Can I define a bounded sequence whose Banach limit is not unique? Banach limit, as a non-constructive object, is not unique. The Banach limit for some sequences, say, convergent sequences, sequences satisfying $a_n = a_{n+m}$ for all $n$ and some $m$, the Banach limit is uniquely determined, but there also exists some bounded sequence $\{a_n\}_{n \in \mathbb N}$ whose Banach limit is not unique. Can we formally construct such a sequence, say, $a_n = f(a_{n-1})$, with a explict function $f$, but its Banach limit is not unique for all Banach limit?
 A: There is a result (IIRC due to Lorenz) which says that for a given sequence $x$ the Banach limits can obtain all values between $m(x)$ and $M(x)$ where
$$
\begin{gather*}
M(x)=\lim_{n\to\infty} \limsup T_n(x),\\
m(x)=\lim_{n\to\infty} \liminf T_n(x),
\end{gather*}
$$
where $T_n(x)$ is the sequence $\left(\frac{x_k+x_{k+1}+\dots+x_{k+n-1}}n\right)_{k=1}^\infty$.
So it suffices to find a sequence such that $m(x)<M(x)$. For this it suffices to take a sequence of zeroes and ones where there are long enough blocks of 0's and long enough blocks of 1's.
See also this answer.
You can also find some more information if you look for almost convergent sequences - they are precisely the sequences for which the Banach limit is uniquely determined. (So your question can be reformulated as looking for an example of a sequence which is not almost convergent.)
A: No, $\sin n$ does not work.  You can use the Lorenz criterion (see Martin's answer).  In fact, if $a_n = \sin \log n$ were almost convergent (to value $L$), then in particular
$$
\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n \sin \log k = L
$$
But
$$
\frac{1}{n}\sum_{k=1}^n \sin \log k - \frac{1}{n}\int_1^{n+1}\sin\log x\;dx
$$
goes to zero, so we may show the limit with integrals does not exist.
This is easy:
$$
\int\sin\log x\;dx = \frac{x\sin\log x - x \cos\log x}{2} +C
$$
and we get
$$
\limsup_{n\to\infty}\; \frac{1}{n}\int_1^{n+1}\sin\log x\;dx  =\frac{1}{\sqrt{2}}
\\
\liminf_{n\to\infty}\; \frac{1}{n}\int_1^{n+1}\sin\log x\;dx = -\frac{1}{\sqrt{2}}
$$
