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院試問題 東大 数理科学研究科 平成31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

(1)
Let \psi be the inverse function of \varphi. By changing variables, we have

(1)   \begin{align*}         ||Tf||^2 &= \int_{0}^{\infty}|f(\varphi(x))|^2dx  \\         &= \int_{0}^{\infty}|f(x)|^2\frac{dx}{\varphi'(\psi(x))} \\         &\leq \frac{||f||^2}{c}      \end{align*}


where c = \inf_{x > 0}\varphi'(x) > 0. Hence ||T|| \leq 1/\sqrt{c}.

(2)
Let \epsilon > 0 and |\lambda| \leq 1/\sqrt{\alpha + \epsilon}.
Define \beta_0 = 0, \beta_1 = \varphi(0) > 0 and \beta_{n+1} = \varphi(\beta_n) for n \geq 1.
Note that since we have \beta_1 > 0 and

(2)   \begin{align*}     \beta_{n+1} - \beta_n = \int_{\beta_{n-1}}^{\beta_n}\varphi'(x)dx \geq \beta_n - \beta_{n-1},  \end{align*}


it follows by induction that \beta_{n+1} \geq \beta_n + \beta_1. Hence the sequence \{\beta_n\}_n is strictly increasing and diverges to \infty.

Define f(x) = \sum_{n = 0}^{\infty}\lambda^n1_{(\beta_n,\beta_{n+1}]}(x).
Then we have

(3)   \begin{align*}     Tf(x) &= \sum_{n = 0}^{\infty}\lambda^n1_{(\beta_n,\beta_{n+1}]}(\varphi(x))  \\     &= \sum_{n = 0}^{\infty}\lambda^n1_{(\beta_{n-1},\beta_{n}]}(x)  \\     &= \lambda f(x).  \end{align*}

We show that f is square-integrable.
Since it holds that\lim_{n \to \infty}\beta_{n} = \infty and \lim_{x \to \infty}\varphi'(x) = \alpha, we have

(4)   \begin{align*}     \beta_{n+1} - \beta_{n} \leq C (\alpha + \epsilon/2)^{n}  \end{align*}


for a constant C > 0 and every n \geq 1.
Then we obtain

(5)   \begin{align*} ||f||^2 &= \sum_{n=0}^{\infty}|\lambda|^{2n}(\beta_{n+1} - \beta_{n})  \\ &\leq C\sum_{n=0}^{\infty}|\lambda|^{2n}(\alpha + \epsilon/2)^n < \infty. \end{align*}


Therefore f is square-integrable and T - \lambda I is not injective.

(3)
Suppose there exists a bounded inverse operator S of T - I.
Then we have for every f \in L^2(\Omega) that

(6)   \begin{align*}     ||f|| = ||S(T - I)f|| \leq ||S|| ||(T - I)f||.  \end{align*}


Hence it follows that ||(T-I)f|| \geq ||f||/||S||. Then it is enough to show that there exists a sequence \{f_n\}_n in L^2(\Omega) such that ||f_n|| = 1 and \lim_{n \to \infty}||(T-I)f_n|| = 0.

For n \in \mathbb{N}, define f_n = (1/ \sqrt{n})1_{(0,1/n]}(x). Then it holds that ||f_n|| = 1.
Then we obtain

(7)   \begin{align*}     ||(T-I)f_n||^2 =& \int_{0}^{\infty}|f_n(\varphi(x))|^2dx  \\     &+ \int_{0}^{\infty}|f_n(x)|^2dx - 2\int_{0}^{\infty}f_n(x)f_n(\varphi(x))dx  \\     =& 1 - n\log (1 + 1/n) \xrightarrow{n \to \infty} 0.  \end{align*}

12

(1)
Since it holds that \mu({x}) = 0 for every x \in I, the function [0,1] \ni t \mapsto \mu(A_t) is continuous. Hence from the intermediate value theorem, we obtain the desired result.

