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A y such that f(x) <= f(y) all x

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    if ##f## is a continuous function, ##f(x) > 0## for all x, and
    [tex]
    \lim_{x\rightarrow +\infty} {f(x)} = 0 = \lim_{x\rightarrow -\infty} {f(x)}
    [/tex]
    Prove: There is a y such that ##f(x) \leq f(y)## for all x

    I'm assuming that the domain of f is ℝ.

    2. Relevant equations
    3. The attempt at a solution

    Since
    [tex]
    \lim_{x\rightarrow +\infty} {f(x)} = 0 = \lim_{x\rightarrow -\infty} {f(x)}
    [/tex]
    -There is an ##N_1 < 0## such that, for all x, if x < ##N_1## then [itex] |f(x)| = f(x) < f(-1)[/itex]
    -There is an ##N_2 > 0## such that, for all x, if x > ##N_2## then [itex] |f(x)| = f(x) < f(+1)[/itex]

    I'm not 100% sure but, this is what I have in mind:
    Let ##a## be any number such that ##a < N_1##, ## - 1## and ##b## any number such that ##b > N_2##, ## 1##.
    Since f is continuous on [a, b], there is a y in [a, b], such that, ##f(x) \leq f(y) ## for all x in [a, b]
    Now, if ##x < a## then ##f(x) < f(-1) \leq f(y) ## and if ##x > b## then ##f(x) < f(1) \leq f(y)##

    So ##f(x) \leq f(y)## for all x
     
  2. jcsd
  3. Feb 26, 2013 #2

    haruspex

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    Why single out -1? What's special about that?
     
  4. Feb 26, 2013 #3
    Well, nothing. I used -1 and 1 because they are simple. The "proof" is basically the same with other numbers.
     
  5. Feb 26, 2013 #4

    jbunniii

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    I think your proof is fine, but it can be decluttered a bit. I don't see why you need both ##f(-1)## and ##f(1)##. Wouldn't it still work if you replaced them both with ##f(0)##? Also, why not just use the interval ##[N_1, N_2]## instead of introducing ##a## and ##b##?
     
  6. Feb 26, 2013 #5
    Of course!!! I can't believe that I didn't do that.

    Thanks.
     
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