Can a linearly homogeneous function be concave?

Question

If a function $f(x)$ ($x$ being a vector) is linearly homogeneous in $x$ (i.e. $k^{\lambda}f(x)=f(kx)\:;\:\: \lambda=1$), then can it also be said to be concave in $x$?

In an answer to this other question, the answerer mentions that a function with degree of homogeneity greater than 1 ($\lambda>1$) is the "convex case". May I conclude that a homogeneous function with $0<\lambda<1$ is the concave case, and that functions with $\lambda=1$ are therefore not concave in $x$?

The motivation for this question is a seminal economics paper which asserts that a certain function is linearly homogeneous in a certain variable, say $w$, and then subsequently "proves" that the same function is also concave in $w$. I'm trying to understand in what sense these two things can both be true.

EDIT: In response to John Hughes' comment, here are the other stated properties of the function in question. Let this function be denoted $C(y, w)$, $y$ being a scalar and $w$ being a vector.

Property 1: $C(y,w)$ is a nonnegative function.

Property 2: $C(y,w)$ is positively linearly homogeneous in $w$ for each fixed $y$.

Property 3: $C(y,w)$ is nondecreasing in $w$ for each fixed $y$.

Property 4: $C(y,w)$ is a concave function of $w$ for each fixed $y$.

My question in this post is essentially "Don't properties 2 and 4 contradict each other?"

Each of these Properties is accompanied by a proof. The paper goes on to prove three more Properties:

Property 5: $C(y,w)$ is a continuous function of $w$ for each fixed $y$.

Property 6: $C(y,w)$ is nondecreasing in $y$ for fixed $w$.

Property 7: For every $w >> 0$, $C(y,w)$ is continuous from below in $y$.

The full treatment can be found on pages 4-6 here.

Answer 1

If $f(x)$ is linearly homogeneous in a single variable $x$, then

$$f(ax)=af(x) \tag{1}$$

Can $f(x)$ be considered concave? Yes, but it can just as well be considered convex.

Proof:

The condition for convexity is

$$f(tx + (1-t)y) \leq tf(x) + (1-t)f(y) \tag{2}$$

$\forall t \in [0,1], \forall x,y \in X$ where $X$ is a convex set (following wikipedia).

Conversely, the condition for concavity is

$$f(tx + (1-t)y) \geq tf(x) + (1-t)f(y) \tag{3}$$

Note that $y$ can be expressed $y=xk$, where $k$ is not something bizarre or ridiculous. Substituting this into $(2)$ gives

$$f(tx + (1-t)xk) \leq tf(x) + (1-t)f(xk)$$

On the left hand side, the $x$ can be factored out

$$f(x(t + (1-t)k))$$

Now, if $f(x)$ is linearly homogeneous in $x$, then, by $(1)$, the factor $t + (1-t)k$ can be factored out of the function, so that this becomes

$$(t + (1-t)k)f(x)$$

Meanwhile, on the right hand side, also by $(1)$, the $k$ can be factored out of the function, i.e.,

$$tf(x) + (1-t)kf(x)$$

and then the $f(x)$ can be factored out, resulting in

$$(t + (1 - t)k)f(x)$$

which is equal to the left hand side.

Therefore, the condition $(2)$ is met, but so is condition $(3)$, meaning $f(x)$ is both convex and concave.

For a function of multiple variables $f(\mathbf{x})$ where $\mathbf{x}$ is a vector, the condition for convexity is

$$\mathbf{x} \cdot H \cdot \mathbf{x} \geq 0 \tag{4}$$

where $H$ is the Hessian matrix of $f(x)$. Conversely,

$$\mathbf{x} \cdot H \cdot \mathbf{x} \leq 0 \tag{5}$$

is the condition for concavity

Meanwhile, it can be shown that functions homogeneous of degree $h$ do this:

$$H \cdot \mathbf{x} = (h - 1) \nabla f$$

By which it is easy to see that if $f(\mathbf{x})$ is linearly homogeneous, i.e. has $h=1$, then

$$H \cdot \mathbf{x} = \mathbf{0}$$

and hence

$$\mathbf{x} \cdot H \cdot \mathbf{x} = 0$$

Therefore, multivariate linearly homogeneous functions satisfy both $(4)$ and $(5)$, and so are both concave and convex.

Answer 2

Take any norm $||.||$ on $\mathbb{R}^n$, the function $x\longmapsto -||x||$ is :

1. positively linearly homogeneous (definition of a norm)
2. concave (triangular inequality)
3. not convex (equality case in triangular inequality)

Answer 3

I just wanted to add something, since I stumbled on this question, and have had a hard time figuring out ben's answer.

Consider the function $f(x,y) = x^a y ^{1-a}$, where $0<a<1$, and $x,y\in \mathbb{R}^+$. It is clear that this is homogenous of degree 1, since $$ \begin{aligned} f(\alpha x,\alpha y) &= (\alpha x)^a (\alpha y) ^{1-a}\\ &= \alpha^a \alpha^{1-a} x^a y^{1-a}\\ &= \alpha x^a y^{1-a}\\ \end{aligned} $$

Economists will recognize this as a CRS production function. Note that this function is NOT both convex and concave. It is concave only.

First, take the Hessian: $$ H = \left( \begin{array}{cc} (a-1) a x^{a-2} y^{1-a} & (1-a) a x^{a-1} y^{-a} \\ (1-a) a x^{a-1} y^{-a} & -(1-a) a x^a y^{-a-1} \\ \end{array} \right) $$

Note that ben is correct to say that a continuous $f$ is weakly concave if and only if the Hessian is negative semidefinite (and the converse for convexity).

Moreover, a symmetric matrix is negative semidefinite if and only if all of its principal minors of an odd order are $\le 0$, and all of its principal minors of an even order are $\ge 0$. In this case, we need therefore only check the entry in the top left is $\le 0$, and the determinant of the entire matrix is $\ge 0$. A symmetric matrix is positive semidefinite if and only if all of its principal minors are $\ge 0$.

Recall that $x,y\ge 0$. One can calculate the determinant and find that it is $0$. However, since $0<a<1$, the top left entry is negative. Thus, this Hessian is negative semidefinite.

So, the function is only concave.

Of course, there exist functions that are homogenous of degree $1$ and are only convex.

Consider, for example, a cone:

$$f(x,y) = \sqrt{x^2 + y^2}$$ Then, this is homogenous of degree 1:

$$ \begin{aligned} f(\alpha x,\alpha y) &= \sqrt{\alpha^2(x^2 + y^2)}\\ &= \alpha\sqrt{x^2 + y^2} \end{aligned} $$

And yet of course a cone is convex, not concave. $$ H = \left( \begin{array}{cc} \frac{1}{\sqrt{x^2+y^2}}-\frac{x^2}{\left(x^2+y^2\right)^{3/2}} & -\frac{x y}{\left(x^2+y^2\right)^{3/2}} \\ -\frac{x y}{\left(x^2+y^2\right)^{3/2}} & \frac{1}{\sqrt{x^2+y^2}}-\frac{y^2}{\left(x^2+y^2\right)^{3/2}} \\ \end{array} \right) $$

Which you can again confirm has determinant of $0$ and has a first principal minor that is $\ge 0$.