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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?

Rodrigo de Azevedo
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user594147
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    Compute the Hessian, and show that it is positive semi-definite will be helpful, since your function is $C^2(\mathbb{R}^2_{++})$. Using definition directly is not quite straightful sometimes in terms of computing. – Mike Oct 06 '20 at 04:49
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    If you do not use Hessian, proceeding along your approach, you may prove that $\frac{1}{\theta^2 x_1y_1 + \theta(1-\theta)(x_1y_2 + x_2y_1) + (1-\theta)^2x_2y_2} \le \frac{\theta}{x_1y_1} + \frac{1-\theta}{x_2y_2}$. After some manipulation, blablabla; By the way, $\frac{1}{\theta^2 x_1y_1 + (1-\theta)^2x_2y_2} \leq \frac{\theta}{x_1y_1} + \frac{1-\theta}{x_2y_2}$ is incorrect. – River Li Oct 06 '20 at 04:50
  • I'm curious what those manipulation steps are... could you elaborate? Although I think I'll go with the Hessian condition approach, it looks cleaner. – user594147 Oct 06 '20 at 04:55
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    @user594147 Sure. I posted it as an answer. – River Li Oct 06 '20 at 07:18
  • Listed – Rodrigo de Azevedo Jun 22 '22 at 06:02

3 Answers3

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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.

daw
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MoneyBall
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  • I'm not familiar with the "all principal submatrices are positive semi-definite so the matrix is positive semi-definite" theorem... I know the other direction is true by setting the corresponding entries of the vectors to 0. Is the proof similar? – user594147 Oct 06 '20 at 05:07
  • Also, not sure if that is actually the case since $\begin{bmatrix}0 & 1\ 1 & 0\end{bmatrix}$ is clearly indefinite, but its principal submatrices (the $0$ entries, unless I mistook what the principal submatrix means) are technically PSD. – user594147 Oct 06 '20 at 05:14
  • @user594147 I'm not sure what you mean by setting entries to 0. You can show it's positive semi definite by showing that $z^THz \ge 0$ for all $z$, where $H$ is the hessian. Since all entries in your hessian are positive, it is positive semi-definite. – MoneyBall Oct 06 '20 at 05:15
  • @user594147 $\begin{bmatrix}0 & 1 \ 1 & 0 \end{bmatrix}$ is not PSD. The matrix has a negative determinant. – MoneyBall Oct 06 '20 at 05:16
  • Sorry, I meant setting the entries of vector $z$ to be 0 like from this post https://math.stackexchange.com/questions/1221790/principal-submatrices-of-a-positive-definite-matrix – user594147 Oct 06 '20 at 05:16
  • Precisely, it is not PSD. But what would the principal submatrices of this example be? I thought they are formed by deleting the $k$-th row and column, so here $k = 1,2$ and we have the $0$ entries, which technically is a PSD $1 \times 1$ matrix. – user594147 Oct 06 '20 at 05:18
  • Thanks very much for clarifying!! – user594147 Oct 06 '20 at 05:22
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Besides using Hessian, there are several approaches:

  1. We have $\frac{1}{x_1x_2} = \mathrm{e}^{-\ln x_1 - \ln x_2}$. Recall that if $g$ is convex, then $\mathrm{e}^g$ is also convex. Since $-\ln x_1 - \ln x_2$ is convex, $\mathrm{e}^{-\ln x_1 - \ln x_2}$ is convex. We are done.

  2. Proceeding along your approach

Let us prove that $$\frac{1}{\theta^2 x_1y_1 + \theta(1-\theta)(x_1y_2 + x_2y_1) + (1-\theta)^2x_2y_2} \le \frac{\theta}{x_1y_1} + \frac{1-\theta}{x_2y_2}.$$ We have \begin{align} &\theta^2 x_1y_1 + \theta(1-\theta)(x_1y_2 + x_2y_1) + (1-\theta)^2x_2y_2\\ =\ & \theta x_1y_1 + (1-\theta)x_2y_2 + \theta(1-\theta)(x_1y_2 + x_2y_1 - x_1y_1-x_2y_2) \end{align} and $$(\theta x_1y_1 + (1-\theta)x_2y_2) \left(\frac{\theta}{x_1y_1} + \frac{1-\theta}{x_2y_2}\right) = 1 + \theta(1-\theta)\left(\frac{x_2y_2}{x_1y_1} + \frac{x_1y_1}{x_2y_2} - 2\right).$$ Thus, it suffices to prove that \begin{align} 1 &\le 1 + \theta(1-\theta)\left(\frac{x_2y_2}{x_1y_1} + \frac{x_1y_1}{x_2y_2} - 2\right)\\ &\qquad + \theta(1-\theta)(x_1y_2 + x_2y_1 - x_1y_1-x_2y_2)\left(\frac{\theta}{x_1y_1} + \frac{1-\theta}{x_2y_2}\right) . \end{align} It suffices to prove that $$0 \le \left(\frac{x_2y_2}{x_1y_1} + \frac{x_1y_1}{x_2y_2} - 2\right) + (x_1y_2 + x_2y_1 - x_1y_1-x_2y_2)\left(\frac{\theta}{x_1y_1} + \frac{1-\theta}{x_2y_2}\right)$$ which is written as \begin{align} 0&\le \left(\frac{x_2y_2}{x_1y_1} + \frac{x_1y_1}{x_2y_2} - 2 + (x_1y_2 + x_2y_1 - x_1y_1-x_2y_2)\frac{1}{x_1y_1}\right)\theta\\ &\qquad +\left(\frac{x_2y_2}{x_1y_1} + \frac{x_1y_1}{x_2y_2} - 2 + (x_1y_2 + x_2y_1 - x_1y_1-x_2y_2)\frac{1}{x_2y_2}\right)(1-\theta)\\ &= \left(\frac{x_1y_1}{x_2y_2} + \frac{y_2}{y_1} + \frac{x_2}{x_1} - 3\right)\theta + \left(\frac{x_2y_2}{x_1y_1} + \frac{x_1}{x_2} + \frac{y_1}{y_2} - 3\right)(1-\theta). \end{align} It is true by AM-GM. We are done.

River Li
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(Same argument given elsewhere:)

We know that $f$ is convex if and only if its supergraph $\{(x', x_n)\ | x_n \ge f(x')\}$ is convex. Now the epigraph equals

$$\{ (x_1, \ldots, x_{n-1}, x_n)\ | \ x_1 \cdot \ldots \cdot x_n \ge 1\}$$

Convexity is really easy: take $x$, $y$ in $(0, \infty)^n$ with $\prod x_i$, $\prod y_i \ge 1$. We have $$\prod_{i=1}^n \frac{x_i + y_i}{2} \ge \prod_{i=1}^n \sqrt{x_i y_i} \ge \sqrt{1\cdot 1} = 1$$

Note that the function $(x_1, \ldots, x_n)\mapsto x_1\ldots x_n$ is not concave, but its superlevel sets are convex. So this function is quasi-concave.

orangeskid
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