[proofplan]
We minimize the positive one-variable function $L_n: (0,\infty) \to (0,\infty)$ defined by
\begin{align*}
L_n(h)=Ah^{2s}+\frac{B}{nh}.
\end{align*}
The optimization uses only the induced constants $A>0$ and $B>0$, together with $s \geq 1$ and $n \geq 1$; the kernel-density hypotheses enter only through the MISE expansion that produces these constants. The derivative has exactly one zero, and the sign of the derivative changes from negative to positive there, so that point is the unique global minimizer. Substituting the minimizer into the two leading terms shows that both the squared-bias term and the variance term scale as $n^{-2s/(2s+1)}$, giving the stated MISE rate.
[/proofplan]
[step:Differentiate the leading MISE approximation and solve the critical point equation]
Fix $n \in \mathbb{N}$. Since $A>0$, $B>0$, and $s \in \mathbb{N}$, define the map $L_n: (0,\infty) \to (0,\infty)$ by
\begin{align*}
L_n(h)=A h^{2s} + \frac{B}{nh} \quad \text{for } h \in (0,\infty).
\end{align*}
This map is differentiable on $(0,\infty)$. Its derivative is
\begin{align*}
L_n'(h)= 2sA h^{2s-1} - \frac{B}{n h^2}.
\end{align*}
Thus $L_n'(h)=0$ is equivalent to
\begin{align*}
2sA h^{2s-1} = \frac{B}{n h^2}.
\end{align*}
Multiplying both sides by $h^2>0$ gives
\begin{align*}
2sA h^{2s+1} = \frac{B}{n},
\end{align*}
and hence
\begin{align*}
h^{2s+1} = \frac{B}{2sA n}.
\end{align*}
Because $2s+1>0$, this equation has the unique positive solution
\begin{align*}
h_* = \left(\frac{B}{2sA n}\right)^{1/(2s+1)} = \left(\frac{B}{2sA}\right)^{1/(2s+1)}n^{-1/(2s+1)}.
\end{align*}
[guided]
Fix $n \in \mathbb{N}$. We are minimizing the function $L_n: (0,\infty) \to (0,\infty)$ defined by
\begin{align*}
L_n(h)=A h^{2s} + \frac{B}{nh} \quad \text{for } h \in (0,\infty).
\end{align*}
The domain is $(0,\infty)$ because bandwidths are positive. The first term $Ah^{2s}$ is the leading squared-bias term, while the second term $B/(nh)$ is the leading variance term.
Since $A>0$, $B>0$, and $s \in \mathbb{N}$, the function $L_n$ is differentiable for every $h>0$. Differentiating term by term gives
\begin{align*}
L_n'(h) = \frac{d}{dh}\left(Ah^{2s}\right) + \frac{d}{dh}\left(\frac{B}{nh}\right).
\end{align*}
Evaluating these two elementary derivatives yields
\begin{align*}
L_n'(h)=2sA h^{2s-1} - \frac{B}{n h^2}.
\end{align*}
A critical point must therefore satisfy
\begin{align*}
2sA h^{2s-1} - \frac{B}{n h^2} = 0.
\end{align*}
Rearranging gives
\begin{align*}
2sA h^{2s-1} = \frac{B}{n h^2}.
\end{align*}
Because $h>0$, multiplying by $h^2$ preserves equivalence and gives
\begin{align*}
2sA h^{2s+1} = \frac{B}{n}.
\end{align*}
Dividing by $2sA>0$ yields
\begin{align*}
h^{2s+1} = \frac{B}{2sA n}.
\end{align*}
The right-hand side is positive because $A>0$, $B>0$, and $n \in \mathbb{N}$. Since the map $h \mapsto h^{2s+1}$ is strictly increasing from $(0,\infty)$ onto $(0,\infty)$, there is exactly one positive solution:
\begin{align*}
h_* = \left(\frac{B}{2sA n}\right)^{1/(2s+1)} = \left(\frac{B}{2sA}\right)^{1/(2s+1)}n^{-1/(2s+1)}.
