Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 5, Problem 6

Problem

Suppose

(a) $f$ is continuous for $x \geq 0$,

(b) $f'(x)$ exists for $x > 0$,

(c) $f(0)=0$,

(d) $f'$ is monotonically increasing.

Put

$$ g(x) = \frac{f(x)}{x}\;\;(x>0) $$

and prove that $g$ is monotonically increasing.

Answer

For $x>0$, $g'(x) = \frac{xf'(x) - f(x)}{x^2}$. Note that there exists $c \in (0,x)$ such that $g(x) = \frac{f(x)}{x} = \frac{f(x) - f(0)}{x- 0} = f'(c)$  $\Rightarrow$  $xf'(x) > xf'(c) = f(x)$  $\Rightarrow$  $g'(x)>0$.

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