Linearity of an additive and linearly homogeneous function on the positive orthant

Question

Let $n$ be a positive integer and $$\mathbb R_{++}^n\equiv\{(x_1,\ldots,x_n)\in\mathbb R^n\,|\,x_i>0\text{ for all $i\in\{1,\ldots,n\}$}\}.$$ Suppose that $\varphi:\mathbb R^n_{++}\to\mathbb R$ is a continuous function that satisfies: \begin{align} \varphi(\mathbf x+\mathbf y)=&\;\varphi(\mathbf x)+\varphi(\mathbf y),\\ \varphi(c\mathbf x)=&\;c\varphi(\mathbf x) \end{align} whenever $\mathbf x,\mathbf y\in\mathbb R_{++}^n$ and $c>0$.

Is $\varphi$ necessarily linear? That is, do there exist $\alpha_1,\ldots,\alpha_n\in\mathbb R$ such that $$\varphi(\mathbf x)=\sum_{i=1}^n\alpha_i x_i\qquad\text{for all $\mathbf x=(x_1,\ldots,x_n)\in\mathbb R_{++}^n$?}$$

The obvious difficulty here is that one cannot consider the unit vectors $(1,0,\ldots,0)$, $(0,1,\ldots,0)$, and so forth, to generate the desired coefficients, since they are not elementwise strictly positive (except when $n=1$), so they cannot be evaluated under $\varphi$. I am thinking about a limiting argument, defining $$\alpha_1^m\equiv \varphi(1,m^{-1},\ldots,m^{-1})$$ for each $m\in\mathbb N$ (and similarly for $i\in\{2,\ldots,n\}$), and then letting $\alpha_1\equiv\lim_{m\to\infty}\alpha_1^m$, but I am unsure how that would work (in particular, whether the limit exists).

I also tried changing bases, but then the problem of negative basis coefficients arises. Any hints would be appreciated.

Answer 1

Yes, $\def\ph{\varphi}\ph$ is necessarily of the form $\alpha\cdot x$ for some vector $\alpha$.

The values of $\ph$ on the orthant of the unit sphere $S:=S^{n-1}\cap \Bbb R_{++}$ determine the value of $\ph$ on all of $\Bbb R_{++}$ via the relation $\ph(cx)=c\ph (x)$. Furthermore, given $x_1,\dots,x_n\in S$, let $$K\equiv K(x_1,\dots,x_n):=\left\{\frac{\sum_i\lambda_ix_i}{\|\sum_i\lambda_ix_i\|}\;\middle|\;\lambda_i\ge 0,\sum_i\lambda_i=1\right\}$$

be the projection of the convex hull of $x_1,\dots,x_n$ onto $S$. Similarly, the images of $\ph(x_1),\dots,\ph(x_n)$ determine that of $\ph(x)$ for all $x\in K$. When $x_1,\dots,x_n$ are linearly independent, this means that $\ph$ restricted to $K$ must be of the form $\alpha\cdot x$.

Now, let $x_i^{(m)}$ be the vector whose $i^\text{th}$ coordinate is $1$, and whose other coordinates are $1/m$, for $m\ge3$. Then $\ph$ restricted to $K_m\equiv K(\dots,x_i^{(m)},\dots)$ is of the form $x\mapsto\alpha_m\cdot x$. Since the sets $K_m$ all have nontrivial intersection, each of these vectors $\alpha_m$ are actually the same $\alpha$. Since the union of $K_m$ is all of $S$, we have that $\ph$ is of the form $\alpha\cdot x$ for all $x\in S$, and therefore for all $x\in \mathbb R_{++}$.

Answer 2

Here is an argument. The idea is to approximate the canonical unit vectors by a sequence of alternative bases consisting of strictly positive vectors, and then take limits.

Fix $i\in\{1,\ldots,n\}$. Define $\mathbf e^i$ to be the $n$-dimensional column vector whose $i$th coordinate is $1$, otherwise $0$. Let $\mathbf 1$ be the $n$-dimensional column vector consisting of ones. Define for each $m\in\mathbb N$ $$\mathbf e^{i,(m)}\equiv\mathbf e^i+m^{-1}\mathbf 1$$ and notice that this vector is in $\mathbb R^n_{++}$, so it can be evaluated under $\varphi$.

The next step is to show that the sequence $(\varphi(\mathbf e^{i,(m)}))_{m\in\mathbb N}$ is Cauchy in $\mathbb R$. Indeed, if $\ell,m\in\mathbb N$ are such that $\ell>m$, then, letting $\delta\equiv m^{-1}-\ell^{-1}>0$, one has $$\varphi(\mathbf e^{i,(m)})=\varphi(\mathbf e^{i,(\ell)}+\delta\mathbf 1)=\varphi(\mathbf e^{i,(\ell)})+\delta\mathbf \varphi(\mathbf1),$$ so that $$\left|\varphi(\mathbf e^{i,(m)})-\varphi(\mathbf e^{i,(\ell)})\right|=\left|\frac{1}{m}-\frac{1}{\ell}\right||\varphi(\mathbf 1)|,$$ which can be made arbitrarily small for large $\ell$ and $m$.

Let $$\alpha_i\equiv\lim_{m\to\infty}\varphi(\mathbf e^{i,(m)})\in\mathbb R.$$ I will show that $$\varphi(\mathbf x)=\sum_{i=1}^n\alpha_ix_i$$ for each $\mathbf x\in\mathbb R_{++}^n$. To this end, define, for each $m\in\mathbb N$, the $n\times n$ matrix $$\mathbf E^{(m)}\equiv[\mathbf e^{1,(m)},\ldots,\mathbf e^{n,(m)}].$$ It is not difficult to compute that $\mathbf E^{(m)}$ is invertible and both $(\mathbf E^{(m)})_{m\in\mathbb N}$ and $((\mathbf E^{(m)})^{-1})_{m\in\mathbb N}$ converge (elementwise) to the identity matrix. (For the record, the diagonal elements of $(\mathbf E^{(m)})^{-1}$ are $1-(m+n)^{-1}$ and the off-diagonal ones are $-(m+n)^{-1}$.)

Fix $\mathbf x\in\mathbb R_{++}^n$ and define for each $m\in\mathbb N$ $$\mathbf z^{(m)}\equiv(\mathbf E^{(m)})^{-1}\mathbf x.$$ Clearly, $\lim_{m\to\infty}\mathbf z^{(m)}=\mathbf x$, and since all elements of $\mathbf x$ are strictly positive, there exists some $M\in\mathbb N$ such that $\mathbf z^{(m)}\in\mathbb R_{++}^n$ for all integers $m$ exceeding $M$. Since $\mathbf x=\mathbf E^{(m)}\mathbf z^{(m)}$, it follows that $$\mathbf x=\sum_{i=1}^nz_i^{(m)}\mathbf e^{i,(m)}.$$ Therefore, if $m$ exceeds $M$, one has that all quantities that are relevant are strictly positive, and \begin{align*} \varphi(\mathbf x)=\varphi\left(\sum_{i=1}^nz_i^{(m)}\mathbf e^{i,(m)}\right)=\sum_{i=1}^nz_i^{(m)}\varphi(\mathbf e^{i,(m)})\to\sum_{i=1}^n\alpha_ix_i\quad\text{as $m\to\infty$.} \end{align*}