[proofplan] The proof separates the likelihood difference into a deterministic log-determinant term and a random quadratic-form term. The revised hypotheses state precisely the two uniform Toeplitz-circulant approximations supplied by the short-memory regularity assumptions. The Gaussian, mean-zero stationary, boundedness, smoothness, and short-memory assumptions are used through those two approximation inputs; once they are available, the remaining argument is deterministic plus a probability estimate. The log-determinant term converges uniformly by the first approximation, while the quadratic-form term converges in probability by the second approximation and Markov's inequality. Adding the two estimates gives the asserted uniform convergence after the common normalising constant cancels. [/proofplan]

[step:Write the likelihood difference as a sum of determinant and quadratic-form errors] Fix a compact set $K\subset\Theta$. Let $(\Omega,\mathcal F,\mathbb P_{\theta_0})$ denote the probability space carrying the process $(X_t)_{t\in\mathbb Z}$ under the true parameter $\theta_0$, and let $\mathbb{E}_{\theta_0}$ denote expectation with respect to $\mathbb P_{\theta_0}$. Let $\mathcal L^1$ denote one-dimensional Lebesgue measure on $[-\pi,\pi]$. For each $n\in\mathbb{N}$, define the random vector \begin{align*} X_n: \Omega &\to \mathbb{R}^n \\ \omega &\mapsto (X_1(\omega),\dots,X_n(\omega))^\top. \end{align*} For $\theta\in K$, define the matrix difference \begin{align*} A_n(\theta):=\Gamma_n(\theta)^{-1}-C_n(\theta)^{-1}\in\mathbb{R}^{n\times n}. \end{align*} The Toeplitz covariance matrix $\Gamma_n(\theta)$ is positive definite because the spectral density $f_\theta$ is bounded below by a positive constant on $[-\pi,\pi]$; indeed, for every non-zero $v\in\mathbb R^n$, the spectral representation gives \begin{align*} v^\top\Gamma_n(\theta)v=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|\sum_{j=1}^{n}v_j e^{ij\lambda}\right|^2 f_\theta(\lambda)\,d\mathcal L^1(\lambda). \end{align*} The trigonometric polynomial $\lambda\mapsto \sum_{j=1}^{n}v_j e^{ij\lambda}$ is not identically zero because $v\neq 0$, and a non-zero trigonometric polynomial has only finitely many zeros on $[-\pi,\pi]$ unless it is identically zero. Hence its squared modulus is positive on a set of positive $\mathcal L^1$-measure. Since $f_\theta$ is bounded below by a positive constant, the displayed integral is strictly positive. The circulant Whittle covariance matrix $C_n(\theta)$ is positive definite by its definition in the theorem statement. Hence both inverses and determinants are well-defined. Using the definitions of the Gaussian likelihood and the Whittle objective in the theorem statement, \begin{align*} -2\ell_n(\theta)&=n\log(2\pi)+\log\det\Gamma_n(\theta)+X_n^\top\Gamma_n(\theta)^{-1}X_n,\\ 2Q_n(\theta)&=n\log(2\pi)+\log\det C_n(\theta)+X_n^\top C_n(\theta)^{-1}X_n. \end{align*} Subtracting these two identities, the common term $n\log(2\pi)$ cancels and gives \begin{align*} \frac{-2\ell_n(\theta)}{n}-\frac{2Q_n(\theta)}{n} =\frac{1}{n}\left(\log\det\Gamma_n(\theta)-\log\det C_n(\theta)\right) +\frac{1}{n}X_n^\top A_n(\theta)X_n. \end{align*} Taking the supremum over $\theta\in K$ and applying the triangle inequality yields \begin{align*} \sup_{\theta\in K}\left|\frac{-2\ell_n(\theta)}{n}-\frac{2Q_n(\theta)}{n}\right| &\leq D_n+R_n, \end{align*} where the deterministic error $D_n\in[0,\infty)$ and the random error $R_n:\Omega\to[0,\infty]$ are defined by \begin{align*} D_n&:=\sup_{\theta\in K}\left|\frac{1}{n}\log\det\Gamma_n(\theta)-\frac{1}{n}\log\det C_n(\theta)\right|,\\ R_n&:=\sup_{\theta\in K}\left|\frac{1}{n}X_n^\top A_n(\theta)X_n\right|. \end{align*} [/step]

