prove the existence of a maximum

• Dec 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 $\displaystyle \mathbb R$, with$\displaystyle \lim_{|x|\rightarrow +\infty} f(x)=0$. Then $\displaystyle |f|$ attains a maximum. Which is the standard way to prove this fact?
• Dec 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 $\displaystyle \mathbb R$, with$\displaystyle \lim_{|x|\rightarrow +\infty} f(x)=0$. Then $\displaystyle |f|$ 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
• Dec 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 $\displaystyle \mathbb{R}$ is not a compact set. However, you know that for some N, $\displaystyle -\epsilon<f(-x)<\epsilon$ and $\displaystyle -\epsilon<f(x)<\epsilon$ for $\displaystyle x>N$. So you just need to make sure that $\displaystyle \epsilon$ is less than the maximum value on [-N,N]. Then whatever happens for x<-N and x>N doesn't affect the maximum.

- Hollywood
• Dec 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 $\displaystyle \mathbb R$, with$\displaystyle \lim_{|x|\rightarrow +\infty} f(x)=0$. Then $\displaystyle |f|$ attains a maximum. Which is the standard way to prove this fact?

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

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

So if $\displaystyle K=\max\{N,M\}$ then $\displaystyle f$ has a maximum on $\displaystyle [-K,K]$ .
• Dec 12th 2012, 11:06 PM
tenderline
Re: prove the existence of a maximum
Quote:

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

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

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

than you Plato for your solution. I have two questions:
1) i think in the second identity we have to replace $\displaystyle x\geq M$ with $\displaystyle x\leq -M$, secondly i think the conclusion is subject to the restriction: maximum on $\displaystyle [-K,K]\geq 1$. 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 $\displaystyle \lim_{|y|\rightarrow +\infty}f(x,y)=0$\$, uniformly with respect to x. Can we conclude that f attains a maximum?
Thanks in advance for any suggestion.