Convergence of a monotone type sequence

I'm stuck with showing the existence of a limit, all I know essentially is that for $g_t \in \mathbb R$, $0 \leq g_{t+C} \leq g_t \leq A \leq \infty$ for all $t \in \mathbb R$ and some $C \in (0,\infty)$, does the limit necessarily exist under this condition? Clearly the sequence has a convergent subsequence, I was hoping that this could be strengthened to the whole sequence.

Also, if this is not true, then what about if $g_t \in \mathbb R$, $0 \leq g_{t+C} \leq g_t \leq A \leq \infty$ for all $t \in \mathbb R$ and for all $C \geq B > 0$.

Thanks.

-
Do you mean to ask whether the limit of $g_t$ as $t \to \infty$ exists, where the limit is over a real $t$ approaching infinity? If so maybe if the map $t \to g_t$ is assumed continuous you could use the bounds on $g_t$ over each "period" $[t,t+C]$ to get convergence. – coffeemath Nov 7 '12 at 12:06
Assuming that you mean the limit for $t\rightarrow \infty$ of $g(t)$ then the answer to your first question is no. Consider for example $g(t):= 1+ \sin(t)$. While for the second question you are essentially cutting out this (periodic) possibility... – Giovanni De Gaetano Nov 7 '12 at 12:16
@coffeemath: I am indeed asking for $\lim_{t \rightarrow \infty} g_t$, essentially, the limit exists and that limit is independent of any sequence $t_j$ that would be used to compute the limit. I do not understand your suggestion, could you explain the connection between the bounds and convergence? – dcs24 Nov 7 '12 at 12:17
@GiovanniDeGaetano Thanks for the first counter example, I don't suppose you have any suggestion on the second part? – dcs24 Nov 7 '12 at 12:19
@dcs24 For the second part let $g := \inf_{t > 0} g_t$. For $\epsilon > 0$, choose $t > 0$ with $g_t < g + \epsilon$, then $g \le g_s < g + \epsilon$ for all $s$ with ... – martini Nov 7 '12 at 12:43

A counterexample for the first part has been given by @Giovanni: If we let $g_t = 1 + \sin(t)$ we have $0 \le g_{t + 2\pi} = g_t \le 2 < \infty$ for all $t \in \mathbb R$, but $\lim_{t\to\infty} g_t$ doesn't exist, as $$\lim_{n\to\infty} g_{n\pi} = 1 \ne 0 = \lim_{n\to\infty} g_{-\frac{\pi}2 + 2n\pi}.$$
For the second part, define $g := \inf_{t \ge 0} g_t$. For arbitrary $\epsilon > 0$ we can choose $t \ge 0$ with $g\le g_t < g + \epsilon$. For any $s \in \mathbb R$ with $s \ge t + B$, we have $s = t + (s-t)$ with $s-t \ge B$, hence $g \le g_s \le g_t < g+\epsilon$. So $g_s \in [g, g+\epsilon)$ for all $s \ge t + B$. As $\epsilon > 0$ was arbitrary $\lim_{t\to \infty} g_t = g$.