院試問題　東大　数理科学研究科　平成30年　専門科目B

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平成31（2019）年度修士課程入学試験について | 東京大学大学院数理科学研究科理学部数学科・理学部数学科

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(4)以外は特段難しくはありません. (1),(2)はBanach空間の問題ですが, (3)からは内積を使う必要が出てきます. 多少頭の切り替えに苦労するかもしれません.　(4)は直交分解に気づけるかが鍵です.

Assumption
It is assumed that for every $g \in L^q(X)$ the following holds:

(1) $\begin{align*} \lim_{n \to \infty}\int_{X}f_ngd\mu = \int_{X}fgd\mu. \end{align*}$

(1)
By Hölder’s inequality, for every $g \in L^q(X)$ s.t. $||g||_q = 1$ , we have

(2) $\begin{align*} \int_{X}fg d\mu \leq ||f||_p. \end{align*}$

Hence $\sup_{||g||_q = 1}\int_{X}fgd\mu \leq ||f||_p$ . Define $h(x) = (1/||f||^{p-1}_p)\mathrm{sgn}(x)|f(x)|^{p-1}$ where

(3) $\begin{equation*} \mathrm{sgn}(x) = \left\{ \begin{aligned} 1, \quad x > 0, \\ -1 \quad x < 0. \end{aligned} \right. \end{equation*}$

Then it holds that $h \in L^q(X), \ ||h||_q = 1$ and

(4) $\begin{align*} \int_{X}fhd\mu = ||f||_p, \end{align*}$

and therefore we obtain the result.

(2)
By Hölder’s inequality, we have for every $g \in L^q(X)$ s.t. $||g||_q = 1$ ,

(5) $\begin{align*} \int_{X}f_ngd\mu \leq ||f_n||_p. \end{align*}$

Then from (1), it follows that

(6) $\begin{align*} \int_{X}fgd\mu \leq \liminf_{n \to \infty}||f_n||_p. \end{align*}$

Then from the problem 1, we obtain the result.

(3)
We denote the inner product of $f,g \in L^2(X)$ as $(f,g)$ . Note that

(7) $\begin{align*} ||f_n - f||^2_2 = ||f_n||^2_2 + ||f||^2_2 - 2(f_n,f). \end{align*}$

From the assumption and (1), we have $\lim_{n \to \infty}||f_n|| = ||f||_2, \ \lim_{n\to \infty}(f_n,f) = ||f||_2$ . Then from (7), we obtain the result.

(4)
We decompose $f = g + h$ where $g \in \overline{V},\ h \in \overline{V}^{\perp}$ . Then it follows that

(8) $\begin{align*} \int_{X}f_nfd\mu = \int_{X}f_ngd\mu \end{align*}$

Then from (1), LHS of (8) converges to $||f||_2$ as $n \to \infty$ , and RHS converges to $(f,g) = ||g||_2$ . Hence it holds that $||f||_2 = ||g||_2$ and therefore $h = 0$ .

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(1)
Assume $A$ is an orthogonal projection. Since $A^2 = A$ , we have

(9) $\begin{align*} P_1P_2 + P_2P_1 = 0. \end{align*}$

Then for every $\xi \in \mathcal{H}$ , we have $\mathrm{Re} \langle P_1\xi,P_2\xi \rangle = 0$ .
Take $\xi \in \mathcal{H}_1 \cap \mathcal{H}_2$ . Then it follows that

(10) $\begin{align*} 0 = \mathrm{Re}\langle P_1\xi,P_2\xi\rangle = ||\xi||^2, \end{align*}$

and hence $\xi = 0$ . Therefore we obtain $\mathcal{H}_1 \cap \mathcal{H}_2 = \{ 0 \}$ .
For every $\xi \in \mathcal{H}$ , it holds from (9) that $P_1P_2\xi \in \mathcal{H}_1 \cap \mathcal{H}_2$ . Hence $P_1P_2 = P_2P_1 = 0$ and from this, $\cos \theta = \pi/2$ easily follows.

