Let $g_1, g_2, \dots$ be continuous functions on $(a, b) \subset {\rm I\!R}$ such that $g_n(x) \rightarrow g(x)$ uniformly for $x$ in any closed subinterval of $(a, b)$. Let $T_n$ be a consistent estimator of $\theta \in (a, b)$. Show that $g_n(T_n)$ is consistent for $\vartheta = g(\theta)$.
I thought of using the basic concept of consistency, where $T_n(X)$ is consistent for $\vartheta$ if and only if $T_n(X) \rightarrow_p \vartheta$. But I'm stuck at this $g_n$. Any hint?
I think you can use the Continuous Mapping Theorem to show this. You need to use the uniform convergence to bound the differences of $g_n(T_m)$ and $g(T_m)$.
Specify $\delta>0$ and $x$ in the support of $T_m$, then we have:
\begin{equation} \begin{split} P[|g_n(x) - \nu| > \delta] & = P[|g_n(x) - g(x) + g(x) - \nu| > \delta]\\ & \leq P[|g_n(x) - g(x)| + |g(x) - \nu| > \delta] \end{split} \end{equation}
With the last inequality following by the triangle inequality.
By uniform convergence, $\forall \phi >0$ there exists $N$ such that $\forall n \geq N$ $|g_n(x) - g(x)|< \phi$ for all $x$.
This gives us that $\forall n\geq N$ $\phi + |g(x) - \nu|>|g_n(x) - g(x)| + |g(x) - \nu| > \delta$. Then, $\forall n > N$:
\begin{equation} \begin{split} P[|g_n(x) - \nu| > \delta] &\leq P[\phi + |g(x) - \nu| > \delta]\\ \Leftrightarrow P[|g_n(x) - \nu| > \delta] &\leq P[|g(x) - \nu| > \delta -\phi] \end{split} \end{equation}
By convergence in probability of $T_m \overset{P}{\rightarrow}\theta$, we have that if $g(x)$ is continuous then $g(T_m) \overset{P}{\rightarrow}\vartheta$ by the Continuous Mapping Theorem. A uniformly convergent sequence of continuous functions has a continuous limit, giving us the previous convergence in probability.
Replacing $x$ by $T_m$ in the previous expression we have:
\begin{equation} P[|g_n(T_m) - \nu| > \delta] \leq P[|g(T_m) - \nu| > \delta -\phi] \end{equation}
$g(T_m) \overset{P}{\rightarrow} \vartheta$ means that $\forall \eta = \delta - \phi >0$ $\forall \epsilon > 0$ $\exists M$ such that $\forall m \geq M$ we have $P[|g(T_m) - \nu| > \eta] < \epsilon$. (Strictly speaking $\eta$ could be less than zero as written, but the absolute value makes this not a problem.)
This means that for for $\epsilon>0$ and $\delta >0$ there exists an $M$ and $N$ such that $\forall m \geq M$ and $\forall n \geq N$:
\begin{equation} P[|g_n(T_m) - \nu| > \delta] <\epsilon \end{equation}
Specifying $K = \max\{M,N\}$ we have that $\forall k \geq K$ $P[|g_k(T_k) - \nu| > \delta] < \epsilon$. This means that $g_k(T_k) \overset{P}{\rightarrow} \vartheta$, so it's a consistent estimator.