[proofplan]
For each $n\in\mathbb N$, the plug-in bandwidth is obtained from the stated AMISE bandwidth by replacing the unknown curvature functional $R(f'')$ with the positive estimator $\widehat{R(f'')}_n:(\Omega,\mathcal F)\to(0,\infty)$. Dividing the two closed-form bandwidths cancels all kernel and sample-size factors, leaving only the ratio
\begin{align*}
\left(\frac{R(f'')}{\widehat{R(f'')}_n}\right)^{1/5}.
\end{align*}
Since $\widehat{R(f'')}_n$ converges in probability to the positive constant $R(f'')$, we prove directly from the definition of convergence in probability that this fifth-root ratio converges in probability to $1$.
[/proofplan]
[step:Cancel the common AMISE factors in the ratio]
Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space from the theorem statement, and define the positive constant
\begin{align*}
\theta:=R(f'')\in(0,\infty).
\end{align*}
For each $n\in\mathbb N$, the theorem statement defines
\begin{align*}
h_{\operatorname{AMISE},n}&=\left(\frac{R(K)}{\mu_2(K)^2\theta n}\right)^{1/5}, &
\hat h_{\operatorname{PI},n}&=\left(\frac{R(K)}{\mu_2(K)^2\widehat{R(f'')}_n n}\right)^{1/5}.
\end{align*}
The assumptions $R(K)>0$, $\mu_2(K)\neq 0$, $\theta>0$, $\widehat{R(f'')}_n:(\Omega,\mathcal F)\to(0,\infty)$, and $n\in\mathbb N$ make both bandwidths positive. Hence division is valid. First,
\begin{align*}
\frac{\hat h_{\operatorname{PI},n}}{h_{\operatorname{AMISE},n}}=\frac{\left(\frac{R(K)}{\mu_2(K)^2\widehat{R(f'')}_n\,n}\right)^{1/5}}{\left(\frac{R(K)}{\mu_2(K)^2\theta\,n}\right)^{1/5}}.
\end{align*}
Since both quantities inside the fifth roots are positive, the quotient of fifth roots equals the fifth root of the quotient. Cancelling the positive factors $R(K)$, $\mu_2(K)^2$, and $n$ gives
\begin{align*}
\frac{\hat h_{\operatorname{PI},n}}{h_{\operatorname{AMISE},n}}=\left(\frac{R(K)}{\mu_2(K)^2\widehat{R(f'')}_n\,n}\cdot\frac{\mu_2(K)^2\theta\,n}{R(K)}\right)^{1/5}=\left(\frac{\theta}{\widehat{R(f'')}_n}\right)^{1/5}.
\end{align*}
[/step]
[step:Show the fifth-root ratio converges in probability to $1$]
For each $n\in\mathbb N$, define the [random variable](/page/Random%20Variable) $Y_n:(\Omega,\mathcal F)\to(0,\infty)$ on the probability space fixed in the theorem statement by
\begin{align*}
Y_n(\omega)=\left(\frac{\theta}{\widehat{R(f'')}_n(\omega)}\right)^{1/5}.
\end{align*}
This map is measurable because $\widehat{R(f'')}_n$ is a positive real-valued random variable and because the function $\varphi:(0,\infty)\to(0,\infty)$ defined by
\begin{align*}
\varphi(t)=\left(\frac{\theta}{t}\right)^{1/5}
\end{align*}
is continuous. We prove $Y_n\xrightarrow{\mathbb P}1$. Fix $\varepsilon>0$. Since $\varphi$ is continuous at $t=\theta$ and satisfies $\varphi(\theta)=1$, there exists $\delta\in(0,\theta/2)$ such that
\begin{align*}
|t-\theta|<\delta
\quad\Longrightarrow\quad
\left|\left(\frac{\theta}{t}\right)^{1/5}-1\right|<\varepsilon
\end{align*}
for every $t\in(0,\infty)$. Therefore
\begin{align*}
\{|Y_n-1|\geq\varepsilon\}
\subset
\{|\widehat{R(f'')}_n-\theta|\geq\delta\}.
