Let $\mu$ be a finite signed measure on the Borel $\sigma$-algebra $\mathcal B(E)$ on a metric vector space $E$ and $x,y\in E$ with $$\mu\ast\delta_x=\mu\ast\delta_y\tag1,$$ where "$\ast$" denotes the convolution operator.
Under which assumptions on $\mu$ can we conclude $x=y$?
I've read that the claim is true when $\mu$ is a probability measure, but I don't know why. We clearly need to assume that $\mu$ is nontrivial, since otherwise the claim is obviously wrong.
It might be useful to note that if $$\tau_z:E\to E\;,\;\;\;w\mapsto w+z$$ denotes the right shift by $z\in E$, then $\mu\ast\delta_z=\tau_z(\mu)$. So, $(1)$ means $$\mu(B-x)=\mu(B-y)\;\;\;\text{for all }B\in\mathcal B(E).$$
Most crucially, if $\mu$ is "translation invariant", which precisely means that $\tau_z(\mu)=\mu$ for all $z\in E$, then the desired implication obviously cannot hold. And the trace of the Lebesgue measure on $\mathcal B(\mathbb R)$ on $[0,1]$ is clearly a translation invariant probability measure ... What am I missing?
The claim can be found in the proof of Corollary 2.3.5 of Linde's Probability in Banach Spaces: Stable and Infinitely Divisible Distributions:
Let $E$ a bounded measurable set with $\mu(E)>0$. You have that $\mu*\delta_{x-y} =\mu$. Let $z=x-y$. If $n$ is big enough, $E\cap (E+nz)=\emptyset$, and by hypothesis $\mu(E+nz)=\mu(E)$. So the $E+knz$, $k\in \mathbf{N}$, are disjoint sets with the same measure, and that's impossible for a finite measure.