Can a linearly homogeneous function be concave? 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.
 A: 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:
$$H \cdot \mathbf{x} = (1-h) \nabla f$$
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.
A: Take any norm $||.||$ on $\mathbb{R}^n$, the function $x\longmapsto -||x||$ is :


*

*positively linearly homogeneous (definition of a norm)

*concave (triangular inequality)

*not convex (equality case in triangular inequality)

