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Let $n\geq 2$ and $\displaystyle f:(x_1,\ldots,x_n) \mapsto \prod_{i=1}^n x_i^{\alpha_i}\quad$ be defined for positive $x_i$

Suppose $\sum_{i=1}^n \alpha_i > 1$. Prove that $f$ is neither concave nor convex.

The Hessian matrix of $f$ is easy to compute. It suffices to prove that it is neither positive semi-definite, neither negative semi-definite at some point.

Let $H(f)(x)$ denote the Hessian matrix of $f$ at $x=(x_1,\ldots,x_n)$. It suffices to find some $x$ such that $\det H(f)(x)<0$ or one may look for $u$ and $v$ such that $u^T H(f)(x)u> 0$ and $v^T H(f)(x)v< 0$.

I've noted that $$\frac{\partial^2 f}{\partial^2 x_i}(x) = \alpha_i(\alpha_i-1)\frac{f(x)}{x_i^2}$$ and $$\frac{\partial^2 f}{\partial x_j \partial x_i}(x) = \alpha_i \alpha_j\frac{f(x)}{x_i x_j}$$

I think the approach with the determinant is the most tractable. $x=(\alpha_1,\ldots, \alpha_n)$ seems like a promising point, although I'm not through with the computations.

EDIT: it seems that the approach with the determinant would require that $n$ is even...

EDIT 2: Provided my computation is correct, at $x=(\alpha_1,\ldots, \alpha_n)$, $$\det H(f)(x)=(f(x))^n\frac{(-1)^{n-1}}{\prod_{i=1}^{n}\alpha_i} \left[\left(\sum_{i=1}^n \alpha_i \right) -1 \right]$$

This solves the problem, with $n$ even. It remains to take care of odd $n$.

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  • $\begingroup$ I imagine that $n>1$, right? If not, the result is wrong. $\endgroup$ – mathcounterexamples.net Mar 9 '17 at 18:45
  • $\begingroup$ @mathcounterexamples.net yes, it's the Cobb-Douglas function in economics. $\endgroup$ – Gabriel Romon Mar 9 '17 at 18:47
  • $\begingroup$ It seems that the claim is not true for $n$ odd. Please see below. $\endgroup$ – mathcounterexamples.net Mar 10 '17 at 10:45
  • $\begingroup$ Below a general proof regarding convexity. $\endgroup$ – mathcounterexamples.net Mar 10 '17 at 14:02
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If $f$ was convex or concave, its graph would sit on a single side of its tangent plane $$\varphi(x_1, \dots, x_n) = 0$$ with $$\varphi(x_1, \dots, x_n)=\alpha_1(x_1-1) + \dots + \alpha_n (x_n-1)$$ at point $U=(1, \dots, 1)$

For the origin $O$ we have $$\varphi(O)-f(O)=- (\alpha_1 + \dots + \alpha_n) < -1 \le 0$$

Now consider $i_0$ for which $\alpha_{i_0} > 0$ and a point $P_t$ whose coordinates are all vanishing except the $i_0$-one which is equal to $t>0$. You have $\varphi(P_t)-f(P_t)= \alpha_{i_0}t - \sum_i \alpha_i > 0$ for $t>\frac{\sum_i \alpha_i}{\alpha_{i_0}}$.

Hence the graph of $f$ is not on a single side of its tangent plane at $U$ proving that $f$ is not convex nor concave.

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  • $\begingroup$ A slight variation of this reasoning may be written as follows: $$\forall x\in (\mathbb R_+^*)^n, \;f(x) \;?\; f(1)+\nabla f(1)\cdot (x-1)$$ Evaluating at $x=0$ (although $f$ may be not be differentiable at that point, you can evaluate it for some vector with small positive coordinates and then use the continuity of $f$), we can replace the $?$ sign with $\geq $ and write $$\forall x, \; f(x) \;\geq\; 1+\alpha_1(x_1-1) +\ldots +\alpha_n(x_n-1)$$ The end of the proof is similar to what you have written. $\endgroup$ – Gabriel Romon Mar 10 '17 at 15:28
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This is a partial answer, proving that the determinant of the Hessian is of constant sign.

Looking at specific cases

Case $n = 2$

At least a partial solution for $n = 2$. I haven't try to extend it for $n>2$.

For $x=(1,1)$ you have $\det H(f)(1,1)=\alpha_1 \alpha_2 (1-\sum \alpha_i)<0$.

And it could be interesting to still look at the value of $\det H(f)(x)$ for $x = (1, \dots,1)$ and $n>2$.

Case $n = 3$

Take $f(x_1,x_2,x_3)=x_1 x_2 x_3$. You have in that case $\det H(f)(x)=2x_1 x_2 x_3 >0$, proving that determinant of the Hessian is always positive in that case.

The general case

Denoting $\alpha^\prime_i = \frac{\alpha_i}{x_i}$ and $\beta_i = \frac{1}{x_i}$, we have: $$ \begin{aligned}\det H(f)(x)&= (f(x))^n \begin{vmatrix} (\alpha^\prime_1 - \beta_1) \alpha^\prime_1 & \alpha^\prime_1 \alpha^\prime_2 & \dots & \alpha^\prime_1 \alpha^\prime_n\\ \alpha^\prime_1 \alpha^\prime_2 & (\alpha^\prime_2 - \beta_2) \alpha^\prime_2 & \dots & \alpha^\prime_2 \alpha^\prime_n\\ \vdots & \ddots & & \vdots\\ \alpha^\prime_1 \alpha^\prime_n & \dots & \alpha^\prime_{n-1} \alpha^\prime_n & (\alpha^\prime_n - \beta_n) \alpha^\prime_n\\ \end{vmatrix}\\ &=(f(x))^n (\alpha^\prime_1 \dots \alpha^\prime_n)^2 \begin{vmatrix} 1-\frac{1}{\alpha_1} & 1 & \dots & 1\\ 1 & 1-\frac{1}{\alpha_2} & \dots & 1\\ \vdots & \ddots & & \vdots\\ 1 & \dots & 1 & 1-\frac{1}{\alpha_n}\\ \end{vmatrix} \end{aligned}$$ proving that the sign of the determinant of the Hessian is independent from the point $x$. And according to your computation for $(x_1, \dots,x_n)=(\alpha_1, \dots,\alpha_n)$ this sign is positive for $n$ odd and negative for $n$ even.

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  • $\begingroup$ Interesting! Thank you for computing the determinant in the general case. However, I'm quite dubious of what you derived. I agree the determinant is $>0$ for odd $n$ regardless of $x$. But this doesn't imply that, for a given $x$ all the eigenvalues share the same sign, let alone that for all $x$, the eigenvalues have the same sign. Unless I've missed something, this does not prove that $f$ is concave or $f$ is convex. $\endgroup$ – Gabriel Romon Mar 10 '17 at 11:14
  • $\begingroup$ You're right. My answer is focused on your statement regarding the sign of the determinant of the Hessian which is constant. This is not sufficient to make a conclusion on the convexity of the function. I'll update my answer accordingly. $\endgroup$ – mathcounterexamples.net Mar 10 '17 at 11:22

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