Anybody have any tips on how to show that the function $\frac{1}{2}\vert x\_1 + x\_2 + x\_3 - x\_1x\_2x\_3\vert^2$ is convex in $\mathbf{x}$, where $\vert x\_i \vert \leq 1$? 


This comes from the following expression, for general N:


\begin{equation}
\frac{1}{2}\left\vert (\begin{array}{cc} 0 & 1 \end{array}) \left(\begin{array}{cc} 1 & -x\_N \\\ x\_N & 1 \end{array}\right) \cdots \left(\begin{array}{cc} 1 & -x\_1 \\\ x\_1 & 1 \end{array} \right) \left(\begin{array}{c} 1 \\\ 0\end{array}\right)\right\vert^2
\end{equation}


It is of course straightforward to calculate the Hessian of this function for N = 3, but it is not readily apparent to me that the Hessian is positive semidefinite. A Monte Carlo simulation over the range of the function does not reveal any $\mathbf{x}$ for which the Hessian has negative eigenvalues. So I believe the above N=3 function is convex. However, what I am hoping is to find a way to show that the function is convex for any N.


Thanks!



Convexity of $\frac{1}{2}\vert x_1 + x_2 + x_3 - x_1x_2x_3\vert^2$
The answer is **NO**. The restriction to the plane $x\_1+x\_2+x\_3=0$ is the function $\frac12(x\_1x\_2x\_3)^2$, that is
$$(x\_1,x\_2)\mapsto\frac12x\_1^2x\_2^2(x\_1+x\_2)^2.$$
**Edit**. This function is not convex at $(\frac12,\frac12,-1)$, for instance. The Jacobian of the above map is negative at that point.


