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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