[proofplan]
Write $X_n := Y_n-\mu$ and first replace the sample-mean autocovariances by oracle autocovariances computed with the true mean. The oracle estimator is split into a deterministic bias term and a stochastic fluctuation term. The bias vanishes because the autocovariance series is absolutely summable, $w$ is continuous at $0$, and $b_N\to\infty$; the fluctuation vanishes by the lag-window variance estimate assumed in the theorem statement together with
\begin{align*}
\frac{b_N}{N}\to0.
\end{align*}
Finally, the correction caused by replacing $\mu$ with $\bar Y_N$ is negligible because the statement assumes
\begin{align*}
\frac{b_N}{\sqrt{N}}\to0,
\end{align*}
while absolute summability of the autocovariances gives a uniform bound on $N\mathbb E[(\bar Y_N-\mu)^2]$.
[/proofplan]
[step:Introduce centered variables and the oracle lag-window estimator]
Define the centered process $(X_n)_{n\ge 1}$ by
\begin{align*}
X_n := Y_n-\mu.
\end{align*}
Then $(X_n)_{n\ge 1}$ is strictly stationary, $\mathbb E[X_1]=0$, $\mathbb E[|X_1|^{4+\delta}]<\infty$, and
\begin{align*}
\gamma_k = \mathbb E[X_1X_{1+|k|}]
\end{align*}
for every $k\in\mathbb Z$.
For $0\le k\le N-1$, define the oracle sample autocovariance
\begin{align*}
\tilde\gamma_k := \frac{1}{N}\sum_{n=1}^{N-k}X_nX_{n+k}.
\end{align*}
Define the oracle lag-window estimator
\begin{align*}
\tilde\sigma_N^2 := \tilde\gamma_0 + 2\sum_{k=1}^{N-1}w\left(\frac{k}{b_N}\right)\tilde\gamma_k.
\end{align*}
It is enough to prove
\begin{align*}
\tilde\sigma_N^2 \xrightarrow{\mathbb P} \sigma_h^2
\end{align*}
and
\begin{align*}
\hat\sigma^2_{\mathrm{SV}}-\tilde\sigma_N^2 \xrightarrow{\mathbb P}0.
\end{align*}
[/step]
[step:Show the deterministic lag-window bias tends to zero]
For each $0\le k\le N-1$, stationarity gives
\begin{align*}
\mathbb E[\tilde\gamma_k] = \frac{N-k}{N}\gamma_k.
\end{align*}
Hence
\begin{align*}
\mathbb E[\tilde\sigma_N^2] = \gamma_0 + 2\sum_{k=1}^{N-1}w\left(\frac{k}{b_N}\right)\frac{N-k}{N}\gamma_k.
\end{align*}
Therefore
\begin{align*}
\mathbb E[\tilde\sigma_N^2]-\sigma_h^2 = 2\sum_{k=1}^{\infty}\left[\mathbb 1_{\{k\le N-1\}}w\left(\frac{k}{b_N}\right)\frac{N-k}{N} - 1\right]\gamma_k.
\end{align*}
For each fixed $k\ge1$, since $b_N\to\infty$ and $w$ is continuous at $0$ with $w(0)=1$,
\begin{align*}
\mathbb 1_{\{k\le N-1\}}w\left(\frac{k}{b_N}\right)\frac{N-k}{N}
\to 1.
\end{align*}
Moreover,
\begin{align*}
\left|\left[\mathbb 1_{\{k\le N-1\}}w\left(\frac{k}{b_N}\right)\frac{N-k}{N} - 1\right]\gamma_k\right| \le 2|\gamma_k|,
\end{align*}
because $|w|\le1$ and $0\le (N-k)/N\le1$. Since $\sum_{k=1}^{\infty}|\gamma_k|<\infty$, dominated convergence for series gives
\begin{align*}
\mathbb E[\tilde\sigma_N^2]\to \sigma_h^2.
