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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.

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    Go for it, it would be wrong for me to take any credit2012-11-07

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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 $\lim_{t\to \infty} g_t = g$.