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My professor posted this question as a supplement to our exercises in Rudin. As a disclaimer, this class has been difficult for me in the past, so forgive me if I've missed any really simple steps... Here is the question:

"Let $f:[0, \infty)\to R$ be given, such that $f$ is continuous on $[0,\infty)$. Suppose $f(0)=0$, where $f$ is differentiable on $(0,\infty)$, and $f'$ is monotonic increasing on $(0,\infty)$. Prove that f is supperadditive on $[0,\infty)$ meaning that $f(x+y)\ge f(x)+f(y)$ for all $x, y \in [0,\infty)$."

What does it mean when the first derivative is monotonic increasing? Does that tell me $f''\ge0$ or is that even useful information?

Where does one begin on problems like this? If not a solution to the problem, general problem solving advice for real analysis proofs would be much appreciated. - Thanks.

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1 Answer 1

If $f$ was twice differentiable, it would be true that $f'' \ge 0$, but you only know that $f$ is differentiable once.

Hint: write the desired inequality as $f(x+y) - f(x) \ge f(0+y) - f(0)$, and turn each side into an integral.

By the way, it can't be for "all real $x$ and $y$", because it's not even defined for negative $x$ or $y$.

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Thanks, I fixed that last part, it should say "for all $x,y \in [0,\infty)$". We only started using integrals today in class, I'm not sure he intended us to use them on this homework. I will give it a try though! –  Lokutus Nov 14 '13 at 7:06
    
If you want to avoid integration, you could look at $\dfrac{\partial}{\partial x} (f(x+y) - f(x))$. –  Robert Israel Nov 14 '13 at 7:48

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