\end{align*}
[guided]
The deterministic bias is the difference between the expected oracle estimator and the infinite autocovariance sum. We first compute the expectation of each oracle autocovariance. Since $(X_n)_{n\ge1}$ is stationary and centered, for every $n$ and every $k\ge0$ with $n+k\le N$,
\begin{align*}
\mathbb E[X_nX_{n+k}] = \gamma_k.
\end{align*}
There are exactly $N-k$ terms in the sum defining $\tilde\gamma_k$, so
\begin{align*}
\mathbb E[\tilde\gamma_k]
=
\mathbb E\left[\frac{1}{N}\sum_{n=1}^{N-k}X_nX_{n+k}\right]
=
\frac{1}{N}\sum_{n=1}^{N-k}\gamma_k
=
\frac{N-k}{N}\gamma_k.
\end{align*}
Substituting this into the oracle estimator gives
\begin{align*}
\mathbb E[\tilde\sigma_N^2] = \gamma_0 + 2\sum_{k=1}^{N-1}w\left(\frac{k}{b_N}\right)\frac{N-k}{N}\gamma_k.
\end{align*}
The target value is
\begin{align*}
\sigma_h^2=\gamma_0+2\sum_{k=1}^{\infty}\gamma_k.
\end{align*}
Thus the bias is
\begin{align*}
\mathbb E[\tilde\sigma_N^2]-\sigma_h^2 = 2\sum_{k=1}^{\infty}\left[\mathbb 1_{\{k\le N-1\}}w\left(\frac{k}{b_N}\right)\frac{N-k}{N} - 1\right]\gamma_k.
\end{align*}
For a fixed lag $k$, the indicator is eventually $1$, the factor $(N-k)/N$ tends to $1$, and
\begin{align*}
\frac{k}{b_N}\to0.
\end{align*}
Since $w$ is continuous at $0$ and $w(0)=1$,
\begin{align*}
w\left(\frac{k}{b_N}\right)\to 1.
\end{align*}
Therefore the bracketed coefficient tends to $0$ for every fixed $k$.
To pass from fixed lags to the whole infinite series, we use the absolute summability assumption. The bound $|w|\le1$ and $0\le (N-k)/N\le1$ imply
\begin{align*}
\left|\left[\mathbb 1_{\{k\le N-1\}}w\left(\frac{k}{b_N}\right)\frac{N-k}{N} - 1\right]\gamma_k\right| \le 2|\gamma_k|.
\end{align*}
The dominating series $\sum_{k=1}^{\infty}2|\gamma_k|$ is finite. Dominated convergence for series therefore yields
\begin{align*}
\mathbb E[\tilde\sigma_N^2]-\sigma_h^2\to0.
\end{align*}
[/guided]
[/step]
[step:Control the stochastic fluctuation of the oracle estimator using the assumed variance estimate]
The theorem statement assumes that there is a constant $C<\infty$, independent of $N$, such that
\begin{align*}
\operatorname{Var}\left(\tilde\gamma_0 + 2\sum_{k=1}^{N-1}w\left(\frac{k}{b_N}\right)\tilde\gamma_k\right) \le C\frac{b_N}{N}.
\end{align*}
By the definition of $\tilde\sigma_N^2$, the [random variable](/page/Random%20Variable) inside the variance is exactly $\tilde\sigma_N^2$. Hence
\begin{align*}
\operatorname{Var}(\tilde\sigma_N^2) \le C\frac{b_N}{N}.
\end{align*}
Since
\begin{align*}
\frac{b_N}{N}\to0,
\end{align*}
we obtain
\begin{align*}
\operatorname{Var}(\tilde\sigma_N^2)\to0.
\end{align*}
By [Chebyshev's inequality](/theorems/1126) applied to the real-valued random variable $\tilde\sigma_N^2$, for every $\varepsilon>0$,
\begin{align*}
\mathbb P\left(\left|\tilde\sigma_N^2-\mathbb E[\tilde\sigma_N^2]\right|>\varepsilon\right) \le \frac{\operatorname{Var}(\tilde\sigma_N^2)}{\varepsilon^2}.
