メモ　2025-02-25.

$(1^\circ)$ Probamus initio supremum in definitione functionis $d_2$ valorem finitum esse. Sint $X,Y$ quemcumque classes ad $\mathcal{Q}_2$ pertinentes. Cum $X$ & $Y$ eadem valores expectandos et latitudines habent, videamus
\begin{align*} &\frac{1}{\xi^2}\left|\varphi_X(\xi)-\varphi_Y(\xi)\right|\\ =\;&\frac{1}{\xi^2}\left|E[e^{iX\xi}-1-iX\xi]-E[e^{iY\xi}-1-iY\xi]\right|\\ \leqslant\;&\frac{1}{\xi^2}\left(E\Bigl[\left|e^{iX\xi}-1-iX\xi\right|\Bigr]+E\Bigl[\left|e^{iY\xi}-1-iY\xi\right|\Bigr]\right)\\ \leqslant\;&\frac{1}{\xi^2}\left(E\Bigl[\frac{\,e\,}{\,2\,}X^2\xi^2\Bigr]+E\Bigl[\frac{\,e\,}{\,2\,}Y^2\xi^2\Bigr]\right)=e \end{align*}
quotiens $0<\left|\xi\right|\leqslant1,$ et
\begin{equation*} \frac{1}{\xi^2}\left|\varphi_X(\xi)-\varphi_Y(\xi)\right|<1+1=2 \end{equation*}
quotiens $\left|\xi\right|>1,$ quae producunt $d_2(X,Y)\leqslant e.$

$(2^\circ)$ Si $X,Y$ quaevis classes ad $\mathcal{Q}_2$ pertinentes repraesentant, facile est perspicere
\begin{equation*} d_2(X,Y)=0\Longleftrightarrow\varphi_X=\varphi_Y\Longleftrightarrow X=Y. \end{equation*}
Adeoque valet $d_2(X,Y)=d_2(Y,X)$.

$(3^\circ)$ Inaequalitas triangularis
\begin{equation*} d_2(X,Z)\leqslant d_2(X,Y)+d_2(Y,Z)\quad(X,Y,Z\in\mathcal{Q}_2) \end{equation*}
derivari potest per illas valoris absoluti $\left|\;\cdot\;\right|$ et supremi.

Itaque $d_2$ est functio quae distantiam definit, q. e. f.

（難しいので省略．後で追記するかも知れません…）

Si classes variabilium $X_1,X_2$ datae sunt, independenterque ad distributionem normalem sequitur, probablitas quod nova classis variabilis $T(X)=(X_1+X_2)/\sqrt{2}$ quemdam numerum $x$ excidat in formulam
\begin{align*} P(T(X)\leqslant x)&=\frac{1}{2\pi}\iint_{(x_1+x_2)/\sqrt{2}\leqslant x}e^{(-x_1^2-x_2^2)/2}dx_1dx_2 \end{align*}
scribitur, quae per substitutionem aliarum variabilium $y_1=(x_1+x_2)/\sqrt{2},$ $y_2=(x_1-x_2)/\sqrt{2}$ in hanc:
\begin{align*} &=\frac{1}{2\pi}\int_{-\infty}^\infty\int_{-\infty}^xe^{(-y_1^2-y_2^2)/2}dy_1dy_2=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-y_1^2/2}dy_1 \end{align*}
abit, quae $=P(X\leqslant x)$ scribi potest. Ergo $T(X)$ etiam distributionem normalem seqeui debet, q. e. d.

