How do I show that $\int_\mathbb{R} \sigma \exp(-\frac{\sigma^2 \xi^2}{2})f(\xi)d\xi\to c<\infty$ as $\sigma\to\infty$? Let $f$ be a bounded probability density function. How do I show that $\int_\mathbb{R} \sigma \exp(-\frac{\sigma^2 \xi^2}{2})f(\xi)d\xi\to c<\infty$ as $\sigma\to\infty$?
I was thinking of doing dominated convergence theorem, but this doesn't work here. Could anyone give me any direction?
 A: With the change of variables $u=\sigma \xi$ we can rewrite your integral as
$$\int_{\mathbb{R}} e^{-u^2/2} f(u/\sigma) \, du.$$
This integrand is dominated by $e^{-u^2/2} (\sup_{x \in \mathbb{R}} f(x))$ which is integrable. If you show that the pointwise limit (as $\sigma \to \infty$) of the integrand is zero almost everywhere (except at $u=0$ where the limit is $f(0)$) then you can conclude by applying the dominated convergence theorem.
A: Hint: $f(\xi)$ is bounded and all terms are positive in the integral. Can you find an upper bound for the integrand?
Let us assume $f(\xi)\leq \dfrac{c}{\sqrt{2\pi}}$.
$$\int_{-\infty}^{\infty}\sigma\exp\left[-\dfrac{\sigma^2\xi^2}{2}\right]f(\xi)d\xi\leq \int_{-\infty}^{\infty}\sigma\exp\left[-\dfrac{\sigma^2\xi^2}{2}\right]\dfrac{c}{\sqrt{2\pi}}d\xi$$
$$=\sigma\dfrac{c}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\exp\left[-\dfrac{\sigma^2\xi^2}{2}\right]d\xi$$
Note, that the upper bound of the integral is gaussian integral. And we have
$$\int_{-\infty}^{\infty}\exp\left[-\dfrac{\sigma^2\xi^2}{2}\right]d\xi = \dfrac{\sqrt{2\pi}}{|\sigma|}.$$
Hence, we obtain
$$\int_{-\infty}^{\infty}\sigma\exp\left[-\dfrac{\sigma^2\xi^2}{2}\right]f(\xi)d\xi\leq\sigma\dfrac{c}{\sqrt{2\pi}}\dfrac{\sqrt{2\pi}}{|\sigma|}=c.$$