\end{align*}
Taking probabilities and using $\widehat{R(f'')}_n\xrightarrow{\mathbb P}\theta$, we obtain
\begin{align*}
0
\leq
\mathbb P(|Y_n-1|\geq\varepsilon)
\leq
\mathbb P(|\widehat{R(f'')}_n-\theta|\geq\delta)
\longrightarrow 0.
\end{align*}
Since $\varepsilon>0$ was arbitrary, $Y_n\xrightarrow{\mathbb P}1$.
[guided]
The only probabilistic input is the consistency assumption
\begin{align*}
\widehat{R(f'')}_n\xrightarrow{\mathbb P}\theta,
\qquad
\theta:=R(f'')>0.
\end{align*}
We want to transfer this convergence through the function $t\mapsto(\theta/t)^{1/5}$. Because the estimator is positive, this function is evaluated only on $(0,\infty)$, where it is well-defined.
Define the function $\varphi:(0,\infty)\to(0,\infty)$ by
\begin{align*}
\varphi(t)=\left(\frac{\theta}{t}\right)^{1/5}.
\end{align*}
The assumption $\theta>0$ is essential here: it places the limiting value inside the domain $(0,\infty)$, away from the singular point $t=0$. The function $\varphi$ is continuous at $\theta$, and
\begin{align*}
\varphi(\theta)=\left(\frac{\theta}{\theta}\right)^{1/5}=1.
\end{align*}
Therefore, for the fixed tolerance $\varepsilon>0$, there exists $\delta\in(0,\theta/2)$ such that every $t\in(0,\infty)$ with $|t-\theta|<\delta$ satisfies
\begin{align*}
\left|\left(\frac{\theta}{t}\right)^{1/5}-1\right|<\varepsilon.
\end{align*}
Now apply this deterministic implication pointwise to the random quantity $\widehat{R(f'')}_n(\omega)$. If
\begin{align*}
|\widehat{R(f'')}_n(\omega)-\theta|<\delta,
\end{align*}
then
\begin{align*}
\left|\left(\frac{\theta}{\widehat{R(f'')}_n(\omega)}\right)^{1/5}-1\right|<\varepsilon.
\end{align*}
Equivalently, the event where the transformed estimator is at least $\varepsilon$ away from $1$ can occur only when the original estimator is at least $\delta$ away from $\theta$:
\begin{align*}
\left\{
\left|\left(\frac{\theta}{\widehat{R(f'')}_n}\right)^{1/5}-1\right|\geq\varepsilon
\right\}
\subset
\{|\widehat{R(f'')}_n-\theta|\geq\delta\}.
\end{align*}
Taking probabilities gives
\begin{align*}
0
\leq
\mathbb P\left(
\left|\left(\frac{\theta}{\widehat{R(f'')}_n}\right)^{1/5}-1\right|\geq\varepsilon
\right)
\leq
\mathbb P(|\widehat{R(f'')}_n-\theta|\geq\delta).
\end{align*}
The right-hand side tends to $0$ by the assumed convergence in probability of $\widehat{R(f'')}_n$ to $\theta$. Hence the transformed sequence converges in probability to $1$.
[/guided]
[/step]
[step:Identify the transformed estimator with the bandwidth ratio]
From the algebraic identity in the first step,
\begin{align*}
\frac{\hat h_{\operatorname{PI},n}}{h_{\operatorname{AMISE},n}}
=
\left(\frac{\theta}{\widehat{R(f'')}_n}\right)^{1/5}
=
Y_n.
\end{align*}
The second step proved $Y_n\xrightarrow{\mathbb P}1$. Therefore
\begin{align*}
\frac{\hat h_{\operatorname{PI},n}}{h_{\operatorname{AMISE},n}}
\xrightarrow{\mathbb P}1,
\end{align*}
which is the desired consistency of the plug-in AMISE bandwidth.
[/step]
