$h_n \to 0 \implies \int_{-\infty}^{+\infty} (\exp(itx)-1-itx/(1+x^2))\frac{1+x^2}{x^2} \nu(dx) \to 0$

Question

Let $\nu$ be a finite Borel measure. I'm trying to prove the following:

$$h_n \to 0 \implies \int_{-\infty}^{+\infty} \left(\exp(ih_nx)-1-\frac{ih_nx}{1+x^2}\right)\frac{1+x^2}{x^2} \nu(dx) \to 0$$

here at $x=0$ the integrand is interpreted to be $-h_n^2/2$.

I guess this is a consequence of the dominated convergence theorem. I see that for every $x \in \mathbb{R}$ the integrand converges to $0$ so if I can find a dominating function, or show that the integrand is bounded ($\nu$ is finite), I'm done. I have trouble doing this though.

For $|x| \geq 1$, I can see that the modulus of the integrand becomes smaller than $4+|h_n|$, which is bounded because $(h_n)_n$ converges. However, for $|x|< 1$ I don't see how to proceed.

Answer 1

• We first consider the case $|x| < 1$. Using the identity

  $$ e^{i\theta} = 1 + i\theta - \theta^2 \int_{0}^{1} (1 - t) e^{i\theta t} \, \mathrm{d}t, $$

  it follows that $e^{i\theta} = 1 + i\theta + \mathcal{O}(\theta^2)$ uniformly in $\theta \in \mathbb{R}$.¹⁾ From this,

  \begin{align*} \biggl( e^{ih_n x} - 1 - \frac{ih_n x}{1+x^2} \biggr) \frac{1+x^2}{x^2} &= \biggl( ih_n x + \mathcal{O}(h_n^2 x^2) - \frac{ih_n x}{1+x^2} \biggr) \frac{1+x^2}{x^2} \\ &= \biggl( \frac{ih_n x^3}{1+x^2} + \mathcal{O}(h_n^2 x^2) \biggr) \frac{1+x^2}{x^2} \\ &= ih_n x + \mathcal{O}(h_n^2(1+x^2)). \end{align*}

• If $|x| \geq 1$, then

  \begin{align*} \left| \biggl( e^{ih_n x} - 1 - \frac{ih_n x}{1+x^2} \biggr) \frac{1+x^2}{x^2} \right| &\leq 4 + |h_n|. \end{align*}

Combining altogether, there exists a uniform constant $C > 0$ such that

\begin{align*} \left| \biggl( e^{ih_n x} - 1 - \frac{ih_n x}{1+x^2} \biggr) \frac{1+x^2}{x^2} \right| &\leq C(1 + |h_n| + |h_n|^2), \qquad \forall x \in \mathbb{R}. \end{align*}

Now assuming that $h_n \to 0$, we can invoke the dominated convergence theorem to conclude:

\begin{align*} &\lim_{n\to\infty} \int_{\mathbb{R}} \biggl( e^{ih_n x} - 1 - \frac{ih_n x}{1+x^2} \biggr) \frac{1+x^2}{x^2} \, \nu(\mathrm{d}x) \\ &= \int_{\mathbb{R}} \lim_{n\to\infty} \biggl( e^{ih_n x} - 1 - \frac{ih_n x}{1+x^2} \biggr) \frac{1+x^2}{x^2} \, \nu(\mathrm{d}x) \\ &= 0. \end{align*}

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^{1) For the purpose of proving the implication in question, this uniform estimate is a bit overshoot. Instead, we may assume that $|h_n| < 1$ and then use the Taylor approximation $e^{iz} = 1 + iz + \mathcal{O}(z^2)$, which holds for any complex $z$ on a given bounded neighborhood of $0$, say the unit disk of radius 1.}