Assume $\cos \theta = \pi/2$ . Then for every $\xi, \eta \in \mathcal{H}$ , we have $\langle P_1\xi,P_2\eta \rangle = 0$ and it easily follows that $P_1P_2 = P_2P_1 = 0$ .

(2)
Note that

(11) $\begin{align*} \mathcal{H} = (\mathcal{H}_1 \ominus \mathcal{H}_2) \oplus (\mathcal{H}_2 \ominus \mathcal{H}_1) \oplus (\mathcal{H}_1 \cap \mathcal{H}_2), \end{align*}$

where $(\mathcal{H}_i \ominus \mathcal{H}_j)$ denotes $\mathcal{H}_i \cap \mathcal{H}^{\perp}_j$ and $\oplus$ does the direct sum of Hilbert spaces.
It is enough to show that $\mathcal{H}_i \ominus \mathcal{H}_j, \ \mathcal{H}_1 \cap \mathcal{H}_2 \subset B\mathcal{H} \ (i,j = 1,2, \ i \neq j)$ .
Let $\xi \in \mathcal{H}_i \ominus \mathcal{H}_j$ . Then we have $B(B\xi) = A\xi = \xi$ .
Hence $\xi \in B\mathcal{H}$ .
Let $\xi \in \mathcal{H}_1 \cap \mathcal{H}_2$ . Then we have $B(B\xi) = A\xi = 2\xi$ , and therefore $\xi \in B\mathcal{H}$ .

(3)
Note that $\langle A\xi, \xi \rangle = 0$ for every $\xi \in \mathcal{L}^{\perp}$ . Then from the problem 2, it is enough to show for every $\xi \in \mathcal{H}$

(12) $\begin{align*} \langle AB\xi, B\xi \rangle \leq (1 + \cos \theta)||B\xi||^2. \end{align*}$

Since it holds that $||B\xi||^2 = ||P_1\xi||^2 + ||P_2\xi||^2$ , we have for $\xi_1 = \frac{P_1\xi}{||P_1\xi||}$ and $\xi_2 = \frac{P_2\xi}{||P_2\xi||}$ ,

(13) $\begin{align*} \langle AB\xi, B\xi \rangle &= ||A\xi||^2 \\ &= (||P_1\xi||^2 + ||P_2\xi||^2) + 2\mathrm{Re} \langle P_1\xi,P_2\xi \rangle \\ &= ||B\xi||^2 + \frac{2||P_1\xi|| ||P_2\xi||}{||P_1\xi||^2 + ||P_2\xi||^2} \mathrm{Re} \langle \xi_1,\xi_2 \rangle ||B\xi||^2 \\ &\leq \left(1 + \frac{2||P_1\xi|| ||P_2\xi||}{||P_1\xi||^2 + ||P_2\xi||^2}\cos \theta \right)||B\xi||^2. \end{align*}$

Since it holds that $\frac{2xy}{x^2 + y^2} \leq 1$ for every $x,y \in \mathbb{R}$ , we obtain the desired result.

14

(1), (2)はかなり容易ですし, 確率論の知識がそれほどなくても解けますが, (3)以降は予備知識なしでは厳しいですね.

Assumption
For every $\epsilon > 0$ and $k \in \mathbb{N}$ , there exists $N \in \mathbb{N}$ and for every $m,n \geq N$ , we have

(14) $\begin{align*}\sup_{x \in \mathbb{R}} | P[X_nS_k > x] - P[X_mS_k > x]| < \epsilon. \end{align*}$

(1)
This is no more than a simple calculation:

(15) $\begin{align*}P[S_k] = 1, \ \text{and} \ P[(S_k - P[S_k])^2] = 1/k. \end{align*}$

(2)
By (15) and Chebyshev’s inequality, we have for every $\epsilon > 0$ ,

(16) $\begin{align*}P[|S_k - 1| > \epsilon] \leq 1/(\epsilon^2k) \xrightarrow{k \to \infty} 0.\end{align*}$

Hence $S_k$ converges to $1$ in probability.

