Arithmetic Mean convergence and concave transformations Suppose $x_i \geq 0$ $\forall i$. If $\frac{1}{n}\sum_{i=1}^nx_i$ converges as $n\to\infty$, then what about $\frac{1}{n}\sum_{i=1}^nx_i^\delta$ where $0<\delta<1$?
By Jensen's inequality we have that $\frac{1}{n}\sum_{i=1}^nx_i^\delta \leq \left(\frac{1}{n}\sum_{i=1}^nx_i\right)^\delta$. This of course implies that $\limsup_{n\to\infty}\frac{1}{n}\sum_{i=1}^nx_i^\delta \leq \left(\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^nx_i\right)^\delta$, but what can we say about its $\liminf$?
Instead of $x_i^{\delta}$ we could have $g(x_i)$ for any concave function $g$.
 A: Here is a negative answer, but with an indirect construction:
Construction. We prepare the setting as follows.


*

*$X$ and $Y$ are non-negative rv's such that $\Bbb{E}X = \Bbb{E}Y = \alpha < \infty$ and $\Bbb{E}X^{\delta} \neq \Bbb{E}Y^{\delta}$.

*Generate independent rv's $X_1, Y_1, X_2, Y_2, \cdots$ such that $X_n \sim X$ and $Y_n \sim Y$.

*$(m_k)$ and $(n_k)$ are non-decreasing sequence of non-negative integers such that $m_k + n_k = k$ and $m_k, n_k \uparrow \infty$.

*Define $Z_1, Z_2, \cdots$ so that
$$ Z_1 + \cdots Z_k = (X_1 + \cdots + X_{m_k}) + (Y_1 + \cdots + Y_{n_k}) $$
for any $k$. As $k$ increases by 1, precisely one of $m_k$ or $n_k$ increases by 1, hence $Z_k$ are well-defined rv's. (In particular, $Z_k$ will be either $X_{m_k}$ or $Y_{n_k}$.)


Then the Strong Law of Large Numbers (SLLN) shows that
$$ \frac{1}{k} \sum_{i=1}^{k} Z_i
= \left( \frac{m_k}{k} \cdot \frac{1}{m_k} \sum_{i=1}^{m_k} X_i + \frac{n_k}{k} \cdot \frac{1}{n_k} \sum_{i=1}^{n_k} Y_i \right)
\rightarrow \alpha \quad \text{a.s.}$$
But at the same time, by SLLN again,
$$ \frac{1}{m_k} \sum_{i=1}^{m_k} X_i^{\delta} \to \Bbb{E}X^{\delta} \quad \text{and} \quad \frac{1}{n_k} \sum_{i=1}^{n_k} Y_i^{\delta} \to \Bbb{E}Y^{\delta} \quad \text{a.s.} $$
So by choosing $m_k$ and $n_k$ so that $(m_k/k)$ oscillates as $k \to \infty$, we can make
$$ \frac{1}{k} \sum_{i=1}^{k} Z_i^{\delta}
= \left( \frac{m_k}{k} \cdot \frac{1}{m_k} \sum_{i=1}^{m_k} X_i^{\delta} + \frac{n_k}{k} \cdot \frac{1}{n_k} \sum_{i=1}^{n_k} Y_i^{\delta} \right)$$
divergent almost surely.
Remark. You see that the probabilistic technicality is not that crucial to our construction. Indeed, all we need are two non-negative sequences $x_k$ and $y_k$ such that 
$$ \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n} x_k = \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n} y_k \quad \text{but} \quad \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n} x_k^{\delta} \neq \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n} y_k^{\delta}. $$
I just wanted to avoid an explicit construction of such sequences, but you may figure out concrete examples.
A: See comment -- does not work (obvious, stupid mistake)
Another counter-example, based on two different (and interleaved) sequences $a,b$ as well:
$a_k = e$ for all $k\geq 1$, while $b_k = \left(1+\frac{1}{k}\right)^k$. Build the sequence $(x_k)_k$ by alternating $a_1,b_1,b_2,a_2,a_3,a_4,1_5,\dots$, i.e. by taking $2^i$ $a_k's$ following by $2^{i+1}$ $b_k$'s.
Since $b_k\xrightarrow[k\to\infty]{} e$, by Césaro the sequence $(S_n)_n$ defined by $S_n=\frac{1}{n}\sum_{k=1}^n x_k$ will converge to $e$ as well. But now, $b_k^\delta \xrightarrow[k\to\infty]{} e^{\delta}$, so Césaro does not apply to $S_{n,\delta}=\frac{1}{n}\sum_{k=1}^n x_k^\delta$; and you can actually show that $S_{n,\delta}$ has two adherence values, namely $e$ and $e^{\delta}$.