(2)
From the problem (1), we can inductively take a sequence \{t_n\}_n in I such that

(8)   \begin{align*}         0 = t_0 < t_1 < t_2 < \cdots, \  \mu(A \cap [t_{n-1},t_n)) = 2^{-n}.       \end{align*}


Define t = \lim_{n \to \infty}t_n, B_1 = A\cap ([t_0,t_1)\cup [t,1]) and B_n = A \cap [t_{n-1},t_n) for n \geq 2. Then it is obvious that B_n satisfies the desired properties.

(3)
If f(x) = \infty \mu-a.e. x, it is clear that \int_{A}f(x)\mu(dx) = \infty for every measurable set A such that \mu (A) > 0.

Assume \int_{A}f(x)\mu(dx) \geq 1 for every measurable set A such that \mu(A) > 0.
If there is some R > 0 such that \mu (f > R) < 1, then we can take some x, y \in I such that 0 < y - x < 1/2R and \mu (\{ f \leq R \}\cap [x,y]) > 0.
Then we have

(9)   \begin{align*}     \int_{\{ f \leq R \}\cap [x,y]}f(x)\mu(dx) \leq 1/2.   \end{align*}


and this contradicts to the assumption.

(4)
This is obvious from Schwarz’s inequality.

(5)
From the problem (3), it is enough to show that

(10)   \begin{align*}         \int_{A}\sum_{n=1}^{\infty}e_n(x)^2\mu(dx) \geq 1      \end{align*}


for every measurable set A such that \mu(A) > 0.
Take such a measurable set A.
Since \{e_n\} is an orthonormal basis, we have

(11)   \begin{align*}         \mu(A) = \sum_{n=1}^{\infty}\left(\int_{A}e_n(x)\mu(dx)\right)^2.      \end{align*}


From the problem (4), it follows that

(12)   \begin{align*}         \left( \int_{A}e_n(x)\mu(dx) \right)^2 \leq \mu(A) \int_{A}e_n(x)^2\mu(dx),      \end{align*}


and therefore from (11) we obtain

(13)   \begin{align*}         \mu(A) \leq \mu(A) \sum_{n=1}^{\infty} \int_{A}e_n(x)^2\mu(dx).      \end{align*}


Then the proof is complete.

13

(1)
It is not difficult to check

(14)   \begin{align*}         \mathbb{E}[X_1^2 - 1] = 0, \mathrm{Var}(X_1^2 - 1) = 2.      \end{align*}


For every \lambda \geq -1, we have

(15)   \begin{align*}         \mathbb{P}[X_1^2 - 1 \leq \lambda] &= \mathbb{P}[|X_1| \leq \sqrt{\lambda + 1}]  \\         &= \frac{2}{\sqrt{2\pi}}\int_{0}^{\sqrt{\lambda + 1}}\mathrm{e}^{-x^2/2}dx,      \end{align*}


and therefore

(16)   \begin{align*}         \frac{d}{d\lambda}\mathbb{P}[X_1^2 - 1 \leq \lambda] = \frac{\mathrm{e}^{-(\lambda + 1)/2}}{\sqrt{2\pi (\lambda +  1)}}.      \end{align*}


Hence the probability density function f of X_1^2 - 1 is

(17)   \begin{align*}         f(x) = \frac{\mathrm{e}^{-(x + 1)/2}1\{x \geq -1\}}{\sqrt{2\pi (x + 1)}}.      \end{align*}



(2)
Since we have S_2 \geq -a_1 - a_2 a.s., the support \sigma_2 of S_2 is contained in [-a_1 - a_2, \infty). For a \geq - a_1 - a_2 and \epsilon > 0, there exists a number \lambda \geq 0 such that a = -\lambda a_1 - (1 - \lambda)(a_1 + a_2) = -a_1 - (1 - \lambda) a_2. Since the support of X_i^2 - 1 is [-1,\infty), we have from the independence of X_1 and X_2,

(18)   \begin{align*}         &\mathbb{P}[|S_2 - a| < \epsilon] \\         \geq& \mathbb{P}[|a_1(X_1^2 - 1) + a_1| < \epsilon / 2]\cdot\mathbb{P}[|a_2(X_2^2 - 1) + (1 - \lambda)a_2| < \epsilon / 2]  \\         >& 0.     \end{align*}


Hence a \in \sigma_2 and it follows that \sigma_2 = [-a_1 - a_2, \infty).

