0
$\begingroup$

As I understand, a strictly convex function $f: D \rightarrow \mathbb{R}$ is one that satisfies the property: $\forall x, y \in D, x \neq y, \forall t \in (0,1), f((1-t)x+ty)<(1-t)f(x)+tf(y)$. A mechanical interpretation of this: a convex function is one whose value at the weighted average of any two domain points is strictly less than the weighted average (using the same weights) of the function at the same two domain points (where the weights are positive and sum to 1). Suppose we look at functions satisfying an analogous property, where the weights sum to, say, 2. That is, suppose a function satisfies $\forall x, y \in D, x \neq y, \forall t \in (0,2), f((2-t)x+ty)<(2-t)f(x)+tf(y)$. Call this Property P. An initial question that arises is whether P is a stronger or weaker (or neither) condition than strict convexity. Using the example $f(x)=x^2, x \in (0,\infty)$, and letting $x=1, y=2, t=1/2$, I found a strictly convex function, $f$, that does not satisfy P, so clearly strict convexity is not stronger than P. However, is P stronger than strict convexity? If not, could you help me find a counterexample (i.e. a function that satisfies P but is not strictly convex)?

$\endgroup$
  • $\begingroup$ What about $f(x)=1$ for all $x \in \mathbb{R}$? This satisfies P. $\endgroup$ – Michael Jan 22 '16 at 2:07
  • $\begingroup$ Or take any (possibly nonconvex) function that satisfies $0.9 \leq f(x) \leq 1$ for all $x \in \mathbb{R}$. Any such function satisfies P. $\endgroup$ – Michael Jan 22 '16 at 2:08
  • $\begingroup$ math.stackexchange.com/questions/2908903/… $\endgroup$ – Blind Sep 13 '18 at 11:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.