If $\|u_n\|_2$ is bounded, then you have the convergence you need.

For fixed $u$ this is the Mann-Wold theorem: convergence in distribution of a vector is equivalent to convergence of all one-dimensional projections.

For $u_n\to u$: vector dot product $(u_n,\hat\theta)\mapsto u^t_n\hat\theta$ is continuous, so use the continuous mapping theorem.

Finally, for bounded $u_n$, take subsequences. Any subsequence of a bounded sequence has a convergent subsubsequence. Along this subsubsequence $\sigma_n^{-1}(u_n\hat\theta-u_n\theta^*)\to N(0,1)$. But that is just to say that $\sigma_n^{-1}(u_n\hat\theta-u_n\theta^*)\to N(0,1)$ for the whole sequence.

If $\|u_n\|_2$ is bounded, then you have the convergence you need.

For fixed $u$ this is the Mann-Wold theorem: convergence in distribution of a vector is equivalent to convergence of all one-dimensional projections.

For $u_n\to u$: vector dot product $(u_n,\hat\theta)\mapsto u^t_n\hat\theta$ is continuous, so use the continuous mapping theorem.

Finally, for bounded $u_n$, take subsequences. Any subsequence of a bounded sequence has a convergent subsubsequence. Along this subsubsequence $\Sigma_n^{-1/2}(u_n\hat\theta-u_n\theta^*)\to N(0,1)$. But that is just to say that $\Sigma_n^{-1/2}(u_n\hat\theta-u_n\theta^*)\to N(0,1)$ for the whole sequence.

If $\|u_n\|_2$ is bounded then you have the convergence you need.

For fixed $u$ this is the Mann-Wold theorem: convergence in distribution of a vector is equivalent to convergence of all one-dimensional projections.

For $u_n\to u$: vector dot product $(u_n,\hat\theta)\mapsto u^t_n\hat\theta$ is continuous, so use the continuous mapping theorem.

Finally, for bounded $u_n$, take subsequences. Any subsequence of a bounded sequence has a convergent subsubsequence. Along this subsubsequence $\sigma_n^{-1}(u_n\hat\theta-u_ntheta^*)\to N(0,1)$. But that is just to say that $\sigma_n^{-1}(u_n\hat\theta-u_n\theta^*)\to N(0,1)$ for the whole sequence.

If $\|u_n\|_2$ is bounded, then you have the convergence you need.

For fixed $u$ this is the Mann-Wold theorem: convergence in distribution of a vector is equivalent to convergence of all one-dimensional projections.

For $u_n\to u$: vector dot product $(u_n,\hat\theta)\mapsto u^t_n\hat\theta$ is continuous, so use the continuous mapping theorem.

Finally, for bounded $u_n$, take subsequences. Any subsequence of a bounded sequence has a convergent subsubsequence. Along this subsubsequence $\sigma_n^{-1}(u_n\hat\theta-u_n\theta^*)\to N(0,1)$. But that is just to say that $\sigma_n^{-1}(u_n\hat\theta-u_n\theta^*)\to N(0,1)$ for the whole sequence.