Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given: $f(px+(1-p)y)\le pf(x)+(1-p)f(y)$

Suppose $f(x)$ is defined on the interval I. If $x_{1}<x_{2}<x_{3}$ are in I and $f(x_{1})<f(x_{2})$ and $f(x_{3})<f(x_{2})$, then show that $f(x)$ is not convex on I.

Not sure where to begin on this one

share|improve this question

closed as off-topic by Jonas Meyer, graydad, John Ma, Claude Leibovici, Macavity Jun 14 at 5:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Claude Leibovici, Macavity
If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer 1

up vote 1 down vote accepted

First express $x_2$ as a weighted linear combination of $x_1$ and $x_3$ i.e. $x_2 = tx_1 + (1-t)x_3$ such that $0 < t <1$. This is always possible since $x_2$ lies on the line segment connecting $x_1$ and $x_3$.

Now if $f(x)$ is convex in the interval $[x_1,x_2]$, then $f(x_2) = f(tx_1 + (1-t)x_3) < tf(x_1) + (1-t)f(x_3)$. But we are given $f(x_1) < f(x_2)$ and $f(x_3) < f(x_2)$. And hence $tf(x_1) + (1-t)f(x_3) < t f(x_2) + (1-t) f(x_2) = f(x_2)$.

So, we get $tf(x_1) + (1-t)f(x_3) < f(x_2)$ which contradicts the fact that $f$ is convex on the interval.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.