Consider sequence $\{ a_i \}_i$ such that $\lim_{n \to +\infty} \frac{1}{n}\sum_{i=1}^{n}a_i = A$. Assume that $f$ is a continous bounded function. Whether the sequence $\{ b_i \}_i$ with $b_i= \frac{1}{i}\sum_{j=1}^{i}f(a_j)$ is also convergent?
I already known that if $\lim_{n \to +\infty} a_n = A$, then $\lim_{n \to +\infty} b_n = f(A)$. But i want to know whether it still true when we only know that the average limit exists.
For example, assume that $0\leq a_i \leq 1$ and $f(x)=x \ln x$.
If we don't require $(a_n)$ to be bounded, then we can use $$a_n = (-1)^n \sqrt{\lfloor n/2\rfloor},$$ for which we note $$0 \leq \sum_{i=1}^n a_i \leq \sqrt{n/2},$$ so that $$\lim_{n\to\infty} \frac1n \sum_{i=1}^n a_i = 0.$$
The benefit here is that $\{a_n : n \geq 1\}$ is a discrete closed set and $i \neq j \implies a_i \neq a_j$, so any bounded function $g : \mathbb{N} \to \mathbb{R}$ gives rise to a bounded continuous function $f : \mathbb{R}\to \mathbb{R}$ such that $f(a_i) = g(i)$.
This transforms your question into the much more basic one:
If $b_n$ is a bounded sequence of real numbers, does $\lim_{n\to\infty} \frac1n \sum_{i=1}^n b_i$ necessarily converge?
for which the answer is "no".
The answer is no. To construct a counterexample, note that if $\{a_i\}$ is any bounded sequence such that $a_{2j}=-a_{2j-1}$ for all $j\ge1$, then $\lim_{n\to\infty} \frac1n \sum_{i=1}^n a_i = 0$ exists.
So take $f(x) = |x|$ (bounded on the interval $[-2,2]$), and take $$ a_i = \begin{cases} (-1)^i, &\text{if $i-1$ has an odd number of digits}, \\ 2(-1)^i, &\text{if $i-1$ has an even number of digits}. \end{cases} $$ Then $\lim_{n\to\infty} \frac1n \sum_{i=1}^n a_i = 0$ as above. However, \begin{align*} \liminf_{n\to\infty} & \frac1n \sum_{i=1}^n |a_i| \le \liminf_{k\to\infty} \frac1{10^{2k+1}} \sum_{i=1}^{10^{2k+1}} |a_i| \\ &= \liminf_{k\to\infty} \frac1{10^{2k+1}} \bigl( 10 + 2(90) + 900 + 2(9{,}000) + \cdots + 2(9\cdot10^{2k-1}) + 9\cdot10^{2k} \bigr) \\ &= \liminf_{k\to\infty} \frac1{10^{2k+1}} \biggl( \frac{12\cdot 10^{2k+1} - 10}{11} \biggr) = \frac{12}{11}, \end{align*} while \begin{align*} \limsup_{n\to\infty} & \frac1n \sum_{i=1}^n |a_i| \ge \limsup_{k\to\infty} \frac1{10^{2k+2}} \sum_{i=1}^{10^{2k+2}} |a_i| \\ &= \limsup_{k\to\infty} \frac1{10^{2k+2}} \bigl( 10 + 2(90) + 900 + 2(9{,}000) + \cdots + 9\cdot10^{2k} + 2(9\cdot10^{2k+1}) \bigr) \\ &= \limsup_{k\to\infty} \frac1{10^{2k+2}} \biggl( \frac{21\cdot 10^{2k+2} - 10}{11} \biggr) = \frac{21}{11}. \end{align*}
(The basic phenomenon is that we have arranged for the sequence $\{|a_i|\}$ to oscillate between ever-lengthening runs of two different values, which is what keeps its sequence of averages from converging.)