Imprimis comparamus
\begin{align*} \varphi_X'(0)&=\left.\frac{d}{d\xi}\int_{-\infty}^\infty e^{ix\xi}f_X(x)dx\right|_{\xi=0}\\ &=\left.\int_{-\infty}^\infty \frac{d}{d\xi}e^{ix\xi}f_X(x)dx\right|_{\xi=0}\\ &=\left.\int_{-\infty}^\infty ixe^{ix\xi}f_X(x)dx\right|_{\xi=0}\\ &=\int_{-\infty}^\infty ixf_X(x)dx\\ &=iE[X]=0,\\ \varphi_X''(0)&=\left.\frac{d^2}{d\xi^2}\int_{-\infty}^\infty e^{ix\xi}f_X(x)dx\right|_{\xi=0}\\ &=\left.\int_{-\infty}^\infty \frac{d^2}{d\xi^2}e^{ix\xi}f_X(x)dx\right|_{\xi=0}\\ &=\left.\int_{-\infty}^\infty -x^2e^{ix\xi}f_X(x)dx\right|_{\xi=0}\\ &=\int_{-\infty}^\infty -x^2f_X(x)dx\\ &=-E[X^2]=-1, \end{align*}
itemque $\varphi_Y'(0)=0$ et $\varphi_Y''(0)=-1.$ Ergo quia theoremate Tayloriano habemus
\begin{equation*} \frac{\left|\varphi_X(\xi)-\varphi_Y(\xi)\right|}{\xi^2}=\frac{\mathrm{O}(\xi^3)}{\xi^2}=\mathrm{O}(\xi) \end{equation*}
ad limitem $\xi\to0,$ existit numerus positivus $\delta$ talis ut inaequalitas
\begin{equation*} \frac{\left|\varphi_X(\xi)-\varphi_Y(\xi)\right|}{\xi^2}<\frac{\,a\,}{\,2\,} \end{equation*}
valeat quotiens $\left|\xi\right|<\delta,$ unde producitur,
\begin{align*} &\frac{\left|E[e^{iT(X)\xi}]-E[e^{iT(Y)\xi}]\right|}{\xi^2}\\ =\;&\frac{\left|E[e^{iX\xi/\sqrt{2}}]-E[e^{iY\xi/\sqrt{2}}]\right|}{\xi^2}\left|E[e^{iX\xi/\sqrt{2}}]+E[e^{iY\xi/\sqrt{2}}]\right|\\ =\;&\frac{\left|E[e^{iX\eta}]-E[e^{iY\eta}]\right|}{\eta^2}\frac{\left|E[e^{iX\eta}]+E[e^{iY\eta}]\right|}{2}\quad(\eta=\xi/\sqrt{2})\\ \leqslant\;&\frac{\,a\,}{\,2\,}\frac{E\Bigl[\left|e^{iX\eta}\right|\Bigr]+E\Bigl[\left|e^{iY\eta}\right|\Bigr]}{2}=\frac{\,a\,}{\,2\,}. \end{align*}
Atque si $\left|\xi\right|\geqslant \delta$ habemus
\begin{align*} &\frac{\left|E[e^{iT(X)\xi}]-E[e^{iT(Y)\xi}]\right|}{\xi^2}\\ =\;&\frac{\left|E[e^{iX\eta}]-E[e^{iY\eta}]\right|}{\eta^2}\frac{\left|E[e^{iX\eta}]+E[e^{iY\eta}]\right|}{2}\quad(\eta=\xi/\sqrt{2})\\ \leqslant\;&a\frac{e^{-\xi^2/2}+1}{2}\\ \leqslant\;&a\frac{e^{-\delta^2/2}+1}{2}. \end{align*}
Itaque
\begin{align*} d_2(X,T(Y))&=d_2(T(X),T(Y))< a, \end{align*}
q. e. f.

略．

Quia summa $X_1+\cdots+X_{2^n}$ divisa per $2^{n/2}$ ac variabilem $T^n(X_1)=T(T\cdots(T(X_1)))$ eandem classem pertinet, oportet conspicere quantum ea prope variabilis $X$ sit, quo $X$ variabilem repraesentat quae ad $\mathcal{N}(0,1)$ sequitur. Series distantiae earum:
\begin{equation*} a_n\ \ =\ \ d_2(X,T^n(X_1)),\qquad n\ \ =\ \ 1,\ \ 2,\ \ 3,\ \ \ldots \end{equation*}
certo limite inferiore praeditum esse, atque propositione 4 descrescens esse patet. Ergo exstat numerus nonnegativus $a$ ad quam haec series convergit, cum subserie convergente
\begin{equation*} Y_n\ \ =\ \ T^{\varphi(n)}(X_1),\qquad n\ \ =\ \ 1,\ \ 2,\ \ 3,\ \ \ldots, \end{equation*}
quod est corollarium propositionis 2. Eam limitem per characterem $Y$ notabimus. Tunc quia $T$ functio continua erat, series
\begin{equation*} Z_n\ \ =\ \ T^{\varphi(n)+1}(X_1),\qquad n\ \ =\ \ 1,\ \ 2,\ \ 3,\ \ \ldots \end{equation*}
limitem $Z=T(Y)$ habet. Si $a>0$ posito ostendimus,
\begin{equation*} a=d_2(X,Z)=d_2(X,T(Y))< d_2(X,Y)=a, \end{equation*}
quod est absurdum. Ergo $a=0$ valet, atque $T^n(X_1)$ ad punctum $X$ convergere patet. Igitur ad omne punctum $\xi\in\mathbb{R}$ tam $\xi\neq0$ quam $\xi=0,$
\begin{equation*} \varphi_{T^n(X_1)}(\xi),\qquad n\ \ =\ \ 1,\ \ 2,\ \ 3,\ \ \ldots \end{equation*}
ad $\varphi_X(\xi)=e^{-\xi^2/2}$ convergere necessaris esse perspicitur, tum Theorema Levianum uti licet ad probandum propositionem.