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Concaveness and continuity

  1. Aug 19, 2012 #1
    Hi all,
    I have the following question: Suppose f: [0, ∞) [itex] \rightarrow [/itex]ℝ and f is concave , nondecreasing and bounded on [ 0, ∞) . Does it follow that f is continuous on [ 0, ∞) ? Thanks in advance, H.
     
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  3. Aug 19, 2012 #2

    tiny-tim

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    hi hermanni! :wink:

    tell us what you think, and then we'll comment! :smile:

    (start by writing out the definition of "concave")
     
  4. Aug 19, 2012 #3

    A step function is non-decreasing, convex (or concave) and we can make it bounded, but won't be continuous. For example
    [tex]f(x)=0\,\,,\,if\,\, x\in [0,1)\,,\,f(x)=1\,\,,\,if\,\, x\geq 1\,[/tex]

    DonAntonio
     
  5. Aug 19, 2012 #4

    micromass

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    How is a step function convex??
     
  6. Aug 19, 2012 #5

    Very simple: it fulfills the definition of concave upwards (if the steps go up) or concave downwards (steps down), just as a constant

    function does as well.

    DonAntonio
     
  7. Aug 19, 2012 #6

    micromass

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    I don't see how. Take the function

    [tex]f:]-1,1[\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{l} 0~\text{if}~x\leq 0\\ 1 ~\text{if}~x>0 \end{array}\right.[/tex]

    This is the function you mean right??

    According to the definition, this function is convex if for all [itex]x,y\in ]-1,1[[/itex] and [itex]t\in [0,1][/itex] holds that

    [tex]f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)[/tex]

    But take x=-1/2, y=1/2 and t=1/4, then

    [tex]f(tx+(1-t)y)=1[/tex]

    while

    [tex]tf(x)+(1-t)f(y)=3/4[/tex]

    and this does not satisfy the inequality.

    Furthermore, it violates the following theorem: http://planetmath.org/ContinuityOfConvexFunctions.html [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. Aug 19, 2012 #7
    No, my function's defined on the convex non-negative ray [itex]\,[0,\infty)\,[/itex]
    Point taken, but eventhough my example indeed doesn't quite fit in the traditional definition, I will change it slightly and still

    will we get a counterexample:

    [tex]f(0)=0\,,\,f(x)=1\,\,,\,\forall\,x>0[/tex]

    The above function is convex upwards in [itex]\,[0,\infty)\,[/itex] but not continuous there.

    DonAntonio
     
    Last edited by a moderator: May 6, 2017
  9. Aug 19, 2012 #8

    micromass

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    OK, but I think the OP cares more about convex downwards...
     
  10. Aug 19, 2012 #9


    No problem: interchange zero and one in my last message's definition.

    DonAntonio
     
  11. Aug 19, 2012 #10

    micromass

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    That's not nondecreasing :frown:
     
  12. Aug 19, 2012 #11

    pwsnafu

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    Hint: suppose you already have a function that is continuous, non-decreasing, concave and bounded on the open interval ##(0,\infty)##. Then continuity on ##[0,\infty)## is completely determined by ##f(0)##.
     
  13. Aug 19, 2012 #12

    haruspex

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    There is perhaps some confusion about the terms convex and concave. I believe that the standard definition is such that for twice differentiable functions convex would mean the second derivative is never negative. I.e. convexity/concavity is as viewed from 'below'.
    This would make DonAntonio's counterexample at post #7 valid. But I suspect the OP was using the terms inversely, so suppose 'convex' was intended. As micromass/pwsnafu point out, the continuity is assured everywhere except at 0 by the convexity alone. The only question is whether the non-decreasing condition makes it continuous at 0+. I believe it does by a simple variation on the proof at the PlanetMath site quoted.
     
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