Assume that $\{P_n: n \in N\} $ and $\{Q_n: n \in N\} $ are sequences of probability measures. Assume that $P_n \stackrel{w}{\to} P. $ Also, assume that $Q_n = \delta_{b_n}, $ the Dirac measure and $b_n \in R. $
If we knew that $P_n\star Q_n $ (the convolution) converges weakly to some probability measure, does it follow that there must exist a $b $ such that $b_n \to b $ as $n \to \infty $ ?
I think that the answer is positive, but I do not seem to be able to make the argument rigorous.
Can anyone help?
Thank you
Maurice
Denoting by $\widehat{\mathbb P_n}$ the Fourier transform of $\mathbb P_n$, we have by the assumptions that for any $t$, $$\left(\widehat{\mathbb P_n}(t)\cdot e^{itb_n}\right)_{n\geqslant 1}\mbox{ is convergent and } \widehat{\mathbb P_n}(t)\to \widehat{\mathbb P}(t). $$
Therefore, denoting by $g(t)$ the limit of the sequence $\left(\widehat{\mathbb P_n}(t)\cdot e^{itb_n}\right)_{n\geqslant 1}$, we have $$\left| \widehat{\mathbb P}(t)\cdot e^{itb_n}-g(t)\right|\leqslant\left| \widehat{\mathbb P}(t)\cdot e^{itb_n}-\widehat{\mathbb P_n}(t)\cdot e^{itb_n}\right|+\left| \widehat{\mathbb P_n}(t)\cdot e^{itb_n}-g(t)\right|\leqslant \left| \widehat{\mathbb P}(t)-\widehat{\mathbb P_n}(t)\right|+\left| \widehat{\mathbb P_n}(t)\cdot e^{itb_n}-g(t)\right|,$$ hence $\lim_{n\to\infty} \widehat{\mathbb P}(t)\cdot e^{itb_n}=g(t)$ for any $t$. For $|t|$ small enough (say than $2\delta$), since $ \widehat{\mathbb P}(0)=1$, we have $\widehat{\mathbb P}(t)\neq 0$. Thus, $$\forall |t|\lt2\delta,\quad \left(e^{itb_n}\right)_{n\geqslant 1}\mbox{ is convergent, say to }f(t). $$ If $b_n=0$ for infinitely many $n$'s, then $f(t)=1$ hence $e^{itb_n}\to 1$ for each $|t|\lt 2\delta$. If it was true for any $t$, then we would readily deduce that $b_n\to 0$, but the general case is not a problem.
In the other case, $b_n\neq 0$ for $n$ large enough. Using the dominated convergence theorem, we get that the sequence $\left(\int_0^x e^{itb_n}\mathrm dt\right)_{n\geqslant 1}$ is convergent for any $|x|\lt \delta$. Choose $x$ such that $f(x)\neq 1$ (it if is not possible, we conclude like in the first case) and $\int_0^xf(t)\mathrm dt\neq 0$ to deduce the convergence of the sequence $(b_n)_{n\geqslant 1}$.