Suppose that $g_n: \mathbb{R}^k \to \mathbb{R}^h$ are random linear mappings such that $$ g_n \stackrel{P}{\to} g, \quad \quad \text{as } n\to\infty, $$ where the non-random limit $g$ is injective. Now assume that $$ g_n(\hat{\beta}_n - \beta_0) \stackrel{P}{\to} 0 , \quad \quad \text{as } n\to\infty. $$ where $\hat{\beta}_n$ is some estimator of $\beta_0$.
Does it follow that $\hat{\beta}_n \stackrel{P}{\to} \beta_0$?
Heuristically I would assume that something like the following steps could be used $$ g_n(\hat{\beta}_n - \beta_0) \stackrel{P}{\to} g(\text{P-lim}\hat{\beta}_n - \beta_0) =0 \iff \text{P-lim}\hat{\beta}_n =\beta_0, $$ as $g$ is injective.
However I have failed to show this formally. How would one argue these steps formally?
My attempt based on the above hints: Note that as $g$ is injective we have that $rank(g) = G$, and as such $rank(g^t g) = rank(g) = G$ so $g^t g$ is invertible. Furthermore by Slutsky's theorem we get that \begin{align*} g_n \stackrel{P}{\to}g \implies g_n^t g_n \stackrel{P}{\to} g^t g \end{align*} as $n\to\infty$, that is for any $\epsilon>0$ \begin{align*} P(||g_n^t g_n-g^t g\| \leq \epsilon) =P(g_n^t g_n \in \overline{B(g^t g,\epsilon)}) \to_n 1. \end{align*} Now note that the set of all non-singular $G\times G$ matrices $NS(G,G)$ is an open subset of all $G\times G$ matrices $M(G,G)$, and as such we know that there exists an $\epsilon>0$ such that \begin{align*} \overline{B(g^t g,\epsilon)} \subset NS(G,G), \end{align*} hence $g_n^t g_n$ will be invertible with probability tending towards 1, that is \begin{align*} P(g_n^t g_n \in NS(G,G)) \underset{n\to\infty}{\to } 1. \end{align*} Let $h_n:\Omega\to NS(G,G)$ be given by \begin{align*} h_n(\omega) = \left\{ \begin{array}{ll} g_n^t(\omega) g_n(\omega), & \text{if } \omega\in (g_n^t g_n \in NS(G,G)) \\ I, & \text{otherwise} \end{array}\right. \end{align*} Realize that $h_n \underset{n\to\infty}{\stackrel{P}{\to}} g^t g$, since for any $\epsilon >0$ \begin{align*} P(\|h_n-g^t g\| >\epsilon ) = &P((\|g_n^t g_n-g^t g\|> \epsilon) \cap (g_n^t g_n \in NS(G,G))) \\ & + P((\|I-g^t g\|> \epsilon )\cap (g_n^t g_n \in NS(G,G))^c) \\ \leq & P(\|g_n^t g_n-g^t g\|> \epsilon) + P(g_n^t g_n \in NS(G,G))^c). \\ \to_n & 0, \end{align*} By continuity of the inverse operator and the continuous mapping theorem we have that $\|h_n^{-1}\|\underset{n\to\infty}{\stackrel{P}{\to}} \|(g^t g)^{-1}\|\in \mathbb{R}$. Furthermore, \begin{align*} \|g_n^t g_n(\hat{\beta}_n-\beta_0) \| \leq \|g_n^t \| \| g_n(\hat{\beta}_n-\beta_0) \| \underset{n\to\infty}{\stackrel{P}{\to}} \|g^t \| \cdot 0= 0, \end{align*} by the assumptions and Slutsky's theorem. Hence for any $\epsilon>0$ \begin{align*} P( \|h_n(\hat{\beta}_n-\beta_0) \| > \epsilon ) =& P( (\|g_n^t g_n(\hat{\beta}_n-\beta_0) \| > \epsilon ) \cap (g_n^t g_n \in NS(G,G)) ) \\ &+ P( (\|\hat{\beta}_n-\beta_0\| > \epsilon) \cap (g_n^t g_n \in NS(G,G))^c) \\ \leq & P( (\|g_n^t g_n(\hat{\beta}_n-\beta_0) \| > \epsilon ) ) + P((g_n^t g_n \in NS(G,G))^c)\\ \to_n & 0. \end{align*} Thus $$ \|\hat{\beta}_n-\beta_0\| = \| h_n^{-1} h_n(\hat{\beta}_n-\beta_0) \| \leq \|h_n^{-1}\| \| h_n(\hat{\beta}_n-\beta_0) \| \underset{n\to\infty}{\stackrel{P}{\to}} \|(g^t g)^{-1}\| \cdot 0=0, $$ by Slutsky's theorem, yielding that $\hat{\beta}_n$ is an consistent estimator of $\beta_0$.