[proofplan] We split $\hat f_n(\omega)-f(\omega)$ into a deterministic bias term and a centred random term. The bias term is controlled by the continuity of $f$ and by the fact that the smoothing window has width $2\pi m_n/n \to 0$. The random term is controlled by computing its variance: the diagonal terms vanish because $\sum_k W_{n,k}^2 \leq \max_k W_{n,k}$, and the off-diagonal terms vanish by the assumed asymptotic uncorrelatedness of separated Fourier ordinates. The variance convergence then gives convergence in probability by the elementary second-moment bound. [/proofplan]

[step:Fix an interior frequency and define the local ordinates] Fix $\omega \in (0,\pi)$. Since $m_n/n \to 0$, there exists $n_0 \in \mathbb{N}$ such that, for every $n \geq n_0$ and every integer $k$ with $|k| \leq m_n$, \begin{align*} 0 < \omega + \frac{2\pi k}{n} < \pi. \end{align*} For $n \geq n_0$ and $|k| \leq m_n$, define the local frequency \begin{align*} \lambda_{n,k} := \omega + \frac{2\pi k}{n} \end{align*} and the local periodogram ordinate \begin{align*} Y_{n,k} := I_n(\lambda_{n,k}). \end{align*} Then \begin{align*} \hat f_n(\omega) = \sum_{k=-m_n}^{m_n} W_{n,k}Y_{n,k}. \end{align*} [/step]

[step:Decompose the estimation error into bias and fluctuation] Define the deterministic bias term $B_n(\omega) \in \mathbb{R}$ and the centred fluctuation term $R_n(\omega)$ by \begin{align*} B_n(\omega) &:= \mathbb{E}[\hat f_n(\omega)] - f(\omega),\\ R_n(\omega) &:= \hat f_n(\omega) - \mathbb{E}[\hat f_n(\omega)]. \end{align*} Then \begin{align*} \hat f_n(\omega)-f(\omega)=B_n(\omega)+R_n(\omega). \end{align*} It is therefore enough to prove $B_n(\omega) \to 0$ and $R_n(\omega) \xrightarrow{\mathbb{P}} 0$. [/step]

[step:Show that the smoothing bias vanishes] Using the definition of $\hat f_n(\omega)$ and the identity $\sum_{k=-m_n}^{m_n} W_{n,k}=1$, we obtain \begin{align*} B_n(\omega) &= \sum_{k=-m_n}^{m_n} W_{n,k}\mathbb{E}[Y_{n,k}] - f(\omega)\\ &= \sum_{k=-m_n}^{m_n} W_{n,k}\left(\mathbb{E}[Y_{n,k}]-f(\lambda_{n,k})\right) + \sum_{k=-m_n}^{m_n} W_{n,k}\left(f(\lambda_{n,k})-f(\omega)\right). \end{align*} Taking absolute values and using non-negativity of the weights gives \begin{align*} |B_n(\omega)| &\leq \max_{|k| \leq m_n} \left|\mathbb{E}[Y_{n,k}]-f(\lambda_{n,k})\right| + \max_{|k| \leq m_n} |f(\lambda_{n,k})-f(\omega)|. \end{align*} The first term tends to $0$ by the assumed local asymptotic unbiasedness. For the second term, observe that \begin{align*} |\lambda_{n,k}-\omega| = \frac{2\pi |k|}{n} \leq \frac{2\pi m_n}{n} \to 0. \end{align*} Since $f$ is continuous at $\omega$, this implies \begin{align*} \max_{|k| \leq m_n}|f(\lambda_{n,k})-f(\omega)| \to 0. \end{align*} Thus $B_n(\omega) \to 0$. [/step]

