Let $f(x)$ has a derivative in point $x = 0$. Then if for sequences of random variables $\left\{\xi_n\right\}_{n = 1} ^ \infty, \left\{\eta_n\right\}_{n = 1} ^ \infty$ we have:
$$
\xi_n\eta_n \xrightarrow{d} \eta\\
\eta_n \xrightarrow{\mathbb{P}} 0
$$
then $\xi_n(f(\eta_n) – f(0)) \xrightarrow{d} f'(0)\eta$.
My solution looks something like this:
By the hereditary convergence theorem for continuous function $h(x) = 1/x$ i can write $\frac{1}{\xi_n\eta_n} \xrightarrow{d} \frac{1}{\eta}$. Since $\eta_n \xrightarrow{\mathbb{P}} 0$, then $\eta_n \xrightarrow{d} 0$. So, i can write by Slutsky's Theorem that $\frac{\eta_n}{\xi_n\eta_n} \xrightarrow{d} 0$. Then i use following theorem.
Let $\xi_n \xrightarrow{d} \xi$. For a function $h(x)$ differentiable at a point and a sequence $b_n \rightarrow 0, b_n \neq 0$, the following holds:
$$
\frac{h(a + \xi_nb_n) – h(a)}{b_n} \xrightarrow{d} h'(a)\xi
$$
where I put $b_n = \frac{1}{\xi_n}$ and $\xi_n = \xi_n\eta_n$
But I'm not sure about this solution because the function $\frac{1}{x}$ is not continuous at zero. I would be very grateful for the hint
Best Answer
Since $\eta_n \to 0$ in probability if and only if every subsequence of $(\eta_n)$ has a further subsequence which converges almost surely to $0$ we can reduce the proof to the case when $\eta_n \to 0$ almost surely.
$\xi_n(f(\eta_n) - f(0))=\xi_n\eta_n \frac {f(\eta_n)-f(0)}{\eta_n}$. Let $X_n=\xi_n\eta_n $ and $Y_n=\frac {f(\eta_n)-f(0)}{\eta_n}$. $X_n$ converges in distribution to $\eta$ and $Y_n \to f'(0)$ almost surely. This implies that $X_nY_n \to f'(0)\eta$ in distribution.
Note: The appearence of $\eta_n$ in the denominator does not cause a problem even though $\eta_n$ can take the value $0$. Simply replace the ratio $\frac {f(\eta_n)-f(0)}{\eta_n}$ by $f'(0)$ whenever $\eta_n=0$.