I have been trying to show that if a function $f: \mathbf{R} \rightarrow \mathbf{R}$ is decreasing and bounded from below then the $\lim\limits_{t \rightarrow \infty}f(t)$ exists. In the case of a sequence which is decreasing & bounded from below, the proof is simple. I think I managed to do it in this case but was a bit complicated and I am wondering if there is a simpler way. I proceeded as follows:
- Let $\{t_{n}\}$ be an increasing sequence with $t_{n} \rightarrow \infty$. Then the sequence $\{f(t_{n})\}$ has a limit, call it $k$.
- Then show that for each $\epsilon$ there are numbers $T(\epsilon)$ and $N_{1}(\epsilon)$ such that $t > T(\epsilon), n > N_{1}(\epsilon)$ implies $\|f(t) – f(t_{n})\| \leq \epsilon$.
- Since the $f(t_{n})$ converge, there is an $N_{2}(\epsilon)$ such that when $n>N_{2}(\epsilon)$ we have $\|f(t_{n}) – k\| \leq \epsilon$.
- For an arbitrary $\epsilon$, let $N = \max\{N_{1}(\epsilon), N_{2}(\epsilon)\}$. Then for $t > T(\frac{\epsilon}{2})$ and an arbitary $n > N$, we have $\|f(t) – k\| \leq \|f(t) – f(t_{n})\| + \|f(t_{n}) – k\| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ and we are done.
Is there a better way to do this? Are there stronger assumptions on $f$ that would simplify this (e.g. continuity)?
Best Answer
I wonder if you like this: the set $A=\{f(x):x\in\mathbb{R}\}$ is bounded from below and thus admits an infimum $L$. We show that $\lim_{t\to\infty} f(t)=L$.
Let $\epsilon>0$ be given. Then because $L+\epsilon>L$, there is some $x_0$ such that $f(x_0)\in[L,L+\epsilon)$. Because $f$ is decreasing, we might increase $x_0$ if necessary to make it strictly positive. Using monotonicity once more, for all $x\geq x_0$, we have $L\leq f(x)\leq f(x_0)<L+\epsilon$. Thus, for given $\epsilon>0$, we have produced $x_0>0$ such that for all $x\geq x_0$, we have $f(x)\in(L-\epsilon,L+\epsilon)$.
Edit: after typing, I realized this is also a hint that Greg Martin left in his comment.