Let $\Omega$ be a Borel space and let $\mathcal P(\Omega)$ be the space of all Borel probability measures on $\Omega$ endowed with the topology of weak convergence. Define the total variation metric on the latter space $$ d(\mu,\nu) :=\sup\left\{\int_\Omega f\;\mathrm d\mu - \int_\Omega f\;\mathrm d\nu:f\in \mathcal B_1(\Omega)\right\} $$ where $\mathcal B_1(\Omega)$ is the space of all Borel functions on $\Omega$ whose absolute value does not exceed $1$.

Clearly, any linear functional $\phi:\mathcal P(\Omega)\to \Bbb R$ of the form $$ \phi(\mu) = \int_\Omega(c+f)\mathrm d\mu \tag{1} $$ for any $f\in \mathcal B_1(\Omega)$ and $c\in \Bbb R$ satisfies the following Lipschitz condition $$ |\phi(\mu) - \phi(\nu)| \leq d(\mu,\nu). \tag{2} $$ Is that true that any linear functional $\phi$ that satisfies $(2)$ can be represented as $(1)$?

Borel space is a topological space homeomorphic to a Borel subset of a complete separable metric space.

The answer below does not seem to be correct according to its author.

  • $\begingroup$ @Thisismuchhealthier: well, scaled version does not satisfy the Lipschitz condition with modulus of continuity $1$, but with that of $c$. Shifting instead does satisfy it as we are dealing with probability measures, unless I'm mistaken. You are right concerning the weak topology - I typed it in automatically as I'm used to ask question about Borel spaces and the topology induced on space of measures. $\endgroup$ – Ilya Jul 11 '14 at 7:11
  • $\begingroup$ I am not sure, why the previous answer was deleted, but anybody is welcome to reply here. $\endgroup$ – Ilya Dec 9 '15 at 8:09
  • $\begingroup$ The answer below does not seem to be correct according to its author. $\endgroup$ – Ilya May 3 '16 at 13:27
  • $\begingroup$ I think that the answer by @Meta was fine. The functional constructed is not weakly continuous, but I don't think that was required by the question. $\endgroup$ – George Lowther May 5 '16 at 0:33

Let $\mu^d$ be the restriction of a measure $\mu\in\mathcal{P}(\Omega)$ to its atoms ($\mu^d(A)$ is the maximum of $\mu(A\cap S)$ over countable $S\subseteq\Omega$). Then a counterexample is given by $$ \phi(\mu)=\mu^d(\Omega)=\max\left\{\mu(S)\colon S\subseteq\Omega\textrm{ is countable}\right\}. $$

  • $\begingroup$ I'm assuming here that, in the question, "linear functionals" are not required to be weakly continuous. $\endgroup$ – George Lowther May 5 '16 at 0:28
  • $\begingroup$ Thanks, this example works perfectly. No weak continuity assumptions, I'm also not sure why did I go for Borel space at the time I've asked this question, as I do not seem to use this structure at all. Do you think weak continuity would make this true, or just harder to find a counterexample (in case you have already thought of that)? $\endgroup$ – Ilya May 6 '16 at 7:34
  • $\begingroup$ Weak continuity would make it true, as the probability measures with finite support are weakly dense in the space of all probability measures. $\endgroup$ – George Lowther May 6 '16 at 11:40

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