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A distance function $d: \mathbb{R} \times \mathbb{R} \rightarrow [0,\infty)$ that is defined by a smooth Riemannian metric on the real line satisfies the following properties:

  1. $d$ is a length metric (a.k.a. intrinsic metric, inner metric, Menger-convex ...);
  2. $d$ is continuous;
  3. for every $x,y \in \mathbb{R}$, the function $t \mapsto d(x+t,y+t)$ is smooth as a function of $t$.

Is the converse true? In other words:

Given a distance function $d: \mathbb{R} \times \mathbb{R} \rightarrow [0,\infty)$ that satisfies these three properties, is it the distance function of a Riemannian metric on the real line?

Remarks.

1. If you eliminate condition (3), the metric may not come from a Riemannian metric no matter how little regularity you're prepared to accept. Take for example $f : \mathbb{R} \rightarrow \mathbb{R}$ to be the Cantor function extended continuously as a constant function outside the interval $[0,1]$ plus the function $x \mapsto x$. The function $f$ is continuous and strictly increasing.Therefore, $d(x,y) = |f(y) - f(x)|$ defines a continuous length metric on the real line, but the distance does not come from any Lebesgue integrable Riemannian metric on the line.

2. The point where I'm stuck: It is not hard to come up with the candidate Riemannian metric by considering $$ \sqrt{g_x(v,v)} := \lim_{t \rightarrow 0^+} d(x,x + tv)/t $$ that can be shown to exist for all $(x,v)$ because $d(x, x + tv)$ is continuous and monotone in $t$ and hence differentiable almost everywhere and because condition (3) implies that if the limit exists for a point, then it will exist for all points.

However, I can't see whether $\sqrt{g_x(\cdot,\cdot)} = \nu(x) |dx|$ is Riemann or Lebesgue integrable and, much less, continuous, etc.

Hopefully, this is all very simple and I'm just overlooking something really obvious.

3. In general, I'm trying to understand the class of distance functions on a manifold that can be uniformly approximated on compact sets by distance functions coming from Riemannian or Finsler metrics. If you have any references on this, I'd be glad to know of them.

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  • $\begingroup$ Let $D(x)= d(0,x)$. Note that $\nu(x) = D'(x)$. By the fundamental theorem of calculus $\int_0^x \nu(t) dt = D(x)$; thus $\nu(x)$ is continuous and integrable (say, if $D(x)$ is in $C^1$). $\endgroup$ – Yury May 16 '14 at 14:29
  • $\begingroup$ @Yury: The fact that $\nu$ is the derivative of an everywhere differentiable function does not mean it is continuous and unless you use the Denjoy-Perron-Henstock-Kurzweil integral you will not have the fundamental theorem of calculus either. The hypotheses do not say that $x \mapsto d(0,x)$ is $C^1$, it is just continuous and the only differentiability hypothesis is (3). $\endgroup$ – alvarezpaiva May 16 '14 at 14:35
  • $\begingroup$ Write $D(x) = D_{ac}(x) + D_s(x) + D_d(x)$, where $D_{ac}$, $D_s$ and $d_d$ are absolute continuous, singular, and jump functions, respectively. Regularity condition (3) means that $D_s$ and $D_d$ are invariant under shifts. Indeed, if $t\mapsto d(x+t,y+t)$ is absolutely continuous (on every segment), then write $d(x+t,y+t) = \alpha_+(t) - \alpha_(t)$ where $\alpha++$ and $\alpha_-$ are non-decreasing absolutely continuous functions. We have $h(t) \equiv (D_{ac}(x+t)+\alpha_+(t))+ D_s(x+t)+D_d(x+t) = (D_{ac}(y+t)+\alpha_-(t))+ D_s(y+t) + D_d(y+t)$. I.e., we get two decompositions of $h(t)$. $\endgroup$ – Yury May 16 '14 at 15:38
  • $\begingroup$ ... in the sum of absolutely continuous, singular and discrete functions. They must coincide: $D_s(x+t) = D_s(y+t)$, $D_d(x+t) = D_d(y+t)$. Thus $D_s = D_h \equiv 0$. Therefore, $D(x)$ is absolutely continuous on every segment, and $\nu(x)$ is Lebesgue integrable. (Here, we only used that $d(x+t,y+t)$ is absolutely continuous on every segment (e.g. it suffices that $d(\cdot,\cdot)$ is Lipschitz on every segment.) $\endgroup$ – Yury May 16 '14 at 15:41
  • $\begingroup$ $D(x) := d(0,x)$ (right?) is continuous. What do you mean by the decomposition? Do you mean the decomposition of the push forward of the Lebesgue measure by $D$? $\endgroup$ – alvarezpaiva May 16 '14 at 17:14
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[Moved from the comments/chat.]

