# Confusion related to a convex optimization problem

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 http://www.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf

Here is the exact screenshot of the page

-

## migrated from stats.stackexchange.comJan 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: math.stackexchange.com/questions/274837. – 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.