\end{align*}
The right-hand side tends to $0$, so
\begin{align*}
\tilde\sigma_N^2-\mathbb E[\tilde\sigma_N^2]\xrightarrow{\mathbb P}0.
\end{align*}
Together with $\mathbb E[\tilde\sigma_N^2]\to\sigma_h^2$, this proves
\begin{align*}
\tilde\sigma_N^2\xrightarrow{\mathbb P}\sigma_h^2.
\end{align*}
[guided]
The stochastic part of the argument is now an application of an explicit hypothesis in the theorem statement, not an uncited external theorem. The statement defines the centered variables $X_n := Y_n-\mu$ and the oracle autocovariances
\begin{align*}
\tilde\gamma_k := \frac{1}{N}\sum_{n=1}^{N-k}X_nX_{n+k}
\end{align*}
for $0\le k\le N-1$, and assumes that there is a constant $C<\infty$, independent of $N$, satisfying
\begin{align*}
\operatorname{Var}\left(\tilde\gamma_0 + 2\sum_{k=1}^{N-1}w\left(\frac{k}{b_N}\right)\tilde\gamma_k\right) \le C\frac{b_N}{N}.
\end{align*}
The expression inside this variance is exactly the oracle lag-window estimator $\tilde\sigma_N^2$ introduced earlier in the proof. Therefore the assumed estimate gives
\begin{align*}
\operatorname{Var}(\tilde\sigma_N^2) \le C\frac{b_N}{N}.
\end{align*}
The bandwidth condition
\begin{align*}
\frac{b_N}{N}\to0
\end{align*}
then implies
\begin{align*}
\operatorname{Var}(\tilde\sigma_N^2)\to0.
\end{align*}
This variance convergence says that the random estimator concentrates around its own mean. To convert this into convergence in probability, apply Chebyshev's inequality to the real-valued random variable $\tilde\sigma_N^2$: for every $\varepsilon>0$,
\begin{align*}
\mathbb P\left(\left|\tilde\sigma_N^2-\mathbb E[\tilde\sigma_N^2]\right|>\varepsilon\right) \le \frac{\operatorname{Var}(\tilde\sigma_N^2)}{\varepsilon^2}.
\end{align*}
The numerator tends to $0$, while $\varepsilon^2$ is fixed and positive, so
\begin{align*}
\tilde\sigma_N^2-\mathbb E[\tilde\sigma_N^2]\xrightarrow{\mathbb P}0.
\end{align*}
The deterministic bias step has already proved
\begin{align*}
\mathbb E[\tilde\sigma_N^2]\to\sigma_h^2.
\end{align*}
Adding a deterministic sequence converging to $\sigma_h^2$ to a random sequence converging to $0$ in probability gives
\begin{align*}
\tilde\sigma_N^2\xrightarrow{\mathbb P}\sigma_h^2.
\end{align*}
[/guided]
[/step]
[step:Show that estimating the mean changes the estimator by a negligible amount]
Let
\begin{align*}
\bar X_N := \frac{1}{N}\sum_{n=1}^{N}X_n.
\end{align*}
Then $\bar Y_N-\mu=\bar X_N$. Absolute summability of $(\gamma_k)_{k\in\mathbb Z}$ gives
\begin{align*}
\mathbb E[\bar X_N^2] = \frac{1}{N^2}\sum_{i=1}^{N}\sum_{j=1}^{N}\gamma_{j-i} = \frac{1}{N}\sum_{|k|\le N-1}\left(1-\frac{|k|}{N}\right)\gamma_k.
\end{align*}
Taking absolute values and using $0\le 1-|k|/N\le1$ for $|k|\le N-1$, we obtain
\begin{align*}
\left|\mathbb E[\bar X_N^2]\right| \le \frac{1}{N}\sum_{k=-\infty}^{\infty}|\gamma_k|.
