# Bounded Lipschitz Metric on Space of Positive Measures

The bounded Lipschitz metric ($d_{BL}$) metrizes the weak convergence of probability measures on $\mathbb{R}$ with respect to bounded continuous test functions $C^0_b(\mathbb{R})$ $$d(\mu, \nu) = \sup_{f \in \text{Lip}(\mathbb{R})} \Big | \int_{\mathbb{R}} f d \nu - \int_{\mathbb{R}} f d \mu \Big |$$ where $$\text{Lip}(\mathbb{R}) = \Big \{ f \in C_b(\mathbb{R}) : \sup_x |f(x) | \leq 1, \sup_{x \neq y} \frac{| f(x) - f(y) |}{|x-y|} \leq 1 \Big \}.$$

Questions:

(1) Does bounded Lipschitz metric also do the same (i.e. metrize) the space of non-negative finite measures ($\mathcal{M}^+(\mathbb{R})$)?

(2) In addition to being a metric does it also define a norm such that $(\mathcal{M}^+(\mathbb{R}),d_{BL})$ is a complete normed space?

Note: As far as I know, $d_{BL}$ induces a norm on the space of finite signed measures ($\mathcal{M}(\mathbb{R})$), however this space is not complete.

Motivation: I need the space of $\mathcal{M}^+(\mathbb{R})$ to be a Banach space so that I can apply classical Parabolic compactness technique by Jacques Simon (1987), which requires the underlying spaces to be Banach spaces, to a problem living on the space of measures.

• For the first question, notice that for a sequence of non-negative measures, convergence in law is the same as $\int fd\mu_n\to \int f\mu$ for all $f$ continuous with compact support and $\mu_n(\mathbb R)\to \mu(\mathbb R)$. For the second question, are you sure that $\mathcal M^+$ is a vector space? Commented Aug 19, 2013 at 21:22
• Thanks @DavideGiraudo for the prompt reply. So $\mathcal{M}^+$ is not a vector space over $(\mathbb{R},+)$. However I am not too sure of it being one over the space $(\mathbb{R}^+,\cdot)$.
– UPS
Commented Aug 20, 2013 at 9:02
• You can consider instead the vector space of finite signed measures. Commented Aug 20, 2013 at 9:10
• The vector space of finite signed measures with the bounded Lipschitz metric is not complete (as far as I know, though I am still looking for a counter example), hence not a Banach space, which is what I need.
– UPS
Commented Aug 20, 2013 at 9:34

The answer is only partially YES. However $$\mathcal{M}^+(\mathbb{\mathbb R})$$ obviously cannot be a vector space due to the positivity constraint. So this rules out both questions as currently written. What is true, though, is that the metric space $$(\mathcal{M}^+(\mathbb{R}),d_{BL})$$ is complete and metrizes the weak convergence. I will only prove rigorously the completeness, see my final remark for how to get the "metrization". The proof below actually works in any dimension, and also in any domain $$\Omega\subset \mathbb R^d$$.
Let $$\{\mu_n\}$$ be a sequence of positive measures, let me denote the mass $$m_n:=\mu_n(\mathbb R)\geq 0$$, and assume that the sequence is Cauchy $$d_{BL}(\mu_p,\mu_q)\to 0 \qquad \mbox{as }p,q\to\infty.$$
1. As a first step, it is easy to see that $$\{m_n\}$$ is a (real) Cauchy sequence: for, testing $$f \equiv 1$$ in the definition of $$d_{BL}$$, we get $$|m_p-m_q|=\left|\int 1 d\mu_p -\int 1 d\mu_q \right|\leq d_{BL}(\mu_p,\mu_q).$$ Since the real line is complete, there is $$m\geq 0$$ such that $$m_n\to m$$. If $$m=0$$ then it is immediate to see that, for any $$f\in \mathcal C_b$$, there holds $$|\int f d \mu_n|\leq ||f||_\infty m_n\to 0$$, which proves that $$\mu_n\to 0$$ weakly (narrowly).
2. If $$m>0$$ then we can assume that $$m/2\leq m_n\leq 2m$$ for $$n$$ large enough, and the renormalized sequence $$\tilde \mu_n:=\frac {\mu_n}{m_n}\in\mathcal P(\mathbb R)$$ is well-defined. I claim that $$\{\tilde\mu_n\}$$ is $$d_{BL}$$-Cauchy as well. Indeed, for $$p,q$$ large enough we have by triangular inequality $$\begin{multline*} \left| \int f d\tilde\mu_p -\int f d\tilde\mu_q\right| = \left| \int f \frac{1}{m_p}d\mu_p -\int f \frac{1}{m_q}d\mu_q \right| \\ \leq \frac 1m \left| \int f d\mu_p -\int f d\mu_q \right| \\ + \left|\left(\frac 1{m_p}-\frac 1m \right)\int f d\mu_p\right| + \left|\left(\frac 1{m_q}-\frac 1m \right)\int f d\mu_q\right| \\ \leq \frac 1m \left| \int f d\mu_p -\int f d\mu_q \right| \\ + \left|\frac 1{m_p}-\frac 1m \right| \|f\|_\infty 2m + \left|\frac 1{m_q}-\frac 1m \right| \|f\|_\infty 2m. \end{multline*}$$ Taking the supremum over $$f$$ such that $$\|f\|,Lip(f)\leq 1$$ gives $$d_{BL}(\tilde\mu_p,\tilde\mu_q)\leq \frac 1md_{BL}(\mu_p,\mu_q) + 2m\left|\frac 1{m_p}-\frac 1m \right| +2m\left|\frac 1{m_q}-\frac 1m \right|$$ and entails my claim.
3. Since $$(\mathcal P(\mathbb R),d_{BL})$$ is complete there is a proabability measure $$\tilde \mu\in \mathcal P(\mathbb R)$$ such that $$d_{BL}(\tilde\mu_n,\tilde \mu)\to 0$$. Because we already proved that $$m_n\to m$$, it is then easy to check that $$\mu_n=m_n\tilde\mu_n$$ converges (in the Bounded-Lipschitz distance) to the limit $$\mu:=m\tilde\mu$$. Indeed for fixed $$f$$ $$\begin{multline*} \left|\int f d\mu_n- \int f d\mu \right| =\left|m_n\int f d\tilde\mu_n- m\int f d\tilde\mu \right| \\ \leq |m_n-m|\cdot \left|\int f d\tilde \mu_n\right| + m\left|\int f d\tilde\mu_n-\int f d\tilde\mu \right| \\ \leq |m_n-m|\cdot\|f\|_\infty+ m\left|\int f d\tilde\mu_n-\int f d\tilde\mu \right|. \end{multline*}$$ Taking one last time the supremum over $$f$$'s gives $$d_{BL}(\mu_n,\mu)\leq |m_n-m| + md_{BL}(\tilde\mu_n,\tilde\mu)\to 0$$ and the proof is complete.
Final remark: following the same lines it is easy to see that $$d_{BL}$$ does indeed metrize the weak convergence. The strategy of proof is identical: show that the masses converge, use this to suitably renormalize $$\tilde\mu_n:=\frac{1}{m_n}\mu$$, and exploit that the statement is already known for probability measures. (The case of vanishing mass $$m_n\to 0$$ must be treated separately.)