Define $D(x)$ by $$ \begin{align*} D(x) = \begin{cases} d(0,x), & \text{ if } x\geq 0;\\ -d(x,0), & \text{ if } x< 0. \end{cases} \end{align*} $$ Function $D(x)$ is non-decreasing. It defines a measure $\nu$ on $\mathbb R$ such that $\nu[a,b] = D(b) - D(a)$ for every $b > a$.

Apply the Lebesgue Decomposition Theorem to $\nu$: $$\nu (x) = \nu_{ac}(x) + \nu_s(x) + \nu_d(x),$$ where $\nu_{ac}$ is an absolutely continuous measure, $\nu_s$ is a singular measure, $\nu_d$ is a discrete measure. Note that the decomposition is unique.

Now we use condition (3). We consider $\alpha(t) = d(x+t,y+t) = D(y+t) - D(x+t)$, which is smooth. Let $\mu(t)$ be the signed measure with density $\alpha'(t)$. Note that $\mu$ is absolutely continuous.

We have $\nu(y+t) - \nu(x+t) = \mu(t)$. Thus, $$(\nu_{ac}(y+t) - - \nu_{ac}(x+t)) + (\nu_s(y+t) - \nu_s(x+t)) + (\nu_d(y+t) - \nu_d(x+t))$$ is an absolutely continuous signed measure. Therefore, its singular and discrete parts are equal to 0: \begin{align*} \nu_s(y+t) - \nu_s(x+t) = 0,\\ \nu_d(y+t) - \nu_d(x+t) = 0. \end{align*} That is, signed measures $\nu_s$ and $\nu_d$ are invariant under shifts by $y-x$ for every $x < y$. Thus, they are identically equal to $0$ (up to normalization, only the standard Lebesgue measure is invariant under all shifts; but it is absolutely continuous). We get that $\nu(x)$ is an absolutely continuous measure. Let $f(x)$ be its Radon—Nikodym derivative w.r.t. to the standard Lebesgue measure. Then $$d(a,b) = \int_a^b f(t) dt,$$ for every $a<b$, which is what we want.

In this proof, we used only that $d(x+t,y+t)$ is absolutely continuous on every segment. E.g., it is sufficient to assume in condition (3) that $d(x+t,y+t)$ is Lipschitz on every segment.

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  • $\begingroup$ Thanks ! For what I need, I really don't need the smoothness of the Riemannian metric and this is quite enough. $\endgroup$ – alvarezpaiva May 16 '14 at 19:22
  • $\begingroup$ Just a remark: your answer made me realize that the Busemann-Pogorelov solution of Hilbert's fourth problem is actually a very natural generalization of the contruction of the Lebesgue-Stieltjes measure of the increasing function $D(x)$ arising from the metric. In other words, construction of the Lebesgue-Stieltjes integral is basically the solution of Hilbert's fourth problem in one dimension. $\endgroup$ – alvarezpaiva May 19 '14 at 12:59
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Yury's answer above reminded me of Z.I. Szabo's paper on Hilbert's fourth problem, where he looks very carefully at all regularity assumptions. There, on page 254 we can find the following lemma.

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function and consider the function $F(x,y) := f(y) - f(x)$ defined for all pairs of real numbers $(x,y)$ with $x \leq y$. If the derivative of $s \mapsto F(x + s, y + s)$ at $s = 0$ exists and is continuous as a function of $(x,y)$, then $f$ is continuously differentiable.

This lemma can then be iterated to yield that

If $d : \mathbb{R} \times \mathbb{R} \rightarrow [0,\infty)$ is a continuous length metric such that for every pair of real numbers the function $s \mapsto d(x + s, y + s)$ is smooth as a function of $s$, then $d$ is the distance function of a smooth Riemannian metric on $\mathbb{R}$.

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