\end{align*}
Since the series on the right is finite, $\mathbb E[\bar X_N^2]\to0$, so $\bar X_N\to0$ in $L^2$, hence in probability.
For $0\le k\le N-1$, expand
\begin{align*}
\hat\gamma_k-\tilde\gamma_k = -\bar X_N\frac{1}{N}\sum_{n=1}^{N-k}(X_n+X_{n+k}) + \frac{N-k}{N}\bar X_N^2.
\end{align*}
Since $w(k/b_N)=0$ whenever $k>b_N$, the number of nonzero lag weights is at most $1+\lfloor b_N\rfloor$. Also $|w|\le1$. The preceding identity gives the bound
\begin{align*}
|\hat\sigma^2_{\mathrm{SV}}-\tilde\sigma_N^2| \le 2|\bar X_N|\left[\frac{1}{N}\sum_{n=1}^{N}|X_n| + 2\sum_{k=1}^{\lfloor b_N\rfloor}\frac{1}{N}\sum_{n=1}^{N}|X_n|\right] + \left(1+2\lfloor b_N\rfloor\right)\bar X_N^2.
\end{align*}
Define the non-negative random variable
\begin{align*}
A_N := \frac{1}{N}\sum_{n=1}^{N}|X_n|.
\end{align*}
Because $\mathbb E[|X_1|]<\infty$, stationarity gives $\mathbb E[A_N]=\mathbb E[|X_1|]$ for every $N$. Hence Markov's inequality implies that the family $(A_N)_{N\ge1}$ is bounded in probability.
Absolute summability of $(\gamma_k)_{k\in\mathbb Z}$ gives the sharper mean-square estimate
\begin{align*}
N\mathbb E[\bar X_N^2]
=
\sum_{|k|\le N-1}\left(1-\frac{|k|}{N}\right)\gamma_k.
\end{align*}
Consequently
\begin{align*}
N\mathbb E[\bar X_N^2]
\le
\sum_{k=-\infty}^{\infty}|\gamma_k|.
\end{align*}
Since $b_N/\sqrt{N}\to0$, Chebyshev's inequality yields
\begin{align*}
b_N\bar X_N\xrightarrow{\mathbb P}0.
\end{align*}
Indeed, for every $\varepsilon>0$,
\begin{align*}
\mathbb P\left(b_N|\bar X_N|>\varepsilon\right) \le \frac{b_N^2}{\varepsilon^2} \mathbb E[\bar X_N^2] \le \frac{b_N^2}{N\varepsilon^2}\sum_{k=-\infty}^{\infty}|\gamma_k| \to0.
\end{align*}
The product of a sequence converging to $0$ in probability and a sequence bounded in probability converges to $0$ in probability, so
\begin{align*}
b_N|\bar X_N|A_N\xrightarrow{\mathbb P}0.
\end{align*}
Also Markov's inequality and the preceding second-moment bound imply
\begin{align*}
b_N\bar X_N^2\xrightarrow{\mathbb P}0.
\end{align*}
Indeed, for every $\varepsilon>0$,
\begin{align*}
\mathbb P\left(b_N\bar X_N^2>\varepsilon\right) \le \frac{b_N}{\varepsilon}\mathbb E[\bar X_N^2] \le \frac{b_N}{N\varepsilon}\sum_{k=-\infty}^{\infty}|\gamma_k| \to0,
\end{align*}
because $b_N/N\to0$. Applying these two convergence statements to the displayed bound for $|\hat\sigma^2_{\mathrm{SV}}-\tilde\sigma_N^2|$ gives
\begin{align*}
\hat\sigma^2_{\mathrm{SV}}-\tilde\sigma_N^2\xrightarrow{\mathbb P}0.
\end{align*}
[guided]
The only difference between $\hat\gamma_k$ and $\tilde\gamma_k$ is that $\hat\gamma_k$ uses the sample mean while $\tilde\gamma_k$ uses the true mean. Define
\begin{align*}
\bar X_N := \frac{1}{N}\sum_{n=1}^{N}X_n.