(3)
Fix $k \in \mathbb{N}$ . From Helly’s selection theorem, there is a subsequence $\{X_{n'}S_k\}_{n'}$ the law of which converges vaguely to a finite measure $\mu_k$ . It is enough to show that $\mu_k$ is a probability measure and the law of $X_nS_k$ converges to $\mu_k$ as $n \to \infty$ . By taking $n$ to $\infty$ along the subsequence ${n'}$ in (14), there exists an integer $N \in \mathbb{N}$ and it holds for every continuity point $x$ of $\mu_k$ , $m \geq N$ and $\epsilon > 0$ ,

(17) $\begin{align*}\sup_{x \in \mathbb{R}}|\mu_k(x,\infty) - P[X_mS_k > x]| \leq \epsilon. \end{align*}$

Fix $\delta > 0$ . By taking $x$ so large that $P[X_NS_k > x] < \delta$ , we have from (17)

(18) $\begin{align*}\mu_k(-\infty,x) \geq 1 - \delta - \epsilon. \end{align*}$

Since $\epsilon$ and $\delta$ can be taken arbitrary small, it follows that $\mu_k(-\infty,\infty) = 1$ .
Again from (17), we have

(19) $\begin{align*}\lim_{m \to \infty}\sup_{x \in \mathbb{R}}|\mu_k(x,\infty) - P[X_mS_k > x]| = 0, \end{align*}$

and hence the law of $X_nS_k$ weakly converges to $\mu_k$ as $n \to \infty$ .

(4)
At first, we show the tightness of the sequence $\{X_n\}_n$ . Fix $\epsilon \in (0,1)$ . Then it holds that for every $R > 0$ and $k \in \mathbb{N}$

(20) $\begin{align*}P[|X_n| > R] &= P[|X_nS_k| > S_kR ] \\&\leq P[|S_k - 1| \geq \epsilon] + P[|X_nS_k| > (1 - \epsilon) R]. \end{align*}$

From the problem 3, it follows that

(21) $\begin{align*}\lim_{R \to \infty}\limsup_{n \to \infty}P[|X_n| > R] \leq P[|S_k - 1| \geq \epsilon]. \end{align*}$

Since $k$ can be taken arbitrary large, we obtain the tightness of $\{X_n\}_n$ from the problem 2.
Suppose that the law of $X_n$ converges to the probability measures $\nu$ and $\nu'$ along the subsequences $\{X_{n'}\}_{n'}$ and $\{X_{n"}\}_{n"}$ , respectively. Since it holds that for every $x$ such that $\mu_k\{x\} = 0 \ (\forall k \geq 0)$ and $\epsilon > 0$ ,

(22) $\begin{align*}&P[X_nS_k > x] - P[X_mS_k > x] \\\leq &P[X_n > x/(1 + \epsilon)] - P[X_m > x/(1 - \epsilon)] + 2P[|S_k - 1| > \epsilon], \end{align*}$

by taking $n$ and $m$ to $\infty$ along ${n'}$ and ${n"}$ , respectively and then taking $k$ to $\infty$ , we have

(23) $\begin{align*}0 \leq \nu[x/(1 + \epsilon), \infty) - \nu'(x/(1 - \epsilon), \infty). \end{align*}$

Hence we have

(24) $\begin{align*}0 \leq \nu[x,\infty) - \nu'(x,\infty), \end{align*}$

and similarly we can show $\nu'[x,\infty) - \nu(x,\infty) \geq 0$ . Then since the point $x \in \mathbb{R}$ such that $\nu\{x\} > 0$ , $\nu'\{x\} > 0$ or $\mu_k\{x\} > 0 \ (\exists k \geq0)$ is at most countable, we obtain $\nu = \nu'$ .

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院試問題　東大　数理科学研究科　平成30年　専門科目B

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