# 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.

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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? – Davide Giraudo Aug 19 '13 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 Aug 20 '13 at 9:02
You can consider instead the vector space of finite signed measures. – Davide Giraudo Aug 20 '13 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 Aug 20 '13 at 9:34