Solution verification of f convex, differentiable and f(0)=0 implies f(x)<=xf'(x)

[Ozma]

Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] convex, differentiable and such that [imath]f(0)=0[/imath]. Prove that [imath]f(x)\le xf'(x)[/imath] for any [imath]x \in \mathbb{R}[/imath].

I tried this: for the mean value theorem, there exists [imath]\xi_x \in (0,x)[/imath], with [imath]x>0[/imath], such that [imath]f(x)-f(0)=xf'(\xi_x)[/imath]. By assumption it is [imath]f(0)=0[/imath], so [imath]f(x)=xf'(\xi_x)[/imath]. By assumption [imath]f[/imath] is differentiable and so [imath]f'[/imath] is increasing, hence being [imath]0< \xi_x <x[/imath] it is [imath]f'(\xi_x) \le f'(x)[/imath]. Since [imath]x>0[/imath], it is [imath]x f'(\xi_x) \le xf'(x)[/imath] and finally [imath]f(x) \le xf'(x)[/imath] for any [imath]x>0[/imath]. A similar strategy works for [imath]x<0[/imath] using the mean value theorem in the interval [imath](x,0)[/imath]. If [imath]x=0[/imath], the inequality becomes [imath]f(0) \le 0 \cdot f'(0)[/imath] which is true because by assumption [imath]f(0)=0[/imath]. So the inequality holds for any [imath]x\in\mathbb{R}[/imath]. Is this correct?

 

[blamocur]

Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] convex, differentiable and such that [imath]f(0)=0[/imath]. Prove that [imath]f(x)\le xf'(x)[/imath] for any [imath]x \in \mathbb{R}[/imath].

I tried this: for the mean value theorem, there exists [imath]\xi_x \in (0,x)[/imath], with [imath]x>0[/imath], such that [imath]f(x)-f(0)=xf'(\xi_x)[/imath]. By assumption it is [imath]f(0)=0[/imath], so [imath]f(x)=xf'(\xi_x)[/imath]. By assumption [imath]f[/imath] is differentiable and so [imath]f'[/imath] is increasing, hence being [imath]0< \xi_x <x[/imath] it is [imath]f'(\xi_x) \le f'(x)[/imath]. Since [imath]x>0[/imath], it is [imath]x f'(\xi_x) \le xf'(x)[/imath] and finally [imath]f(x) \le xf'(x)[/imath] for any [imath]x>0[/imath]. A similar strategy works for [imath]x<0[/imath] using the mean value theorem in the interval [imath](x,0)[/imath]. If [imath]x=0[/imath], the inequality becomes [imath]f(0) \le 0 \cdot f'(0)[/imath] which is true because by assumption [imath]f(0)=0[/imath]. So the inequality holds for any [imath]x\in\mathbb{R}[/imath]. Is this correct?

Looks good to me. A minor quibble: [imath]f^\prime[/imath] is increasing because [imath]f[/imath] is convex, not just because it is differentiable.

 

[Ozma]

@blamocur: thank you for the help, yes it was a typo of course I meant "since [imath]f[/imath] convex, it is [imath]f'[/imath] increasing", thanks for spotting it.