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Math Help - prove the existence of a maximum

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    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?
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    Re: prove the existence of a maximum

    Quote Originally Posted by tenderline View Post
    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
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    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.


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    Re: prove the existence of a maximum

    Quote Originally Posted by tenderline View Post
    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] .
    Last edited by Plato; December 13th 2012 at 04:20 AM.
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    Re: prove the existence of a maximum

    Quote Originally Posted by Plato View Post
    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.
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