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Spring 2011 - Problem 3

measure theory

Let $\mu$ be a Borel probability measure on $\R$ and define $f\p{t} = \int e^{itx} \diff\mu\p{x}$ . Suppose that

\lim_{t\to0} \frac{f\p{0} - f\p{t}}{t^2} = 0.

Show that $\mu$ is supported at $\set{0}$ .

Solution.

Observe that by taking real parts,

\Re\p{\frac{f\p{0} - f\p{t}}{t^2}} = \int \frac{1 - \cos{tx}}{t^2} \,\diff\mu\p{x},

Since $\cos{x} \leq 1$ the integrand is non-negative, so by Fatou's lemma,

\begin{aligned} 0 &= \liminf_{t\to0} \int \frac{1 - \cos{tx}}{t^2} \,\diff\mu\p{x} \\ &\geq \int \liminf_{t\to0} \frac{1 - \cos{tx}}{t^2} \,\diff\mu\p{x} \\ &= \int \frac{x^2}{2} \,\diff\mu\p{x} \\ &= \int_{\set{x^2 > 0}} \frac{x^2}{2} \,\diff\mu\p{x}. \end{aligned}

Suppose $\mu$ were supported away from $\set{0}$ . Since $\mu$ is Borel, this means that there exists a closed interval $I$ not containing $0$ such that $\mu\p{I} > 0$ . Thus, $x^2 \geq \delta > 0$ for some $\delta > 0$ on $I$ since $x^2$ is increasing, and so

0 \geq \int_{\set{x^2 > 0}} \frac{x^2}{2} \,\diff\mu\p{x} \geq \int_I \frac{\delta}{2} \,\diff\mu\p{x} = \frac{\delta\mu\p{I}}{2} > 0,

which is impossible. Thus, $\mu$ must have been supported on $\set{0}$ to begin with.