\[
\varphi(\theta)
:= \frac{1}{d}\sum_{k=1}^d \cos \theta_k,
\qquad \theta = (\theta_1,\dots,\theta_d)\in \mathbb{R}^d.
\]
一次微分（勾配）
各成分について
\[
\partial_{\theta_j}\varphi(\theta)
= \partial_{\theta_j}\left(\frac{1}{d}\sum_{k=1}^d \cos\theta_k\right)
= -\frac{1}{d}\sin\theta_j.
\]
したがって勾配の二乗ノルムは
\[
|\nabla \varphi(\theta)|^2
= \sum_{j=1}^d \left(-\frac{1}{d}\sin\theta_j\right)^2
= \frac{1}{d^2}\sum_{j=1}^d \sin^2\theta_j.
\]
二次微分とラプラシアン
\[
\partial_{\theta_j}^2 \varphi(\theta)
= \partial_{\theta_j}\left(-\frac{1}{d}\sin\theta_j\right)
= -\frac{1}{d}\cos\theta_j.
\]
よってラプラシアンは
\[
\Delta \varphi(\theta)
= \sum_{j=1}^d \partial_{\theta_j}^2 \varphi(\theta)
= -\frac{1}{d}\sum_{j=1}^d \cos\theta_j
= -\varphi(\theta).
\]
合成関数の公式
\[
\Delta(f^n)
= n(n-1)f^{\,n-2}|\nabla f|^2
+ n f^{\,n-1}\Delta f
\]
を \(f=\varphi\) に適用すると
\[
\Delta \varphi^n(\theta)
= n(n-1)\varphi^{n-2}(\theta)\,
\frac{1}{d^2}\sum_{j=1}^d \sin^2\theta_j
+ n\varphi^{n-1}(\theta)\,(-\varphi(\theta)).
\]
整理して
\[
\boxed{
\Delta \varphi^n(\theta)
= n(n-1)\varphi^{n-2}(\theta)
\left(\frac{1}{d^2}\sum_{j=1}^d \sin^2\theta_j\right)
– n\varphi^n(\theta)
}
\]
$d$ 次元変数 $\theta \in \mathbb{R}^d$ に対して、
\[
G_n(\theta) = \exp\left(- \frac{n}{2d} |\theta|^2 \right), \quad |\theta|^2 = \sum_{j=1}^d \theta_j^2
\]
を考える。
任意の滑らかな関数 $f:\mathbb{R}^d \to \mathbb{R}$ に対して、
\[
\Delta e^f = (\lvert \nabla f \rvert^2 + \Delta f) e^f
\]
が成り立つ。ここで
\[
\nabla f = \left( \frac{\partial f}{\partial \theta_1}, \dots, \frac{\partial f}{\partial \theta_d} \right), \quad
\Delta f = \sum_{j=1}^d \frac{\partial^2 f}{\partial \theta_j^2}.
\]
$f(\theta) = – \frac{n}{2d} |\theta|^2$ とすると、
\[
\frac{\partial f}{\partial \theta_j} = – \frac{n}{d} \theta_j \quad \Rightarrow \quad
|\nabla f|^2 = \sum_{j=1}^d \left( -\frac{n}{d} \theta_j \right)^2 = \frac{n^2}{d^2} |\theta|^2,
\]
\[
\frac{\partial^2 f}{\partial \theta_j^2} = – \frac{n}{d} \quad \Rightarrow \quad
\Delta f = \sum_{j=1}^d \frac{\partial^2 f}{\partial \theta_j^2} = -n.
\]
上記の式を用いると、
\[
\Delta G_n(\theta) = (\lvert \nabla f \rvert^2 + \Delta f) e^f
= \left( \frac{n^2}{d^2} |\theta|^2 – n \right) \exp\left(- \frac{n}{2d} |\theta|^2 \right)
= \left( \frac{n^2}{d^2} |\theta|^2 – n \right) G_n(\theta).
\]
\[
\text{スケーリング: } \theta = \frac{\alpha}{n}
\]
${\Delta \varphi}_n$ の展開:
\[
\varphi\Big(\frac{\alpha}{n}\Big) = 1 – \frac{|\alpha|^2}{2dn} + O(n^{-2})
\]
\[
\varphi^{\,n-2}\Big(\frac{\alpha}{n}\Big) \approx e^{-|\alpha|^2 / 2d}
\]
\[
\sum_{j=1}^d \sin^2\Big(\frac{\alpha_j}{n}\Big) = \sum_{j=1}^d \left(\frac{\alpha_j^2}{n^2} + O(n^{-4})\right) = \frac{|\alpha|^2}{n} + O(n^{-2})
\]
\begin{align}
\phi\!\left(\frac{\alpha}{n}\right)
&= 1 – \frac{|\alpha|^2}{2d n} + O(n^{-2}), \\
\phi^n\!\left(\frac{\alpha}{n}\right)
&= \left( 1 – \frac{|\alpha|^2}{2d n} + O(n^{-2}) \right)^n
\approx e^{- \frac{|\alpha|^2}{2d}} \left( 1 + O(n^{-1}) \right).
\end{align}
これらを $\Delta {\varphi}_n$ に代入すると：
\[
\Delta \varphi_n\Big(\frac{\alpha}{n}\Big)
\approx n(n-1) \cdot e^{-|\alpha|^2 / 2d} \cdot \frac{d}{2} \cdot \frac{|\alpha|^2}{n} – n \, e^{-|\alpha|^2 / 2d}
\]
\[
= \left(n \frac{d}{2} |\alpha|^2 – n\right) e^{-|\alpha|^2 / 2d} + O(1)
\]
${\Delta G}_n$ の展開:
\[
\Delta G_n\Big(\frac{\alpha}{n}\Big) = \left(\frac{d}{2} n^2 \frac{|\alpha|^2}{n} – n\right) e^{-|\alpha|^2 / 2d} = \left(n \frac{d}{2} |\alpha|^2 – n\right) e^{-|\alpha|^2 / 2d}
\]
差のオーダーは：
\[
\Delta \varphi_n\Big(\frac{\alpha}{n}\Big) – \Delta G_n\Big(\frac{\alpha}{n}\Big) = O(1)
\]
である。
主要な $O(n) $項はキャンセルされ、残るのは $O(1) $項のみである。
結果、
\[
\begin{aligned}
|x|^2E(n,x)
&= (2\pi)^{-d}
\int_{[-\pi,\pi]^d}
e^{-i x \cdot \theta} \,
\Biggl[
-\Delta_\theta
\Bigl( \varphi_n(\theta) – G_n(\theta) \Bigr)
\Biggr]
\, d\theta \\
&\quad \overset{\theta = \alpha / \sqrt{n}}{=}
(2\pi)^{-d}
\int_{[-\pi \sqrt{n}, \pi \sqrt{n}]^d}
e^{-i x \cdot (\alpha / \sqrt{n})} \,
\Biggl[
-\Delta_\alpha
\Bigl( \varphi_n(\alpha / \sqrt{n}) – G_n(\alpha / \sqrt{n}) \Bigr)
\Biggr]
\, n^{-d/2} \, d\alpha
= O(n^{-d/2})
\end{aligned}
\]
参考:
Intersections of Random Walks (Modern Birkhäuser Classics) (English Edition)
