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I got this question:

Let $f : [0, \infty) \to \mathbb{R}$ be a function such that $ \lim_{x \to \infty}f(x) = 0$ such that $f$ is monotonically decreasing on $[0,\infty)$, must it be the case that for all $x \in [0,\infty), f(x)>0$ ?

From an intuitive point of view it sounds true but I tried to prove it but failed so far.

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  • $\begingroup$ True with $f(x)\geq 0$ $\endgroup$ – Sylvain Biehler Mar 19 '14 at 15:21
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We have $f(x)\ge f(y) $ when $x \le y$. Hence $f(x) \ge \lim_{y \to \infty} f(y) = 0$ for all $x$.

If the relationship is strict, then so is the conclusion, for the same reason.

Elaboration: Let $\epsilon>0$, then there is some $M$ such that if $y \ge M$, then $-\epsilon < f(y) < \epsilon$. If we take $y \ge \max(M,x)$ we get (by monotonicity) $f(x) \ge f(y) > -\epsilon$. Hence $f(x) > -\epsilon$ for all $\epsilon>0$. From this we conclude $f(x) \ge 0$.

If the relationship is strict, then $f(x) > f(x+1)$. By the above reasoning, we have $f(x+1) \ge 0$, hence $f(x) > 0$.

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    $\begingroup$ this captures the intuition perfectly. much proof. such satisfying. many beautiful. so $\lim$. wow. $\endgroup$ – Guy Mar 19 '14 at 15:25
  • $\begingroup$ The inequality should be strict! Anyway the proof is nice. $\endgroup$ – user63181 Mar 19 '14 at 15:31
  • $\begingroup$ Nice solution, but how do I prove that for all $x \in [0,\infty), f(x) \geq \lim_{y \to \infty} f(y)$ by using the definition of limit? $\endgroup$ – MathNerd Mar 19 '14 at 15:37
  • $\begingroup$ I will add an elaboration. $\endgroup$ – copper.hat Mar 19 '14 at 15:38
  • $\begingroup$ Surely your proof (although very nice) does not work for strict inequalities, as taking the limit only preserves regular inequality, not strict inequality. $\endgroup$ – Joshua Pepper Mar 19 '14 at 15:45
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Assume that there's $x_0\ge 0$ such that $f(x_0)\le0$ so for $x_1>x_0$ we have $$f(x_1)<0$$

Let $\epsilon=-f(x_1)>0$, from the definition of the limit there's $A>0$ such that for all $x>A$ we have $$f(x)\ge-\epsilon= f(x_1)$$ and the last inequality gives an obvious contradiction. We conclude that $$f(x)>0\quad \forall x\ge0$$

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