How can I prove that f + g is convex?

[mohdhm]

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

 

[HallsofIvy]

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, $tf(x)+ (1-t)f(y)\ge f(tx+ (1-t)y)$. 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
$$t(f(x)+ g(x))+ (1-t)(f(y)+ g(y))\ge f(tx+ (1-t)y)+ g(tx+ (1-t)y$$
and you can assume that is true of f(x) and g(x) separately. Looks like straightforward algebra to me.