月別アーカイブ: 2019年4月

院試問題 京大 数学・数理解析専攻 平成30年 専門科目

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過去の入試問題 | Department of Mathematics Kyoto University

6

ストーン-ワイエルシュトラスの定理を使うのはすぐに思いつきますが, 仮定が奇数にしか成り立たないため, そのまま定理を適用することはできません.
指数関数\mathrm{e}^{-x}を測度の密度だと思うことで仮定が偶数について成り立つと思うことができ, ストーン-ワイエルシュトラスの定理 の仮定を満たす部分代数が定義できます. 後半はリース-マルコフ-角谷の表現定理を使います.
ストーン-ワイエルシュトラスの定理とリース-マルコフ-角谷の表現定理 という2つの一般的な定理を使うという珍しい問題です.

Assumption
For every positive odd number n, it holds that

(1)   \begin{align*} \int_{[0,\infty)}\mathrm{e}^{-nx}d\mu(x) = \int_{[0,\infty)}\mathrm{e}^{-nx}d\nu(x).  \end{align*}


(1)
Define d\mu'(x) = \mathrm{e}^{-x}d\mu(x) and d\nu'(x) = \mathrm{e}^{-x}d\nu(x).
From (1), we have for every non-negative even integer n,

(2)   \begin{align*}     \int_{[0, \infty)}\mathrm{e}^{-nx}d\mu'(x) = \int_{[0, \infty)}\mathrm{e}^{-nx}d\nu'(x).  \end{align*}


Note that the set \mathcal{A} = \{ \sum_{k = 1}^{n}a_k\mathrm{e}^{-2kx} \mid n \in \mathbb{N}, a_k \in \mathhbb{R} \} comprises a subalgebra in C_0([0,\infty)) := \{ f: [0,\infty) \to \mathbb{R} \mid \lim_{x \to \infty}f(x) = 0 \} equipped with the sup norm. Then from the Stone-Weierstrass theorem, the subalgebra \mathcal{A} is dense in C_0([0,\infty)). Therefore from (2), for every f \in C_0([0,\infty)), it holds that

(3)   \begin{align*}     \int_{[0,\infty)}f(x)d\mu'(x) = \int_{[0,\infty)}f(x)d\nu'(x).  \end{align*}


Then from the Riesz–Markov–Kakutani representation theorem, we obtain d\mu' = d\nu' and therefore d\mu = d\nu.

7

この問題は最初方針を誤り, コンパクト作用素の特異値分解を使って, ゴリゴリと計算したのですが一向に解けませんでした.
方針転換をして単位球の像が相対コンパクトであることを利用することを考えたら割合すぐにできました. 一般に強収束する作用素の列\{ A_n \}とコンパクト作用素Bに対し, A_nBは作用素ノルムの意味で収束することが言え, この答案ではこの事実を証明しつつ使っています.
これは個人的に感じていることに過ぎないのですが, この手の証明ではコンパクト性を用いる際に収束部分列の存在を使うよりも全有界性を利用する方が証明が書きやすいです. 部分列をとると, そのまた部分列をとる必要が出てきたりして記法がややこしくなり, 答案が書きにくいことがあるからです.

(1)
Define S_k = f_k \otimes e_k where (x\otimes y)z := (z,y)x for x,y,z \in H.
Then it holds that T_k = S_kT. Note that S := \sum_{n=1}^{\infty}S_n converges in the strong operator topology. Indeed, since \{e_n\}_n is an orthonormal basis, we have \sum_{n=1}^{\infty}|(x,e_n)|^2 = \|x\|^2, and therefore we have

(4)   \begin{align*}     \left\|\sum_{k=n}^{m}S_kx \right\|^2 = \sum_{k=n}^{m}|(x,e_k)|^2 \xrightarrow{n,m \to \infty} 0.  \end{align*}


