2

Show that the function $f : \mathbb{R}^n_{++} \to \mathbb{R}$, where $\mathbb{R}^n_{++} = \left\{x \in \mathbb{R}^n \mid x_i > 0 \right\}$, $$ f(x) = \ln \left(\sum_{i=1}^n \frac{1}{x_i} \right)$$ is convex.

My approach

In the case of one dimension ($n=1$), the function $\ln(1/x)$ is convex by using the fact that

A twice differentiable function of one variable is convex on an interval if and only if its second derivative is non-negative.

I am struggling to prove the case of two dimensions ($n=2$) and higher. I tried to prove that

$$ \lambda f(x) + (1-\lambda) f(y) \geq f(\lambda x + (1-\lambda) y)$$ for all $x,y \in \mathbb{R}^2_{++}, 0 \leq \lambda \leq 1$, but the natural log function was in the way, and I am not sure if there is an equality that I can use to deal with it.

Rodrigo de Azevedo
  • 23,223
AquaBlaze0010
  • 33

2 Answers2

3

The function $$ f(x) = \ln \left(\sum_{i=1}^n \frac{1}{x_i}\right) $$ is convex if $$ g(x) = \sum_{i=1}^n \frac{1}{x_i} $$ is log-convex, and that is true because

Martin R
  • 128,226
1

Just commenting on the great answer of @Martin R.:

Why is the positive linear combination of log-convex functions again log-convex?

Let $\phi_1$, $\ldots$, $\phi_m$ convex functions, $c_1$, $\ldots$, $c_m > 0$. We are to show that $\log (\sum_{i=1}^m c_i \exp \phi_i)$ is again convex. Now this follows from :

  1. $$\phi= (\phi_1, \ldots, \phi_m)$$ is convex

  2. $$(y_1, \ldots, y_m) \mapsto \log (\sum_{i=1}^m c_i \exp y_i)$$ is convex and increasing (convex is equivalent to Hölder's inequality)

  3. The composition of a convex and increasing function with a convex function is again convex.

orangeskid
  • 56,630
  • 1
    Convex function (as I know it) are real-valued functions. What does “$\phi= (\phi_1, \ldots, \phi_m)$ is convex” mean in this context? – Martin R Jul 19 '23 at 06:47
  • 1
    What you could say (I think) is: If $\phi_1, \ldots, \phi_m$ are convex functions and $h:\Bbb R^n \to \Bbb R$ is convex and increasing in each argument then $h(\phi_1, \ldots, \phi_n)$ is convex. Perhaps that would be a good addition to https://math.stackexchange.com/q/665768/42969 (which I used in my answer). You would still have to show that $h(y_1, \ldots, y_n) = \log (\sum_{i=1}^m c_i \exp y_i)$ is convex (which might be trivial or not). – Martin R Jul 19 '23 at 07:02
  • @Martin R: that would be a definition overloading... we have the standard result that $f\circ g$ is convex when $f$ convex increasing, and $g$ convex; and it is useful to extend that to $g$ having values in higher dimensions...( some people study ordered linear spaces) about $h$, it is convex due to C-B-S ineq or,trully, Hölder, worth checking it :-) – orangeskid Jul 19 '23 at 16:15
  • @MartinR: your solution really got me thinking... and probably there are other angles to it... say you want to show that if $f_k$ are positive and log-convex then so is $\sum f_k$. Now log convex ( for fn of 1 var) means some matrix (with the derivatives of $f$ ) is positive semidefinite and that behaves OK under positive linear combinations. What about for several variables? – orangeskid Jul 19 '23 at 17:04