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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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    @JonasMeyer The effort put in the question **is** minimal since the question is merely copied from PMA. (I know the same remark does not apply to the *answer*.)2012-06-30

4 Answers 4

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

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    This is the way a proof should look like :-) (+1)2012-06-30
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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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    @did. Ok. I meant that convex functions are those whose incremental ratios increase. This is a particular case.2012-06-30
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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 ${f(x) \over x} \leq {f(y) \over y}$