Showing that $H_n G_n \overset{d}{\rightarrow} \mathcal{N}(0,I)$ if $H_n H_n' = I$ and $G_n \overset{d}{\rightarrow} \mathcal{N}(0,I)$ Suppose $A_n,B_n$ are invertible symmetric square matrices. Define $C_n := A_nB_nA_n$ so $C_n$ is symmetric and invertible as well. Define $C_n^{-1/2}$ to be such that $C_n^{-1} = C_n^{-1/2} C_n^{-1/2}$, where $C_n^{-1/2}$ is symmetric (note that this is always possible by diagonalizing $C_n^{-1}$ since it is symmetric).
Define $H_n:=C_n^{-1/2} A_n B_n^{1/2}$, noting that $H_n H_n' = C_n^{-1/2} A_n B_n^{1/2} B_n^{-1/2}A_nC_n^{-1/2} = C_n^{-1/2} C_n C_n^{-1/2} = I$, where $I$ is the identity matrix. Suppose $G_n \overset{d}{\rightarrow} \mathcal{N}(0,I),$ $\textbf{How do I show that $H_n G_n \overset{d}{\rightarrow} \mathcal{N}(0,I)$?}$
 A: Denote the distributions of $ H_nG_n $ by $ P_n=P_{H_nG_n} $. To prove
\begin{equation*}
 P_n\Rightarrow_n N(0,I), \quad (\;\text{i.e.}, H_nG_n\overset{d}{\to}_nN(0,I)), \tag{1}
\end{equation*}
we use the sufficient condition of following Theorem (cf. P. Billingsley, Convergence of Probability Measures, 2ndEd., 1999. \S2, Th2.6, p.20).
Theorem 2.6. A necessary and sufficient condition for $ P_n\Rightarrow_n P $ is that each subsequence $ \{ P_{n_i}\} $ contain a further subsequence
$ \{ P_{n_{i(m)}}\} $ converging weakly $ (m\to\infty) $ to $P$.
Since $ H_nH'_n=I $, all elements of $H_n$ are bounded, so each subsequence $\{ H_{n_i} \}$ contain a further subsequence  $ \{ H_{n_{i(m)}}\} $
converging(elementwizely) $ (m\to\infty) $ to $H$ and $HH'=I$.
Furthermore, $P_{n_{i(m)}}\Rightarrow_m N(0,HIH')=N(0,I) $. Now (1) follows from Theorem 2.6 above.
A: This follows from Slutsky's Theorem and the continuous mapping theorem:

If $X_n\overset{d}{\longrightarrow} X$, $Y_n\overset{\mathrm P}{\longrightarrow} Y$, $n\to\infty$, then $(X_n,Y_n)\overset{d}{\longrightarrow} (X,Y)$, $n\to\infty$. In particular, for any continuous function $f(x,y)$, $f(X_n,Y_n)\overset{d}{\longrightarrow} f(X,Y)$, $n\to\infty$.

