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So limit goes to infinity, and I have to show that there exists a finite infimum. how do i show this?

Dafty
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2 Answers2

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Think of a chain you hold by both ends very high. It has to ''drop down'' somewhere. That's roughly the idea.

So let's hold that rope ''very high''. For every $M > 0$, there exists $K > 0$ such that for every $x > K$, $f(x) > M$ and for every $x < -K$, $f(x) > M$ ; in other words, for every $x$ satisfying $|x| > K$, $f(x) > M$. Assuming $M$ is large enough so that $f^{-1}(]-\infty,M])$ is non-empty, then we know the infimum of $f$ is inside $[-K,K]$ because outside this interval, $f(x) > M$.

Now $[-K,K]$ is a compact interval. Since $f$ is continuous we can find the minimum on it. :)

For the example of the function $g$, you have very simple examples : for instance, take

$$ g(x) \overset{def}= \begin{cases} -x-2 & \text{ if } x < -1 \\ \\ \frac 1x & \text{ if } 0 < |x| < 1 \\ \\ 0 & \text{ if } x = 0 \\ \\ x & \text{ if } x > 1. \end{cases} $$ I'm basically just taking what happens with $\frac 1x$, satisfying your conditions, and then "holding the function with a rope" (the two linear parts).

Hope that helps,

Patrick Da Silva
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    Thank you Patrick. I can understand now... And I really appreciate for the example!!! – Dafty Oct 18 '13 at 07:01
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Hint: On each $[-n, n]$, the function attains a minimum, since $[-n, n]$ is compact and $f$ is continuous.

Since $f(x) \to \infty$ as $x \to \pm \infty$, there exists an $N$ such that

$$|x| > N \implies f(x) > f(0)$$

Can you combine these two facts?

  • hmmm.. I am almost getting there. but I am not sure how to combine? – Dafty Oct 18 '13 at 06:54
  • This shows that $f$ must be large on $[-N, N]^c$, and so the minimum would have to be between $-N$ and $N$. –  Oct 18 '13 at 06:56
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    A-ha....... wait so just let me try to understand. Since absolute value of x is bigger than N, f must be large on [-N,N]^c is that correct? – Dafty Oct 18 '13 at 06:58
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    @Dafty : Large enough to be bigger than the minimum, since the minimum exists on $[-N,N]$ and must be smaller than $f(0)$. – Patrick Da Silva Oct 18 '13 at 07:01
  • Or it is $f(0)$ itself.Then you are ready too off course – drhab Oct 18 '13 at 07:03
  • @drhab : I didn't say strictly smaller. – Patrick Da Silva Oct 18 '13 at 07:13
  • @PatrickDaSilva Sorry, I am still not used to this mathematical way of speaking (proper, strictly). Small categories are large, you can say. In the 'real' world people think that something is wrong with you if you say things like that. – drhab Oct 18 '13 at 07:16
  • @drhab : In the real world, if you said that, I would answer exactly in the same way. It's the same thing as "positive" means $>0$, but sometimes people say it to mean $\ge 0$, and then from the context, if a $\ge$ is on the blackboard then you know positive meant $\ge$, even though the person meant non-negative. These things happen. Don't worry about it. But I assure you that "smaller" usually refers to $\le$ and strictly smaller to $<$. – Patrick Da Silva Oct 18 '13 at 08:49