[step:Use the uniform Toeplitz determinant approximation] By the first uniform Toeplitz-circulant approximation in the theorem statement, applied on the fixed compact set $K\subset\Theta$, we have \begin{align*} D_n=\sup_{\theta\in K}\left|\frac{1}{n}\log\det\Gamma_n(\theta)-\frac{1}{n}\log\det C_n(\theta)\right|\to 0. \end{align*} Since $D_n$ is deterministic, this convergence also implies $D_n\xrightarrow{\mathbb P}0$. [/step]

[step:Convert the uniform quadratic-form expectation bound into convergence in probability]The random variable $R_n$ is non-negative by definition. The second uniform Toeplitz-circulant approximation in the theorem statement gives \begin{align*} \mathbb{E}_{\theta_0}[R_n] =\mathbb{E}_{\theta_0}\left[\sup_{\theta\in K}\left|\frac{1}{n}X_n^\top\left(\Gamma_n(\theta)^{-1}-C_n(\theta)^{-1}\right)X_n\right|\right]\to 0. \end{align*} Let $\varepsilon>0$. Applying [Markov's Inequality](/page/Markov%27s%20Inequality) to the non-negative random variable $R_n$ gives \begin{align*} \mathbb{P}_{\theta_0}(R_n>\varepsilon)\leq \frac{\mathbb{E}_{\theta_0}[R_n]}{\varepsilon}\to 0. \end{align*} Therefore $R_n\xrightarrow{\mathbb P}0$.[/step]

[guided]The term $R_n$ is the only random part of the likelihood approximation error. We have defined \begin{align*} R_n=\sup_{\theta\in K}\left|\frac{1}{n}X_n^\top\left(\Gamma_n(\theta)^{-1}-C_n(\theta)^{-1}\right)X_n\right|, \end{align*} so $R_n\geq 0$. The theorem statement assumes exactly the uniform quadratic-form approximation \begin{align*} \mathbb{E}_{\theta_0}[R_n]\to 0. \end{align*} To pass from an $L^1$ estimate to convergence in probability, fix $\varepsilon>0$ and apply [Markov's Inequality](/page/Markov%27s%20Inequality) to the non-negative random variable $R_n$: \begin{align*} \mathbb{P}_{\theta_0}(R_n>\varepsilon)\leq \frac{\mathbb{E}_{\theta_0}[R_n]}{\varepsilon}. \end{align*} The denominator $\varepsilon$ is fixed and positive, while the numerator tends to $0$. Hence \begin{align*} \mathbb{P}_{\theta_0}(R_n>\varepsilon)\to 0. \end{align*} This is precisely the definition of $R_n\xrightarrow{\mathbb P}0$.[/guided]

[step:Combine the two error bounds to obtain the uniform Whittle approximation] Let $\varepsilon>0$. Since $D_n\to 0$, there exists $N\in\mathbb{N}$ such that $D_n\leq \varepsilon/2$ for every $n\geq N$. For such $n$, the bound from the first step gives \begin{align*} \left\{\sup_{\theta\in K}\left|\frac{-2\ell_n(\theta)}{n}-\frac{2Q_n(\theta)}{n}\right|>\varepsilon\right\} \subseteq \left\{R_n>\frac{\varepsilon}{2}\right\}. \end{align*} Taking probabilities with respect to the true law $\mathbb{P}_{\theta_0}$ and using $R_n\xrightarrow{\mathbb P}0$, we obtain \begin{align*} \mathbb{P}_{\theta_0}\left(\sup_{\theta\in K}\left|\frac{-2\ell_n(\theta)}{n}-\frac{2Q_n(\theta)}{n}\right|>\varepsilon\right) \leq \mathbb{P}_{\theta_0}\left(R_n>\frac{\varepsilon}{2}\right)\to 0. \end{align*} Because $\varepsilon>0$ was arbitrary, this proves \begin{align*} \sup_{\theta\in K}\left|\frac{-2\ell_n(\theta)}{n}-\frac{2Q_n(\theta)}{n}\right|\xrightarrow{\mathbb P}0. \end{align*} [/step]