From Problem 3.16(c) of Boyd & Vandenberghe's Convex Optimization:
Determine if $f(x_1,x_2) = \frac{1}{x_1x_2}$ is convex, quasiconvex,
concave, or quasiconcave on $\mathbb{R}^2_{++}$.
From this post Determining whether $\alpha$ sublevel sets are convex I tried to plot it with mesh in Matlab, and I think it is convex on $\mathbb{R}^2_{++}$. But I'm stuck trying to prove it with the definition. My current attempt:
We have for $(x_1, y_1) \in \mathbb{R}^2_{++}$ and $(x_2, y_2) \in \mathbb{R}^2_{++}$, i.e., $x_1, x_2, y_1, y_2 > 0$, and $0 \leq \theta \leq 1$,
\begin{align*}
f\left(\theta(x_1,y_1) + (1-\theta)(x_2,y_2)\right) &= f\left(\theta x_1 + (1-\theta)x_2, \theta y_1 + (1-\theta)y_2\right) \\
&= \frac{1}{(\theta x_1 + (1-\theta)x_2)(\theta y_1 + (1-\theta)y_2)} \\
&= \frac{1}{\theta^2 x_1y_1 + \theta(1-\theta)(x_1y_2 + x_2y_1) + (1-\theta)^2x_2y_2} \\
&< \frac{1}{\theta^2 x_1y_1 + (1-\theta)^2x_2y_2} \\
\end{align*}
I know ultimately I need to show that $$\frac{1}{\theta^2 x_1y_1 + (1-\theta)^2x_2y_2} \leq \frac{\theta}{x_1y_1} + \frac{1-\theta}{x_2y_2}$$ but I'm not sure how to proceed or whether my approach is wrong. I don't think partial fraction expansion strategy applies here because we have different variables. Any hints?
Best Answer
To show that it is convex, it would be easier if you show that the Hessian of the function is positive semi-definite. In fact, the Hessian is
$$\begin{bmatrix} {2 \over x_1^3 x_2} & {1 \over x_1^2 x_2^2} \\ {1 \over x_1^2x_2^2} & {2 \over x_1 x_3^3} \end{bmatrix}$$
The matrix and all of its principal submatrices are positive semi-definite, so it is convex, hence quasiconvex as well.
Or you can show PSD by showing $z^T H z \ge 0$, and we get
$$ {2 \over x_1^3 x_2}z_1^2 + {1 \over x_1^2 x_2^2}z_2z_1 + {1 \over x_1^2x_2^2}z_1z_2 + {2 \over x_1 x_3^3}z_2^2 \ge 0$$
The above is non-negative since you can complete the square.