Suppose $W : \Bbb{R}^n \to \Bbb{R}_+$ is a continuous, positive function, with exactly $n$ zeros $\alpha_1,...,\alpha_n$. Define the following 'distance':

$$ d(\alpha_i,\alpha_j)=\inf\{\int_0^1 \sqrt{W(\gamma(t))}| \gamma'(t)|dt : \gamma \in C^1([0,1];\Bbb{R}^n), \gamma(0)=\alpha_i,\ \gamma(1)=\alpha_j\}$$

Suppose I have a set of real, positive numbers $\sigma_{ij}>0,\ i \neq j$ with the property that $\sigma_{ij}=\sigma_{ji}$ and $\sigma_{ij} \leq \sigma_{ik}+\sigma_{kj},i,j,k=1,...,n$.

My question is:

Can we find $\alpha_i, i=1..n$ and $W$ with the desired properties, such that $d(\alpha_i,\alpha_j)=\sigma_{ij}$?

I feel that the fact that we can choose $W$ and the zeros of $W$, $\alpha_1,...,\alpha_n$ gives enough freedom for us to solve this system. Thank you.

I should say why I need to know if this result is true. I am studying an article of Baldo: Minimal interface criterion for phase transitions in mixtures of Cahn-Hilliard fluids, where he proves that the following functional $$ \mathcal{F}(E_1,...,E_n) = \sum_{1\leq i < j\leq n} d(\alpha_i,\alpha_j) \mathcal{H}^{N-1}(\partial^*E_i \cap \partial^*E_j \cap \Omega) $$ is a $\Gamma$-limit of certain functionals, and therefore it is lower semicontinuous, where $d(\alpha_i,\alpha_j)$ is defined as above. I was wondering if it is possible to prove that for any $\sigma_{ij}$ which satisfy the triangle inequality (which is a necessary physical condition), the lower semicontinuity, and therefore the existence of a minimum for the given energy, $$ \mathcal{F}(E_1,...,E_n) = \sum_{1\leq i < j\leq n} \sigma_{ij} \mathcal{H}^{N-1}(\partial^*E_i \cap \partial^*E_j \cap \Omega) $$ still holds.


Your question is somewhat poorly phrased: The condition $\gamma>0$ in the definition of the "distances" $d(\alpha_i,\alpha_j)$ is meaningless and should be dropped. Once you do this, the object you are looking at is a singular conformally-Euclidean Riemannian metric $g= \rho(x) g_0$, where $g_0$ is the flat metric on ${\mathbb R}^n$ and $\rho(x)=W^2(x)$ is a continuous function which (for some unexplained reasons) is required to vanish at the points $\alpha_1,...,\alpha_n$. For such singular metrics one defines the usual notion of length of a piecewise-smooth path and distance function $d(x,y)$ (by infimizing lengths of paths). Then your numbers $d(\alpha_i,\alpha_j)$ are nothing but the distances between the singular points $\alpha_i,\alpha_j$. Thus, a better title of your question is probably "Singular conformally-Euclidean metrics" and right tag is differential geometry.

Now, your question has positive but not pretty answer. Furthermore, you can even do this with $n$ prescribed singular points in ${\mathbb R}^2$. Here is a sketch (in ${\mathbb R}^n$ as you requested). Start with $n$ points $\alpha_1,...,\alpha_n$ in ${\mathbb R}^n$ within unit distance from each other. Let $\Gamma$ be the complete graph that these points span (connect points by straight-line segments). On each segment $E_{ij}:=[\alpha_i, \alpha_j]$ choose a piecewise-linear function $W_{ij}$ (with one break-point, say, at the mid-point $m_{ij}$ of $E_{ij}$) which vanishes at the end-points and whose integral over $E_{ij}$ equals $\sigma_{ij}$. Thus, we defined a singular Riemannian metric on the graph $\Gamma$ and the corresponding distance function. The triangle inequalities that you impose on the numbers $\sigma_{ij}$ will ensure that the distances between points $\alpha_i, \alpha_j$ on $\Gamma$ are exactly $\sigma_{ij}$. Next, you will extend $W$ to small conical neighborhoods $C_{ij}$ of the segments $E_{ij}$, where each $C_{ij}$ is the union of two cones of revolution with vertices $\alpha_i, \alpha_j$, small vertex angles, axes $[\alpha_i, m_{ij}], [\alpha_j, m_{ij}]$ and equal bases orthogonal to $E_{ij}$. Let $C$ denote the union of all the "double cones" $C_{ij}$. Let $p_{ij}$ denote the orthogonal projection of $C_{ij}$ to $E_{ij}$. Extend $W_{ij}$ to ${\mathbb R}^n$ by the formula: $W_{ij}(x)=(1+ c d_{ij}(x)) W_{ij}(p_{ij}(x))$. Here $d_{ij}(x)$ is the distance from $x$ to $E_{ij}$ and $c\ge 0$ is a constant. Now, restrict $W_{ij}$ to $C_{ij}$ and let $W$ denote the resulting function on $C$. Then it is easy to see that for every choice of $c\ge 0$, for the singular Riemannian metric $g=W^2 g_0$ on $C$, the distances between the points $\alpha_i, \alpha_j$ are exactly $\sigma_{ij}$. (Use the fact that each $p_{ij}$ will contract the metric $g$.)

Lastly, you have to extend the function $W$ to the rest of ${\mathbb R}^n$. This is where things will become ugly especially near the points $\alpha_i$. It will be up to you to do the dirty part (assuming that you really need it). If $\epsilon>0$ is given and $c$ is large enough, then outside of Euclidean $\epsilon$-neighborhoods of the points $\alpha_k$ you can take $W$ equal $$ W(x)= \min_{i\ne j} W_{ij}(x) $$ where $W_{ij}$ are defined on the entire ${\mathbb R}^n$ as above. The hard part is taking care of Euclidean $\epsilon$-balls centered at the points $\alpha_k$. It might be easier to do the extension if you allow each piecewise-linear function $W_{ij}$ to have two break-points so that slopes of $W_{ij}$ (on $E_{ij}$) are the same at all points $\alpha_k$.

  • $\begingroup$ Thank you very much for your guidelines and for your answer. I edited my question using your suggestions, and I will say why I needed to know if this result holds in a next edit of my question. $\endgroup$ Mar 2 '12 at 10:39

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