If $f(0)=0$ and $f'$ is increasing, then $\frac{f(x)}{x}$ is increasing.

Problem: Suppose $f$ is continuous for $x\ge 0$, differentiable for $x>0$, $f(0)=0$, and $f'$ is monotonically increasing.

Define $g(x)=\frac{f(x)}{x}$ for $x>0$. Prove that $g$ is monotonically increasing.

Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 6.

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– Did Jun 30 '12 at 6:27
@did, Potato: I think this discussion on meta is quite relevant here: User flooding the site with questions - more than 6/day. – Rahul Jun 30 '12 at 6:40
I don't think that Rahul's link is very relevant, because there were different factors beyond the number of questions being posted that caused much of the concern there. – Jonas Meyer Jun 30 '12 at 6:52
@JonasMeyer, I think many of the responses in the meta discussion remain relevant, for example: "The questions should be judged on their individual merits. The site should be filled with good questions." and "What is the difference if one person posts 10 monthly problems in one day, and if 10 different people post one monthly problem each in different months?" vs. "What bothers me about a large number of questions in a short period of time is that typically the effort put in by the poster is minimal." and "We don't need a list of textbook questions and answers on this site." – Rahul Jun 30 '12 at 7:18
"...typically the effort put in by the poster is minimal."--I don't think that applies here. – Jonas Meyer Jun 30 '12 at 7:21

$${f(x)\over x}=\int_0^1 f'(t\, x)\ dt\qquad(x>0)\ .$$

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This is neat. Side note: There could be problems applying the FTC, or even with the derivative not being integrable in general (e.g., see this MathOverflow question), which is a reason the mean value theorem is often preferred for problems of this sort. However, it follows from the hypotheses of the problem that $f'$ is continuous, so there is no difficulty here applying FTC. – Jonas Meyer Jun 30 '12 at 8:15
This is the way a proof should look like :-) (+1) – I'm an artist Jun 30 '12 at 9:17

Note that $g$ is differentiable everywhere it is defined. Fix $x, y\in\mathbb{R}$ with $y>x$. By the mean-value theorem, for some $t\in(x,y)$, we have

$$\frac{g(y)-g(x)}{y-x} = g'(t)= \frac{tf'(t)-f(t)}{t^2}.$$

Because $y-x$ is positive, it suffices to show that $$h(x)=\frac{xf'(x)-f(x)}{x^2}$$ is nonnegative. This reduces to showing that

$$f'(x)\ge \frac{f(x)}{x}$$

for $x>0$.

Consider the mean-value theorem applied to $[0,x]$ For some $t\in(0,x)$,

$$\frac{f(x)-f(0)}{x}=f'(t),$$

and the result follows because $f(0)=0$ and $f'$ is monotonically increasing.

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By assumption, $f$ is a convex function. Then $$g(x)=\frac{f(x)-f(0)}{x-0}$$ must be increasing in $x$, by a standard property of convex functions of one variable.

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Replace characterization by property. – Did Jun 30 '12 at 10:55
@did. Ok. I meant that convex functions are those whose incremental ratios increase. This is a particular case. – Siminore Jun 30 '12 at 12:20

The condition that $f'$ is increasing implies that $f$ is convex. Geometrically, the statement to be proved is that for $0 < x < y$ the slope of the chord from $(0,0)$ to $(x,f(x))$ is less than or equal to the slope of the chord from $(0,0)$ to $(y,f(y))$. Thus we expect to use the definition of convexity for the three points $0,x$, and $y$: $$f({1 -x \over y}0 + {x \over y}y) \leq {1 -x \over y}f(0) + {x \over y}f(y)$$ Since $f(0) = 0$, this is the same as $$f(x) \leq {x \over y} f(y)$$ This in turn is the same as $${f(x) \over x} \leq {f(y) \over y}$$

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