1

Let $f:(a,b) \to \mathbb{R}$ be a differentiable and strictly convex function

I tried to explore some of the properties of such a function. For all $x,y \in (a,b)$ with $x \neq y$ I could apply the mean value theorem to such a function and I would obtain that: $$f(x) > f(y) + f'(y)(x-y) \tag{*} $$ because by the MVT $$\frac{f(x)-f(y)}{x-y}= f'(\xi), \text{ where } \xi \in (y,x) \\ \implies f(x)-f(y)=f'(\xi)(x-y) > f'(y) (x-y), \text{ because } f \text{ is strictly convex} $$ making use of $f$ strictly convex implying that $f'$ is strictly monotone increasing.

(*) is nice because now if $x_0 \in (a,b)$ such that $f'(x_0)=0$ this would show that $x_0$ is a minimum of $f$ on $(a,b)$ and also due to the strict convexity of $f$ such a minimum would be unique.

My Problem: Assuming that the first derivative $f':(a,b) \to \mathbb{R}$ is also continuous shouldn't it then be possible by using (*) to also show that $f$ is strictly monotone increasing, monotone decreasing or both when dividing $(a,b)$ into two subintervals with unique root $x_0$ such that $f$ is strictly decreasing $(a,x_0)$ and strictly increasing on $(x_0,b)$. This seems possible because the convexity of a function is a very strong condition.

I would have to check four three cases to check, namel $f'>0$, $f'<0$ for all $(a,b)$ or the mixed case with the unique root $x_0$. (Edited as suggest by Michael Grant)

Note: I am aware that for a continuous function $f:(a,b) \to \mathbb{R}$ with $f'>0$ on said interval implies that $f$ is strictly monotone increasing. My idea was to show this property only using (*) which may or may not even be possible. For example for $f'<0$ on $(a,b)$ I cannot use $(*)$ to show that $f(x)<f(y)$ for $y<x$

Spaced
  • 3,499
  • Of course, $f'\equiv 0$ is not possible since you're assuming strict convexity. – Michael Grant Nov 29 '14 at 19:01
  • @MichaelGrant thanks for pointing that out, you're indeed right about that! However there would still be 3 missing cases, namely $f'$ being negative (and still strictly increasing) over the interval, same with positive and the mixed case. – Spaced Nov 29 '14 at 20:56
  • Yes, I agree with you. – Michael Grant Nov 29 '14 at 20:58

1 Answers1

1

My idea was to show this property only using $(*)$ which may or may not even be possible. For example for $f'<0$ on $(a,b)$ I cannot use $(*)$ to show that $f(x)<f(y)$ for $y<x$

You can: just swap the roles of $x$ and $y$. $$f(y) > f(x) + f'(x)(y-x)$$ Since $y-x<0$ and $f'(x)<0$, it follows that $f(x)<f(y)$.

Thus, considering the three possible behaviors of $f'$ (always positive, always negative, and changing sign) you get the three monotonicity patterns of $f$.

By the way, you don't need any assumptions on the derivative: every strictly convex function has this property. Indeed, convexity implies the existence of one-sided derivatives, which satisfy $$f_-'(x)\le f_+'(x)\le f_-'(y)\le f_+'(y), \quad x<y \tag{1}$$

Property $(1)$ implies that one of the following holds:

  1. both one-sided derivatives are nonnegative on $(a,b)$
  2. both one-sided derivatives are nonpositive on $(a,b)$
  3. there is $x_0\in (a,b)$ such that both one-sided derivatives are nonnegative on $(x_0,b)$ and nonpositive on $(a,x_0)$

The proof of the existence of one-sided derivatives also yields an appropriate version of $(*)$; it uses the derivative at $x$ on the same side as $y$ is with respect to $x$. From here you can get monotonicity of $f$ as above. Strict convexity is not needed until now; only when you want to replace monotonicity by strict monotonicity, you need to invoke the fact that $f$ cannot be constant on any subinterval.

Community
  • 1