# Translation invariant metric

Under what conditions can a metric vector space be given an equivalent metric that is translation invariant?

I was wondering if the probability measures on real line can be embedded in vector space of bounded measures on real line, so that the Lévy–Prokhorov metric is extended in the natural manner: If $\mu$ and $\nu$ are two measures on real line,

$$d (\mu, \nu) := \inf \left\{ \varepsilon > 0 \mid \mu(A) \leq \nu (A^{\varepsilon}) + \varepsilon \ \text{and} \ \nu (A) \leq \mu (A^{\varepsilon}) + \varepsilon \ \text{for all} \ A \in \mathcal{B}(\mathbb{R}) \right\}.$$

I want to see if this creates topological vector space. I could not prove that the metric is translation invariant.

• What exactly do you mean by "equivalent metric"? If you only want to induce the same topology by a translation invariant metric then the answer is "always". (Every first-countable topological group admits an invariant metric by the Birkhoff-Kakutani theorem). – t.b. Aug 7 '12 at 22:02
• Thanks. Yes, I wanted a metric which induced the same topology. So, the answer to my question is - the vector space with the metric may have to be a topological group under vector addition. – Mayank Aug 8 '12 at 3:54
• @user224863: Um, every topological vector space is a topological group under addition. – Nate Eldredge Aug 8 '12 at 4:20
• @NateEldredge: Actually, I am thinking of a vector space that has a metric defined over it, that may not make translations continuous. Sorry about the tag, I am trying to prove that a vector space is a topological vector space. – Mayank Aug 8 '12 at 14:33
• Then in that case, this property is sufficient: $d(x+z,y+z) \leq d(x,y) \forall x,y,z$, – Mayank Aug 8 '12 at 14:39