We denote B as the closed unit ball of H. Since T is a compact operator, \overline{TB} is compact. Fix \epsilon > 0. Then there exists y_1, \cdots, y_N \in TB such that \overline{TB} \subset \cup_{i=1}^{N}B_{\epsilon}(y_i), where B_{\epsilon}(x) = \{ y \in H \mid \| y - x \| < \epsilon \}. Then for every x \in B , there is i \in \{1,\cdots,N\} such that \|Tx - y_i \| < \epsilon, and therefore we have

(5)   \begin{align*}     \left\| \sum_{k=n}^{m}S_kTx  \right\| &\leq \left\| \sum_{k=n}^{m}S_ky_i \right\| + \| S \| \|Tx - y_i\|   \\     &\leq \sup_{1 \leq j \leq N}\left\| \sum_{k=n}^{\infty}S_ky_j \right\| + \|S\| \epsilon. \label{} \end{align*}


Since (5) does not depend on x \in B, we have

(6)   \begin{align*}     \left\| \sum_{k=n}^{m}S_kT \right\| \leq \sup_{1 \leq j \leq N}\left\| \sum_{k=n}^{\infty}S_ky_j \right\| + \|S\| \epsilon \xrightarrow{n \to \infty} \| S \|\epsilon.  \end{align*}


Since we can take \epsilon > 0 arbitrary small, we obtain the desired result.

院試問題 京大 数学・数理解析専攻 平成29年 専門科目

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過去の入試問題 | Department of Mathematics Kyoto University

6

特にひねったところのないストレートな問題です. 測度の計算に慣れていれば容易でしょう.

(1)
For every n \in \mathbb{N}, there exists an element f_n \in \mathcal{C} such that \int f_nd\mu > M - 1/n. Define \varphi(x) = \sup_n f_n(x). We show \varphi satisfies the desired properties. Define \varphi_n(x) := \max_{1 \leq k \leq n}f_n(x). Then from the assumption (A), we have \varphi_n \in \mathcal{C}. Therefore it holds that M - 1/n \leq \int\varphi_nd\mu \leq M. Since \varphi_n is non-decreasing and converges to \varphi almost everywhere, we obtain from the monotone convergence theorem,

(1)   \begin{align*}     \int\varphi d\mu = \lim_{n \to \infty}\int \varphi_nd\mu = M.  \end{align*}


Assume there exists an element f \in \mathcal{C} such that f(x) > \varphi(x) on A \in \mathcal{F} with \mu(A) > 0. Define \psi_n(x) = \max\{ f(x), \varphi_n(x) \}. Then \psi_n \in \mathcal{C}, and we have

(2)   \begin{align*}     \liminf_{n \to \infty}\int \psi_nd\mu \geq \int_{A}f(x)d\mu + \int_{A^{c}}\varphi d\mu > M,  \end{align*}


and this contradicts to the definition of M. Hence for every f \in \mathcal{C}, we have f(x) \leq \varphi(x) \ \mathrm{a.e.}

(2)
Take an element A \in \mathcal{F} such that \mu(A) > 0. From the problem 1, it is obvious the following holds:

(3)   \begin{align*}     \sup_{f \in \mathcal{C}} \mathrm{ess \ sup}_A \ f \  \leq \mathrm{ess \ sup}_A\ \varphi.  \end{align*}


We show the inverse inequality holds. Take \epsilon > 0. Then there exists a subset B \subset A such that \mu(B) > 0 and \varphi(x) > \mathrm{ess \ sup}_A\varphi - \epsilon holds on B. Since it holds that \lim_{n \to \infty}\varphi_n(x) = \varphi(x) \ \mathrm{a.e}, we have

(4)   \begin{align*}     \lim_{n \to \infty}\mu\{x  \in B \mid \varphi_n(x) > \varphi(x) - \epsilon \} = \mu(B).  \end{align*}


Therefore we can take an integer N such that \mu\{x  \in B \mid \varphi_N(x) > \varphi(x) - \epsilon \} > 0.
Then it follows that for x \in B,

(5)   \begin{align*}     \mathrm{ess \ sup}_A \ \varphi < \varphi(x) + \epsilon < \varphi_N(x) + 2\epsilon. \end{align*}


Hence it holds that

(6)   \begin{align*}     \mathrm{ess \ sup}_A\varphi < \sup_{f \in \mathcal{C}}\mathrm{ess \ sup}_A f + 2 \epsilon,  \end{align*}


and since \epsilon > 0 is arbitrary, the proof is complete.

