a sequence of nonnegative real numbers. Does
$$
\frac{1}{n}\sum_{i=1}^nx_i\to L
$$
for some finite $L$ imply
$$
\frac{1}{n}\max_{1\leq i\leq n}x_i\to 0?
$$
I have been thinking about this question for the past few hours because of here but I can't come up with anything. If you can help me, please feel free to head there to resolve that post as well.
 A: Yes. 
Assume for contradiction that for some $\varepsilon > 0$ there are arbitrarily large $n$ such that 
$$\frac{1}{n} \max_{1 \leqslant i \leqslant n} x_i \geqslant \varepsilon,$$ 
i.e. 
$$\max_{1 \leqslant i \leqslant n} x_i \geqslant n \varepsilon.$$ 
From there it follows (it takes some proof though) that there are arbitrarily large $n$ such that $x_n \geqslant n \varepsilon$. Therefore 
$$ \frac{x_1 + \ldots + x_n}{n} \geqslant \frac{x_1 + \ldots + x_{n-1}}{n} + \varepsilon = \frac{n-1}{n} \cdot \frac{x_1 + \ldots + x_{n-1}}{n-1} + \varepsilon$$
so $L \geqslant 1 \cdot L + \varepsilon$ and we have a contradiction. 
A: Based on the proof of Adayah, I came up with the following direct proof. I post it as a community wiki post, because the original idea is due to Adayah:
Let $\varepsilon > 0$. Take $n_0 \in \Bbb{N}$ such that
$$
\left| \frac{1}{n}\sum_{i=1}^n x_i - L \right| < \varepsilon \text{ and } \left| L - \frac{n}{n+1} L \right| < \varepsilon.
$$
This yields
\begin{align*}
\left|\frac{x_{n+1}}{n+1}\right| & =\left|\frac{1}{n+1}\sum_{i=1}^{n+1}x_{i}-\frac{1}{n+1}\sum_{i=1}^{n}x_{i}\right|\\
 & \leq\left|\frac{1}{n+1}\sum_{i=1}^{n+1}x_{i}-L\right|+\left|L-\frac{n}{n+1}L\right|+\frac{n}{n+1}\left|L-\frac{1}{n}\sum_{i=1}^{n}x_{i}\right|\\
 & <3\varepsilon
\end{align*}
for all $n \geq n_0$ and hence
$$
\frac{1}{n} \max_{1 \leq i \leq n} x_i \leq \frac{1}{n} \max_{1 \leq i \leq n_0} x_i + \max_{n_0 + 1 \leq i \leq n} \frac{x_i}{n} \leq \frac{1}{n} \max_{1 \leq i \leq n_0} x_i + \max_{n_0 + 1\leq i \leq n} \frac{x_{i}}{i} < \varepsilon + \frac{1}{n} \max_{1 \leq i \leq n_0} x_i \to \varepsilon,
$$
which establishes $\limsup_n \frac{1}{n} \max_{1 \leq i \leq n} x_i = 0$, because $\varepsilon > 0$ was arbitrary.
