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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

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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
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1 Answer

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.

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