# If there is a $T$ such that $V(t)<V(t-T) \ \forall t$, does that imply $V(t) \to 0$?

Let $V(t)$ denote a continuous scalar function $\mathbb{R} \mapsto \mathbb{R}$. Assume that we can find a constant $T \in \mathbb{R}$ such that $V(t)<V(t-T)$ for all $t$. Does that imply that $V(t) \to 0$ as $t \to \infty$?

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Any strictly decreasing function would satisfy your condition for any $T>0$. –  David Mitra Jan 29 '13 at 17:19
indeed, but the class of functions that satisfy the mentioned condition includes much more functions than strictly decreasing functions. –  mike Jan 30 '13 at 13:59
So, one can then find a function, $V$, which satisfies the condition, but such that $V(t)$ does not converge to $0$ as $t\rightarrow\infty$ (for example, $v(t)=-t$). –  David Mitra Jan 30 '13 at 14:02
Thank you, yes that is a counter example. what can be said if we constrain $V$ such that $V(t)>0 \ \forall t$? –  mike Jan 30 '13 at 14:24
No, pick any number $a \in \mathbb R$ and let $V(t) = e^{-t} + a$. This function is monotonically decreasing so for all $t$ we have $V(t) < V(t - 1)$ but $V(t) \to a$ as $t \to \infty$.
and if $V$ is bounded below, e.g. $V(t)>0 \forall t$? –  mike Jan 30 '13 at 19:56
I believe it still doesn't have to be the case that $V(t)$ converges as $t \to \infty$ but the counterexamples become harder to describe with non-obvious details to check. –  Jim Jan 30 '13 at 20:26