\end{align*}
Then $\bar Y_N-\mu=\bar X_N$. Absolute summability of the autocovariance sequence gives the second-moment identity
\begin{align*}
\mathbb E[\bar X_N^2] = \frac{1}{N}\sum_{|k|\le N-1}\left(1-\frac{|k|}{N}\right)\gamma_k.
\end{align*}
Taking absolute values yields
\begin{align*}
\mathbb E[\bar X_N^2] \le \frac{1}{N}\sum_{k=-\infty}^{\infty}|\gamma_k|.
\end{align*}
This is the quantitative form of $\bar X_N=O_{\mathbb P}(N^{-1/2})$, but we do not need to introduce that notation.
Expanding the product in the definition of $\hat\gamma_k$ gives, for $0\le k\le N-1$,
\begin{align*}
\hat\gamma_k-\tilde\gamma_k = -\bar X_N\frac{1}{N}\sum_{n=1}^{N-k}(X_n+X_{n+k}) + \frac{N-k}{N}\bar X_N^2.
\end{align*}
Since $w(k/b_N)=0$ when $k>b_N$ and $|w|\le1$, only at most $1+\lfloor b_N\rfloor$ lag weights contribute. Therefore
\begin{align*}
|\hat\sigma^2_{\mathrm{SV}}-\tilde\sigma_N^2| \le 2|\bar X_N|\left[\frac{1}{N}\sum_{n=1}^{N}|X_n| + 2\sum_{k=1}^{\lfloor b_N\rfloor}\frac{1}{N}\sum_{n=1}^{N}|X_n|\right] + \left(1+2\lfloor b_N\rfloor\right)\bar X_N^2.
\end{align*}
Let
\begin{align*}
A_N := \frac{1}{N}\sum_{n=1}^{N}|X_n|.
\end{align*}
Because $\mathbb E[|X_1|]<\infty$ and the sequence is stationary, $\mathbb E[A_N]=\mathbb E[|X_1|]$ for all $N$. Markov's inequality shows that $(A_N)_{N\ge1}$ is bounded in probability.
The bandwidth condition $b_N/\sqrt N\to0$ and Chebyshev's inequality give $b_N|\bar X_N|\to0$ in probability, since
\begin{align*}
\mathbb P\left(b_N|\bar X_N|>\varepsilon\right) \le \frac{b_N^2}{N\varepsilon^2}\sum_{k=-\infty}^{\infty}|\gamma_k| \to0.
\end{align*}
Hence $b_N|\bar X_N|A_N\to0$ in probability. Markov's inequality also gives
\begin{align*}
\mathbb P\left(b_N\bar X_N^2>\varepsilon\right) \le \frac{b_N}{N\varepsilon}\sum_{k=-\infty}^{\infty}|\gamma_k| \to0,
\end{align*}
because $b_N/N\to0$. Substituting these two estimates into the displayed bound proves
\begin{align*}
\hat\sigma^2_{\mathrm{SV}}-\tilde\sigma_N^2\xrightarrow{\mathbb P}0.
\end{align*}
[/guided]
[/step]
[step:Combine the oracle convergence and the mean-correction estimate]
We have proved
\begin{align*}
\tilde\sigma_N^2\xrightarrow{\mathbb P}\sigma_h^2
\end{align*}
and
\begin{align*}
\hat\sigma^2_{\mathrm{SV}}-\tilde\sigma_N^2\xrightarrow{\mathbb P}0.
\end{align*}
By Slutsky's theorem, equivalently by the triangle inequality and the definition of convergence in probability,
\begin{align*}
\hat\sigma^2_{\mathrm{SV}} = \tilde\sigma_N^2+\left(\hat\sigma^2_{\mathrm{SV}}-\tilde\sigma_N^2\right) \xrightarrow{\mathbb P} \sigma_h^2.
\end{align*}
This is the asserted consistency of the lag-window spectral variance estimator.
[/step]
