1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper

    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook