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Math Help - concaveness and continuity

  1. #1
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    concaveness and continuity

    Hi all,
    I have the following questions:
    1. Suppose  f : [0, \infty ) \mapsto R is bounded, concave and nondecreasing. Does it follow that f is continuous on [0, \infty)?
    2. Suppose  f_{n} : [0, \infty ) \mapsto R are bounded, concave , continuous and nondecreasing. Let f be the pointwise limit of f_{n}, does it follow that f is continuous?
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  2. #2
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    Re: concaveness and continuity

    For #1, the answer is certainly no, since continuity could easily fail at x=0. Ex: Define f by f(0)=0, and f(x)=1+arctan(x) for x>0. Then f satisfies all those conditions, but isn't continuous at x=0.

    However, it looks to be necessarily continuous at every point except at x=0:
    From bounded and non-decreasing, the left and right limits for f exist at each x>0. Fix a>0. Suppose f(x)->m1 as x->a-, and f(x)->m2 as x->a+. Since f non-decreasing, must have m1<=f(a)<=m2. Let x1, x2, such that 0<x1<a and x2>a. Since f concave, for x1<=x<=x2, have f(x) >= the line through (x1,f(x1)) & (x2,f(x2)) evaluated at that x. Since f(x2)>=m2, the line through (x1,f(x1)) & (x2,f(x2)) evaluated at that x is >= the line through (x1,f(x1)) & (x2,m2) evaluated at that x, since those lines intersect at (x1,f(x1)) and the first line has the greater slope (one could work it out easily enough algebraically).
    Thus, for x1<=x<=x2, have f(x) >= [ (m2-f(x1)) / (x2-x1) ] * (x-x1) + f(x1). The that (closed) inequality holds as take the limit x->a- (since we know the limit x->a- of f(x) exists and = m1). Get m1 >= [ (m2-f(x1)) / (x2-x1) ] * (a-x1) + f(x1). Since that (closed) inequality holds for all x2>a, it holds when take the (existing) limit as x2->a+.
    Get: m1 >= [ (m2-f(x1)) / (a-x1) ] * (a-x1) + f(x1) = m2.
    Therefore m1<=f(a)<=m2 and m2<=m1. Therefore m1=f(a)=m2. Thus both one-sided limits of f exist at x=a, and are equal (m1=m2), so the full limit exists at x=a and equals that value (which is not only m1 and m2, but also f(a)). Thus, as x->a, f(x)->f(a). Thus f is continuous at x=a.
    Last edited by johnsomeone; September 6th 2012 at 10:04 PM.
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