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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