1

Let $A \subset \mathbb{R}$ be an open set. A differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ is strictly pseudoconvex in $A$ iif for all $x,y \in A$ $$ \frac{df}{dx} (y-x) \geq 0 \mbox{ we have } f(y) - f(x) > 0. $$ If we replace $f(y) - f(x)$ by its second order approximation of Taylor serie around $x \in A $. Thus we have the condition named (1)
$$ \mbox(1 ) \quad \frac{df}{dx} (y-x) \geq 0 \mbox{ we have } \frac{df}{dx}(y-x) + \frac{1}{2}\frac{df^2}{dx^2}(y-x)^2 + h_3(y)(y-x)^3 > 0. $$ We assume that $h_3(y)(y-x)^3$ is sufficient small in $A$ such that $|h_3(y)(y-x)^3| <\frac{df}{dx}(y-x) + \frac{1}{2}\frac{df^2}{dx^2}(y-x)^2 $ then the condition $(1)$ states pseudoconvexity for $f$ in neighborhood of $x$. Now as $\frac{df}{dx} (y-x) \geq 0 $ in the antecedent of the implication, we can replace $\frac{df}{dx} (y-x) $ by $|\frac{df}{dx} (y-x)|$ the consequent and drop $h_3(y)(y-x)^3$. We obtain the condition $(2)$: for all $x,y \in A$ $$ \mbox(2) \quad \frac{df}{dx} (y-x) \geq 0 \mbox{ we have } |\frac{df}{dx}(y-x)| + \frac{1}{2}\frac{df^2}{dx^2}(y-x)^2 > 0. $$ Now if we drop the antecedent in the condition $(2)$, we obtain the following condition: for all $x,y \in A$ $$ \mbox(3)\quad |\frac{df}{dx}(y-x)| + \frac{1}{2}\frac{df^2}{dx^2}(y-x)^2 > 0. $$ My question is: Is $(3)$ a sufficient condition for the strict pseudoconvexity of $f$ in a neighborhood $A$ of $x$?.

Thanks in advance.

jaogye
  • 963

1 Answers1

1

(a) Note that the followings are sufficient conditions :

(i) $f'(x)>0$ for all $x$

(ii) $f'(x)<0$ for all $x$

(iii) $f'(x)=0$, $f'(y)>0$ for $y>x$ and $f'(y)<0$ for all $y<x$

(b) Now consider the condition (3) : Assume that $f'(x)=f'(y)=0$ where $x <y$.

Hence $f''(x),\ f''(y)>0$ by condition $(3)$. They are strict local minimum points. Hence there is a point $z\in (x, y)$ of local maximum s.t. $f'(z)=0$.

By (3) $f''(z) >0$ so that it is a contradiction. Hence it is sufficient.

HK Lee
  • 19,964