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Mean value theorem

  1. Jan 13, 2005 #1
    Prove: If f : U → R is continuous on U, and E ⊂ U is closed and bounded, then
    f attains an absolute minimum and maximum on E.

    I have no idea even where to start on this. Intuitively it's so obvious that i don't know what to do. The definitions given by the teachers that I have to work with are as follows:

    A set F ⊆ Rn is closed if, for every convergent sequence {xi}(from i=1 to infinity) ⊆ F,
    we have limxn(as n goes to infinity)⊆F
    (in other words it contains its limit points)

    A bounded set is one for which there exists r such that the set is contained in Nr(0).
    (in other words some ball around the origin of any size contains the set.)

    F continuous on u means for all c in F, lim(as x approaches c) exists and equals F(c).

    A compact set is one which is closed and bounded, so E in this proof is compact.

    I know that what needs to be shown is that there exists xm and xM such that:
    f(xm)<=f(x)<=f(xM) for all x in E.

    any advice? hints? etc.. on how to start this? I suspect that the mean value theorem might have something to do with it, but I'm not sure how to incorporate it. Thanks for any help.
     
  2. jcsd
  3. Jan 13, 2005 #2

    Galileo

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    Hint: Consider the supremum of f(E).
     
  4. Jan 13, 2005 #3

    matt grime

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    Depending on your familiarity with these ideas:

    Let K be the image of E, under F. Let C be a covering of K by open sets. Consider the inverse image of the sets in C by f. What do you know about compact sets? Can you use this here, what about mapping back to K? Is K thus compact?
     
    Last edited: Jan 13, 2005
  5. Jan 13, 2005 #4
    Unfortunately im not familiar with most of these terms. What's the supremum? What does a covering of K by open sets mean? I'm pretty sure I get the image part. Does the image of E under F simply mean the range of F if the domain were restriced to E?
     
  6. Jan 13, 2005 #5
    An upper bound for a set F (subset of R) is a real number M such that for all x in F, x <= M.

    Example: Let F=(-oo,3]. Then 666 is an upper bound for F, as are 3 and 5 and any number not less than 3.

    The supremum of a set F (subset of R) is the least upper bound of F.

    Example: Let F=(-oo,3]. Then 3 is the supremum of F. 3 is also the supremum of (-oo,3).

    A set G is open if for all x in G, there is an interval (x-d, x+d) fully contained in G. In other words, a d>0 such that (x-d,x+d) is a subset of G.

    A covering of K by open sets means a family of sets, W, such that each member of W is open and K is a subset of the union of all sets in W. (The union of all sets in W is the set of all points in some element of W.)

    For example, Let W={(-oo,n):n &isin; Z}. W is an open covering of the set K=[36,666].

    The image of E under f, denoted f(E), is the set of elements in the range mapped to by some element of E. So y is in f(E) iff f(x)=y for some x in E.
     
  7. Jan 14, 2005 #6

    matt grime

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    If you're not familiar with those things (and it would take too long to learn for the difficulty of the question, and you'll learn them later anyway) then it can be done from more "basic" analysis, ie the epsilon delta arguments.

    There are several variations though (do you know about cauchy sequences?) here's some of the ways to do it. Warning it is a solution, but it goes quite slowly so you can read bits of it and see if you can predict the rest. It starts with two proofs that use compactness and then gives one that doesn't presume you know about it

    http://www.dpmms.cam.ac.uk/~wtg10/bounded.html
     
  8. Jan 14, 2005 #7
    Thank you all, I'll give it another try. I'm pretty sure it's the epsilon delta type proof he's looking for. If I'm still stuck before it's due I'll take a look at the link.
     
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