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Every function on [0,1] attains its maximum

  1. Nov 14, 2011 #1
    1. The problem statement, all variables and given/known data
    True or False: Every function on [0,1] attains its maximum

    I was working on a practice midterm, and this question stumped me

    3. The attempt at a solution
    I know that by "every function on [0,1]" means that dom f = [0,1] but what exactly does it mean by "attains its maximum" simply that it has a limit, as in, it doesn't approach +∞?

    If this is what it means, it would be false, correct? Because you could have a function that approaches ±∞ as it approaches 1 from below.

    For instance, take f(x)=[itex]\frac{1}{1-x}[/itex]
  2. jcsd
  3. Nov 14, 2011 #2


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    The question presumes the function has a maximum value, and asks if that maximum value is in the function's image.

    It's a bit oddly phrased, since a maximum is, by definition, in the function's image. It's either a 'trick' question, or they meant "has a maximum" or "attains its supremum". Or, you learned a slightly non-standard definition of those terms in class.

    That's not a function on [0,1].
  4. Nov 14, 2011 #3


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    The main point is that [0, 1] is a closed interval.

    It's certainly not true that "Every function on (0, 1) attains its maximum."
  5. Nov 14, 2011 #4
    I think the question really means every bounded function on [0,1] attains its least upper bound. But I may be wrong.
  6. Nov 14, 2011 #5


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    Funny nobody has mentioned the word 'continuous' yet. All continuous functions on [0,1] have an upper bound and attain it. The given problem didn't say 'continuous'. So it's false.
  7. Nov 14, 2011 #6
    I would take that to mean there is a [itex]c[/itex] in [itex][0, 1][/itex] such that [itex]\forall x[/itex] in [itex][0, 1], f(x)\leq f(c)[/itex].

    Certainly this is true for all continuous functions (Use the fact that if K is compact, f(K) is also compact for continuous f, and [0, 1] is compact. It can be shown that compact sets have a maximum and minimum value).

    For discontinuous functions though, take
    [itex]\begin{align*}f(x)&=\frac{1}{x} \text{ when }x≠0\\ &=0 \text{ when } x=0\end{align*}[/itex]

    Then [itex]f(x)[/itex] does not attain it's maximum on [itex][0, 1][/itex]
  8. Nov 15, 2011 #7
    Thanks guys! This helped a lot!
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