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Math Proofs (convex function)

  1. Sep 18, 2008 #1
    hey everyone:

    Use a definition to work forward from each of the following statements.

    b. for functions f anf g the function f + g is convex, where f + g is the function whoes value at any point x is f(x) + g(x).

    Definition of a convex function: http://en.wikipedia.org/wiki/Convex_function (the equation)

    So guys, i'm really stumped, what are we asked to do here. I have a hunch that we have to add g(x) of a convex function to the f(x) of one, but we only have the function of f to work with, and also in the definition we get a y variable, that was not even mentioned in the question... what is up with that. If someone can at least guide me in the right direction, i'd be eternally grateful
     
  2. jcsd
  3. Sep 18, 2008 #2

    HallsofIvy

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    No wonder you are stumped! The "theorem" as you state it is false. Did you not include the condition that f(x) and g(x) are convex themselves?

    I have absolutely no idea what you mean by "g(x) of a convex function"- either g(x) itself is a convex function or it is not, but there is no "g(x) of a convex function".

    A function is convex if and only if, for x and y any two points in its domain, [itex]tf(x)+ (1-t)f(y)\ge f(tx+ (1-t)y)[/itex]. In other words, the straight line between (x,f(x)) and (y, f(y)) is above the point on the graph (tx+ (1-t)y, f(tx+ (1-t)y)).

    Okay, if the function is f(x)+ g(y) that means you want to prove that
    [tex]t(f(x)+ g(x))+ (1-t)(f(y)+ g(y))\ge f(tx+ (1-t)y)+ g(tx+ (1-t)y[/tex]
    and you can assume that is true of f(x) and g(x) separately. Looks like straightforward algebra to me.
     
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