prove the existence of a maximum

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December 12th 2012, 09:34 AM
tenderline

prove the existence of a maximum

I apologize if my question is too elementary for this forum, but i can't find a rigorous proof of the following obvious fact: let f be a real function of one real variable, defined and continous on $\mathbb R$ , with $\lim_{|x|\rightarrow +\infty} f(x)=0$ . Then $|f|$ attains a maximum. Which is the standard way to prove this fact?
December 12th 2012, 10:44 AM
earboth

Re: prove the existence of a maximum

Quote:

Originally Posted by tenderline

I apologize if my question is too elementary for this forum, but i can't find a rigorous proof of the following obvious fact: let f be a real function of one real variable, defined and continous on $\mathbb R$ , with $\lim_{|x|\rightarrow +\infty} f(x)=0$ . Then $|f|$ attains a maximum. Which is the standard way to prove this fact?

For a start: Have a look here: Rolle's theorem - Wikipedia, the free encyclopedia

You'll find additional information here: Mean value theorem - Wikipedia, the free encyclopedia
December 12th 2012, 04:43 PM
hollywood

Re: prove the existence of a maximum

You might also need to know that a continuous function on a compact set has a maximum value. But $\mathbb{R}$ is not a compact set. However, you know that for some N, $-\epsilon<f(-x)<\epsilon$ and $-\epsilon<f(x)<\epsilon$ for $x>N$ . So you just need to make sure that $\epsilon$ is less than the maximum value on [-N,N]. Then whatever happens for x<-N and x>N doesn't affect the maximum.

- Hollywood
December 12th 2012, 06:00 PM
Plato

Re: prove the existence of a maximum

Quote:

Originally Posted by tenderline

I apologize if my question is too elementary for this forum, but i can't find a rigorous proof of the following obvious fact: let f be a real function of one real variable, defined and continous on $\mathbb R$ , with $\lim_{|x|\rightarrow +\infty} f(x)=0$ . Then $|f|$ attains a maximum. Which is the standard way to prove this fact?

If $\lim _{x \to \infty } f(x) = 0$ then $\left( {\exists N} \right)\left[ {x \geqslant N \Rightarrow f(x) \le 1} \right]$

If $\lim _{x \to -\infty } f(x) = 0$ then $\left( {\exists M} \right)\left[ {x \le -M \Rightarrow f(x) \le 1} \right]$

So if $K=\max\{N,M\}$ then $f$ has a maximum on $[-K,K]$ .
December 12th 2012, 11:06 PM
tenderline

Re: prove the existence of a maximum

Quote:

Originally Posted by Plato

If $\lim _{x \to \infty } f(x) = 0$ then $\left( {\exists N} \right)\left[ {x \geqslant N \Rightarrow f(x) \le 1} \right]$

If $\lim _{x \to -\infty } f(x) = 0$ then $\left( {\exists M} \right)\left[ {x \geqslant M \Rightarrow f(x) \le 1} \right]$

So if $K=\max\{N,M\}$ then $f$ has a maximum on $[-K,K]$ .

than you Plato for your solution. I have two questions:
1) i think in the second identity we have to replace $x\geq M$ with $x\leq -M$ , secondly i think the conclusion is subject to the restriction: maximum on $[-K,K]\geq 1$ . Is it right what i'm saying?
2) Can we extend this argument to the plane? I mean, let f be a continous (differentiable if necessary) function of two real variables x,y, such that $\lim_{|y|\rightarrow +\infty}f(x,y)=0$$ , uniformly with respect to x. Can we conclude that f attains a maximum?
Thanks in advance for any suggestion.