1

My problem is to determine if the following set is convex: \begin{align*} \sum_{i=1}^N \frac{1}{x_i} \leq 1 \quad \textrm{for } \mathbf{x} \in \mathbb{R}_{++}^N, \end{align*} where $N \in \mathbb{N}$.

I have tried the case for $N=2$, I think it can be done by \begin{align*} \frac{1}{x_1}+\frac{1}{x_2} \leq 1 \iff (x_1-1)(x_2-1) \geq 1 \end{align*} But this method isn't applied to the case for higher dimension.

My intuition is Yes! However, when I try to prove the general case by definition, \begin{align*} \sum_{i=1}^N \frac{1}{\theta x_i + (1-\theta) y_i} \leq 1, \quad 0\leq\theta\leq1 \end{align*} I can't derive that the convex combination is upper bounded by 1, since all variables are in denominator part. I can only find the bound $\frac{\theta}{1-\theta}+\frac{1-\theta}{\theta}$. Does anyone have some intuitive interpretation or give me some hints?

Thank you!

Asaf Karagila
  • 393,674
Yanni99
  • 13

1 Answers1

0

The function $t \to 1/t$ is convex for $t > 0$. It follows for each $i$, $$ \frac{1}{\theta x_i + (1-\theta) y_i} \le \theta \frac{1}{x_i} + (1-\theta) \frac{1}{y_i} \, , $$ and adding this inequality over all components gives the desired conclusion.

In other words: The function $f: \Bbb R_+^N \to \Bbb R$ defined by $$ f(\mathbf{x}) = \sum_{i=1}^N \frac{1}{x_i} $$ is a sum of convex functions, and therefore convex. It follows that the sublevel set $$ \{ \mathbf{x} \in \Bbb R_+^N\mid f(\mathbf{x}) \le 1 \} $$ is a convex set.

Martin R
  • 113,040