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$f:\mathbb{R}\longrightarrow \mathbb{R}$ is continuous and $\lim_{|x|\rightarrow\infty}f(x)=\infty$. Show that there is $x_0$ such that $f(x_0)\leq f(x)$ for every $x\in \mathbb{R}$.

My idea

According to the definition, there is some $M>0$ such that $f(x)>1$ for every $|x|>M$. Now, since $f$ takes its minimum in $[-M, M]$, we are done as it would be the minimum of the function.

Am I right?

Majid
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  • Just because the limit on each side approaches $\infty$, each side may grow at different rates so I would suggest looking to using an $M_1, M_2$ rather than $-M, M$ – wjmccann Jul 29 '19 at 00:34
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    Yes this is the idea. I have redacted it in details here https://math.stackexchange.com/questions/2240269/coercive-continuous-function-on-a-closed-subset-has-a-global-minimum-proof/2240390#2240390 – zwim Jul 29 '19 at 00:35

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I believe that I could solve this question. So, I am adding the solution here.

As the function is defined on $\mathbb{R}$, one can consider $A = |f(0)|+1$, and then since $\lim_{|x|\rightarrow\infty}f(x)=\infty$, there is some $M>0$ such that for each $x\notin [-M, M]$, $f(x)>A>f(0)$.

Now, as $f$ is continuous, there is some point $x_0\in [-M, M]$ such that $f(x_0)\leq f(x)$, for each $x\in [-M, M]$. So, as $0\in [-M, M]$, we have

$$f(x_0)\leq f(0)<f(x),\ \forall x\notin [-M, M]$$

So, we are done.

Majid
  • 3,357