Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was going through Stephen Boyd's lecture on convex optimization. However, I am a bit confused about a problem

Given Minimize $f(x) = x_1^2+x_2^2$

subject to $f_1(x) = \frac{x_1}{1+x_2^2} \leq 0$

How come $f_1(x)$ is not convex?

I was going through Stephen Boyd's book related to convex optimization

Here is the exact screenshot of the page

enter image description here

share|cite|improve this question

migrated from Jan 10 '13 at 15:40

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

I cant understand you question, what do you want to know? Do you want to know if $f_1$ is convex? – Tomás Jan 10 '13 at 16:01
Closely related: – whuber Jan 10 '13 at 16:19
The constraint only says $x_1 \le 0$. So looks like min is $0$ when $(x_1,x_2)=(0,0)$. – coffeemath Jan 10 '13 at 17:23
Compare $f(1,0)$ with $\frac12\big(f_1(1,1)+f_1(1,-1)\big)$. Or fix $x_1=1$ and check the sign of $\frac{\mathrm d^2}{\mathrm dx_2^2}f(1,x_2)$. – Rahul Jan 10 '13 at 19:38
@Tomás. Yeah I want to know why $f_1(x)$ is not convex? – user34790 Jan 10 '13 at 19:43

Look at $f_1$ along any "vertical" line $x_1=c$ where $c\neq0$. For positive $c$, you get a failure of convexity near the $x_1$-axis, and for negative $c$ you get a failure of convexity far from the $x_1$-axis.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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