1
$\begingroup$

Let $\Omega\subset\mathbb{R}^n$ be a open convex set and $A=\{x_1,...,x_k\}\subset\Omega$ a finite set. Suppose $f:\Omega\rightarrow\mathbb{R}$ is a $C^1$ function in $\Omega\setminus A$ and continuous in $\Omega$ such that $(f'(u)-f'(v))(u-v)\geq 0,\ \forall\ u,v\in\Omega\setminus A$

Can I conclude that $f$ is convex?

Edit. It was missing the hypothesis of continuity.

  • 0
    I agree @GEdgar, and im trying to solve by using this approach but with no sucess.2012-12-24

1 Answers 1

2

First: to show that a function defined on an open convex set in $\mathbb R^n$ is convex, it suffices to show that its restriction to any line is convex. Therefore we may assume $n=1$.

So: let $\Omega$ be an open interval in $\mathbb R$, let $A \subset \Omega$ be a finite set, let $f : \Omega \to \mathbb R$ be a function. Assume $f$ is continuous on $\Omega$ and differentiable on $\Omega \setminus A$. Assume: for $u,v \in \Omega \setminus A$, if $u \le v$ then $f'(u) \le f'(v)$. We must show that $f$ is convex on $\Omega$. That is: Let $a,b,c \in \Omega$, $a. Then $ f(b) \le \frac{c-b}{c-a}f(a) + \frac{b-a}{c-a}f(c) . \tag{$*$} $ Now, if we prove ($*$) whenever $f$ is differentiable at $b$, it follows for general $b$ by continuity. So let's assume $f$ is differentiable at $b$.

Claim 1. If $u,v \in \Omega$ and $b \le u < v$, then $ f'(b) \le \frac{f(u)-f(v)}{u-v} . \tag{1}$ Claim 2. If $u,v \in \Omega$ and $u < v \le b$, then $ \frac{f(u)-f(v)}{u-v} \le f'(b) . \tag{2}$ Note: Claims 1 and 2 are enough to finish the proof: Indeed, from Claim 1 we have $f(c) \ge f(b)+(c-b) f'(b)$, and from Claim 2 we have $f(a) \ge f(b) - (b-a) f'(b)$. Multiply the first inequality by $(b-a)$, the second inequality by $(c-b)$ and add to get $ (b-a)f(c)+(c-b)f(a) \ge (c-a)f(b), $ as required.

Proof for Claim 1. First, if $(u,v) \cap A = \varnothing$, regardless of whether $u$ or $v$ belong to $A$, then $(f(v)-f(u))/(v-u)$ is $f'$ evaluated at some point between $u$ and $v$ by the Mean Value Theorem. So ($1$) is true. For the general case: Let $A \cap (u,v) = \{x_1,x_2,\dots,x_n\}$ with $x_1 < x_2 < \dots < x_n$. Then $ \begin{align} (x_1-u)f'(b) &\le f(x_1)-f(u), \\ (x_2-x_1)f'(b) &\le f(x_2)-f(x_1), \\ (x_3-x_2)f'(b) &\le f(x_3)-f(x_2), \\ ... \\ (x_n-x_{n-1})f'(b) &\le f(x_n)-f(x_{n-1}), \\ (v-x_{n})f'(b) &\le f(v)-f(x_{n}). \end{align} $ Add to get $ (v-u)f'(b) \le f(v)-f(u) $ as claimed.

The proof for Claim 2 is similar.