7

(1)は微積分レベルの計算をするだけです. 積分記号下での微分の定理を使ってもいいですが, 普通に計算してもそれほど労力は変わらないでしょう.
(2)は多少テクニカルです. 一致の定理まではストレートですが, そこからフーリエ変換が消えていることが導かれることは少し気づきにくいかもしれません.

答案ではL^2のフーリエ変換とそのユニタリ性を利用しましたが, 実はL^1の意味でのフーリエ変換が0になることから元の関数が0であることが従います. これは例えば黒田成俊『関数解析』(共立出版)の5章でGauss総和法を用いて示されています. よって, この答案では少し余計なことをしているといえばそうなのですが, L^2のフーリエ変換のユニタリ性の方がメジャーかと思いましたのでこのような証明を採用しました.
(2)でthe identity theorem とありますが, これは一致の定理のことです.

(1)
Fix z \in \mathbb{C} and \delta \in (0,1). Then for every h \in \mathbb{C} such that |h| < \delta, we have

(7)   \begin{align*}     &\left|\frac{F(z + h) - F(z)}{h} - \int_{-\infty}^{\infty}x\mathrm{e}^{-x^2 + zx}dx\right|  \\     =& \left| \int_{-\infty}^{\infty} \mathrm{e}^{-x^2 + zx}\left( \frac{\mathrm{e}^{hx} - 1}{h} - x \right)dx \right|  \\     \leq&  |h|\int_{-\infty}^{\infty}x^2 \mathrm{e}^{-(1 - \delta)x^2 + |z|x}dx \xrightarrow{|h| \to 0} 0, \end{align*}


where we used an elementary estimate \frac{|\mathrm{e}^{hx} - 1 - hx|}{|h|} \leq |h|x^2\mathrm{e}^{|h|x}. Hence F is holomorphic on \mathbb{C}.

(2)
We denote the closure of linear spans of \{f_n\}_n as M and its orthogonal complement as M^{\perp}.
Let g \in M^{\perp} and define F(z) = \int_{-\infty}^{\infty}\mathrm{e}^{-x^2 + zx}g(x)dx for z \in \mathbb{C}. Then for every n \in \mathbb{N}, we have F(1/n) = 0.
From the problem 1, the function F is holomorphic on \mathbb{C}, and therefore from the identity theorem, we obtain F(z) = 0 for every z \in \mathbb{C}. In particular, we have for every \xi \in \mathbb{R},

(8)   \begin{align*}     \int_{-\infty}^{\infty}\mathrm{e}^{-x^2}g(x)\mathrm{e}^{i\xi x}dx = 0.  \end{align*}


This means that the Fourier transform in the sense of L^1(\mathbb{R}) of the function \mathrm{e}^{-x^2}g(x) is 0. Since \mathrm{e}^{-x^2}g(x) \in L^1(\mathbb{R})\cap L^2(\mathbb{R}), the Fourier transform in the sense of L^1(\mathbb{R}) and L^2(\mathbb{R}) coincide. Since the Fourier transform in the sense of L^2(\mathbb{R}) is unitary, we have \mathrm{e}^{-x^2}g(x) = 0, and because the function \mathrm{e}^{-x^2} is always positive, we obtain g = 0 in L^2(\mathbb{R}).
Hence M^{\perp} = {0} and therefore M is dense.

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

9

(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.

11

(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'.