# 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

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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

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.
A set is convex if contains every convex combination of points within the set. If you plot $$f_1$$, then you get the following image.
If you want, you can also use this image to choose points x and y as well as $$\lambda \in [0, 1]$$ such that $$f(\lambda x + (1- \lambda)y) \not\leq \lambda f(x) + (1 - \lambda) f(y)$$ -- this will prove that the function is not convex using the definition of a convex function.