I want to show$$\lim\limits_{t\to 0} \sup\limits_{n}\int (1-\cos tx)dM_n(x) = 0 \ \ \ \ \ \ \ \text{ implies } \ \ \ \ \ \ \ \sup\limits_{n}M_n\{ x : \varepsilon \leq |x| \} < \infty.$$
$M_n$ are positive measures and I need to show the above statement holds for any $\varepsilon > 0$. I'm having a surprisingly hard time doing so.
I know that $1-\cos tx \geq 0$, and decreasing $t$ increases the period. I thought about choosing two small $t_1, t_2$ with $\frac{t_1}{t_2}$ irrational so that the function $$f(x) = \max\{1-\cos t_1x, 1 - \cos t_2x\} > 0 $$ on $\{x: \varepsilon \leq |x|\}$. Then perhaps I could try using some estimate like $$M_n\{x: \varepsilon \leq |x|\} \cdot\inf_{\varepsilon \leq |x|} f(x) \leq \int (1-\cos t_1x)dM_n(x) + \int (1-\cos t_2x)dM_n(x). $$ But unfortunately, the infimum occuring on the left of the inequality could be $0$. So I don't know what to do. Perhaps I'm missing something easy.
Please help me, I'm struggling :|
Edit: If it helps, this is at the top of page 88 of Varadhan's notes.
I should probably have added: The $M_n$ are positive measures on $\mathbb{R}$, each assumed to satisfy $$\int \frac{x^2}{1+x^2} dM_n(x) < \infty. $$ The book calls them Levy measures.
Fix $T$ so that $\sup_n\int (1-\cos(tx)\,dM_n(x)<1$ for all $t\in [-T,T]$. For fixed $n$, integrate $\int 1-\cos(tx)\,dM_n(x)$ with respect to $t$ on $[-T,T]$, interchange integrals, and divide by $2T$ to find that $$\int 1-\frac{\sin Tx}{Tx}\,dM_n(x)< 1$$ uniformly in $n$. You may check that $1-\frac{\sin Tx}{Tx}\geq C_{T,\delta}>0$ for all $x\geq \delta>0$ from which the claim follows.
Full disclosure: I found this argument by adapting parts of the proof of theorem 3.19 in the linked pdf.