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Real Analysis: Coninuity

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data

    1. f is a continuous real valued function on a closed interval. f assumes minimum fvalues on [a,b], there exit x_o,y_o in [a,b] such that f(x_o) <= f(x) <=f(y_o) for all x in [a,b]. Show that the function -f assumes its maximum at x_o in [a,b], then f assumes its minimum at x_o.

    2. Suppose that f is continuous on [0,2] and that f(0) = f(2). Prove that there exist x,y in [0,2] such that |x-y| = 1 and f(x) = f(y). Hint: consider g(x) = f(x+1) - f(x) on [0,1]


    2. Relevant equations



    3. The attempt at a solution

    1. clearly since f(y_o) <= -f(x) <=f(x_o)

    multiplying this inequality through by - we get

    -f(y_o) <= f(x) <=-f(x_o)

    or

    f(x_o) <= f(x) <=f(y_o) for all x in [a,b]

    2. Hint: consider g(x) = f(x+1) - f(x) on [0,1]

    since:

    g(0) = f(1) - f(0)
    g(1) = f(2) - f(1)

    since: f(2) = f(0)

    g(0) = f(1) - f(0)

    g(1) = f(0) - f(1)

    this shows that

    g(0) + g(1) = 0

    or

    g(0) = -g(1)

    then

    f(1) - f(0) = -[f(0) - f(1)]

    f(1) - f(0) = f(1) - f(0)



    therefore f(1) = f(0) and 0,1 are in [0,2] and satisfy |y-x| = 1

    is this anywhere close? any help would be greatly appreciated for I spent much time trying to figure out this one. Thanks again!
     
  2. jcsd
  3. Nov 12, 2008 #2
    This conclusion doesn't make any sense.

    Take it from here.
    By the intermediate value theorem, you know that there exists [itex]x_0\in[0,1][/itex] with [itex]g(x_0)=0[/itex].
    How can you go on from here?
     
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