[step:Bound the variance of the centred fluctuation] Let \begin{align*} M(\omega) := \sup_{n \in \mathbb{N}}\max_{|k| \leq m_n}\operatorname{Var}(Y_{n,k}). \end{align*} By hypothesis, $M(\omega)<\infty$. Since $R_n(\omega)$ is centred, \begin{align*} \mathbb{E}[R_n(\omega)^2] = \operatorname{Var}(\hat f_n(\omega)). \end{align*} Expanding the variance of the weighted sum gives \begin{align*} \operatorname{Var}(\hat f_n(\omega)) &= \sum_{k=-m_n}^{m_n} W_{n,k}^2\operatorname{Var}(Y_{n,k})\\ &\quad+ \sum_{\substack{|k|,|\ell| \leq m_n\\ k \neq \ell}} W_{n,k}W_{n,\ell}\operatorname{Cov}(Y_{n,k},Y_{n,\ell}). \end{align*} For the diagonal part, since $0 \leq W_{n,k} \leq \max_{|j|\leq m_n}W_{n,j}$ and $\sum_k W_{n,k}=1$, \begin{align*} \sum_{k=-m_n}^{m_n} W_{n,k}^2\operatorname{Var}(Y_{n,k}) &\leq M(\omega)\sum_{k=-m_n}^{m_n} W_{n,k}^2\\ &\leq M(\omega)\max_{|k| \leq m_n} W_{n,k} \sum_{k=-m_n}^{m_n} W_{n,k}\\ &= M(\omega)\max_{|k| \leq m_n} W_{n,k} \to 0. \end{align*} For the off-diagonal part, define \begin{align*} C_n(\omega) := \max_{\substack{|k|,|\ell| \leq m_n\\ k \neq \ell}} |\operatorname{Cov}(Y_{n,k},Y_{n,\ell})|. \end{align*} By hypothesis, $C_n(\omega)\to0$. Therefore \begin{align*} \left| \sum_{\substack{|k|,|\ell| \leq m_n\\ k \neq \ell}} W_{n,k}W_{n,\ell}\operatorname{Cov}(Y_{n,k},Y_{n,\ell}) \right| &\leq C_n(\omega) \sum_{\substack{|k|,|\ell| \leq m_n\\ k \neq \ell}} W_{n,k}W_{n,\ell}\\ &\leq C_n(\omega) \left(\sum_{k=-m_n}^{m_n} W_{n,k}\right)^2\\ &= C_n(\omega) \to 0. \end{align*} Combining the diagonal and off-diagonal bounds yields \begin{align*} \operatorname{Var}(\hat f_n(\omega)) \to 0. \end{align*} [/step]

[step:Convert the variance bound into convergence in probability] Let $\varepsilon>0$. Since $R_n(\omega)$ is centred, the elementary second-moment estimate gives \begin{align*} \mathbb{P}\left(|R_n(\omega)|>\varepsilon\right) \leq \frac{\mathbb{E}[R_n(\omega)^2]}{\varepsilon^2} = \frac{\operatorname{Var}(\hat f_n(\omega))}{\varepsilon^2}. \end{align*} The variance convergence proved above implies \begin{align*} \mathbb{P}\left(|R_n(\omega)|>\varepsilon\right)\to0. \end{align*} Hence $R_n(\omega)\xrightarrow{\mathbb{P}}0$. Finally, $B_n(\omega)\to0$ deterministically, so for every $\varepsilon>0$ and all sufficiently large $n$, \begin{align*} |B_n(\omega)|\leq \frac{\varepsilon}{2}. \end{align*} Therefore \begin{align*} \mathbb{P}\left(|\hat f_n(\omega)-f(\omega)|>\varepsilon\right) &= \mathbb{P}\left(|B_n(\omega)+R_n(\omega)|>\varepsilon\right)\\ &\leq \mathbb{P}\left(|R_n(\omega)|>\frac{\varepsilon}{2}\right) \to0. \end{align*} This proves \begin{align*} \hat f_n(\omega)\xrightarrow{\mathbb{P}} f(\omega), \end{align*} as required. [/step]