\end{align*}
This is the only candidate for an interior minimizer.
[/guided]
[/step]
[step:Show that the critical point is the unique global minimizer]
For $h>0$, rewrite the derivative as
\begin{align*}
L_n'(h)= \frac{1}{h^2}\left(2sA h^{2s+1} - \frac{B}{n}\right).
\end{align*}
Since $h^2>0$, the sign of $L_n'(h)$ is the sign of
\begin{align*}
2sA h^{2s+1} - \frac{B}{n}.
\end{align*}
The map $h \mapsto 2sA h^{2s+1} - B/n$ is strictly increasing on $(0,\infty)$ and vanishes exactly at $h_*$. Therefore
\begin{align*}
L_n'(h) < 0 \quad \text{for } 0<h<h_*,
\end{align*}
and
\begin{align*}
L_n'(h) > 0 \quad \text{for } h>h_*.
\end{align*}
Since $L_n$ is differentiable on $(0,\infty)$, its restrictions to $(0,h_*)$ and $(h_*,\infty)$ are differentiable. By the monotonicity criterion for differentiable functions, the inequalities $L_n'(h)<0$ on $(0,h_*)$ and $L_n'(h)>0$ on $(h_*,\infty)$ imply that $L_n$ is strictly decreasing on $(0,h_*)$ and strictly increasing on $(h_*,\infty)$. Therefore $h_*$ is the unique global minimizer of $L_n$ on $(0,\infty)$. Hence
\begin{align*}
h_{\mathrm{MISE}} = h_* =
\left(\frac{B}{2sA}\right)^{1/(2s+1)}n^{-1/(2s+1)}.
\end{align*}
[/step]
[step:Evaluate the leading MISE at the optimal bandwidth]
Substituting
\begin{align*}
h_{\mathrm{MISE}}
= \left(\frac{B}{2sA}\right)^{1/(2s+1)}n^{-1/(2s+1)}
\end{align*}
into the squared-bias term gives
\begin{align*}
A h_{\mathrm{MISE}}^{2s}
&= A\left(\frac{B}{2sA}\right)^{2s/(2s+1)}
n^{-2s/(2s+1)}.
\end{align*}
Substituting the same bandwidth into the variance term gives
\begin{align*}
\frac{B}{n h_{\mathrm{MISE}}}
= \frac{B}{n}
\left(\frac{B}{2sA}\right)^{-1/(2s+1)}
n^{1/(2s+1)}.
\end{align*}
Equivalently,
\begin{align*}
\frac{B}{n h_{\mathrm{MISE}}}
= B\left(\frac{2sA}{B}\right)^{1/(2s+1)}
n^{-2s/(2s+1)}.
\end{align*}
Therefore
\begin{align*}
L_n(h_{\mathrm{MISE}})= A h_{\mathrm{MISE}}^{2s} + \frac{B}{n h_{\mathrm{MISE}}}.
\end{align*}
Using the two evaluated terms gives
\begin{align*}
L_n(h_{\mathrm{MISE}})=\left[A\left(\frac{B}{2sA}\right)^{2s/(2s+1)} + B\left(\frac{2sA}{B}\right)^{1/(2s+1)}\right] n^{-2s/(2s+1)}.
\end{align*}
Defining
\begin{align*}
C_{s,A,B}:= A\left(\frac{B}{2sA}\right)^{2s/(2s+1)} + B\left(\frac{2sA}{B}\right)^{1/(2s+1)},
\end{align*}
we have $C_{s,A,B}>0$ and
\begin{align*}
L_n(h_{\mathrm{MISE}})=C_{s,A,B}n^{-2s/(2s+1)}.
\end{align*}
Thus the leading MISE is of order $n^{-2s/(2s+1)}$, completing the proof.
[/step]