(3)
From the independence of X_i, we have for m > n

(19)   \begin{align*}         \mathbb{E}|S_n - S_m|^2 &= \sum_{k=n+1}^{m}a_k^2 \mathrm{Var}(X_k^2 - 1)   \\         &= 2\sum_{k=n+1}^{m}a_k^2 \xrightarrow{m,n \to \infty} 0.       \end{align*}


Hence the sequence \{ S_n \}_n is Cauchy in L^2(\mathbb{P}) and converges to some random variable S in L^2(\mathbb{P}).

(4)
We denote the support of S as \sigma.
We show that a necessary and sufficient condition for \sigma = \mathbb{R} is

(20)   \begin{align*}         \sum_{n=1}^{\infty}a_i = \infty.      \end{align*}


Note that from the same argument in the problem (2), the support \sigma_n of S_n is [-\sum_{k=1}^{n}a_k, \infty).
Suppose (20) holds. Then for every a \in \mathbb{R}, N \in \mathbb{N} and \epsilon > 0, we have

(21)   \begin{align*}         \mathbb{P}[|S - a| < \epsilon] \geq \mathbb{P}[|S_N - a| < \epsilon / 2]\cdot \mathbb{P}[ |\tilde{S}_N| < \epsilon / 2 ]      \end{align*}


where \tilde{S}_N = \sum{k=N+1}^{\infty}a_k(X_k^2 - 1). Since \mathbb{P}[|S_N - a| < \epsilon / 2] > 0 for every large N, we obtain a \in \sigma if we can show \mathbb{P}[ |\tilde{S}_N| < \epsilon / 2 ] > 0 for large N.
From the Chebychev’s inequality, we have

(22)   \begin{align*}     \mathbb{P}[ |\tilde{S}_N| < \epsilon / 2 ] &= 1 - \mathbb{P}[ |\tilde{S}_N| \geq \epsilon / 2 ]  \\     &\geq 1 - \frac{8}{\epsilon^2}\sum_{k=N+1}^{\infty}a_k^2  \\     &\xrightarrow{N \to \infty} 1.  \end{align*}


Hence we obtain a \in \sigma.

Suppose (20) does not hold, that is, \alpha := \sum_{n=1}^{\infty}a_n < \infty. Then for every a < -\alpha and N \in \mathbb{N}, we take \epsilon > 0 so that a + 2 \epsilon < -\alpha, we have

(23)   \begin{align*}     &\mathbb{P}[|S - a| < \epsilon] \\      \leq& \mathbb{P}[|S - a| < \epsilon, |S - S_N| < \epsilon] + \mathbb{P}[|S - S_N| \geq \epsilon]  \\     \leq& \mathbb{P}[|S_N - a| < 2\epsilon] + \mathbb{P}[|S - S_N| \geq \epsilon]  \\     =& \mathbb{P}[|S - S_N| \geq \epsilon]  \\     \xrightarrow{N \to \infty}& 0.  \end{align*}


Hence it follows that a \not\in \sigma.

院試問題 京大 数学・数理解析専攻 平成31年 専門科目

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

6

(1)はルベーグの収束定理を使うためにうまい優関数を見つける必要があります. やること自体はシンプルなのですが何となく馴染みのない計算で少し苦労しました. (2)は特段詰まるところのない問題です.

(1)
Define f(x) = (1 + x)^{1/x}.
From an elementary calculation, we have f(x) \geq 2 for x \in (0,1]. Therefore it follows that for 1/n \leq t \leq n,

(1)   \begin{align*} \left( 1  + \frac{t}{n}  \right)^{-n}t^{x-1}  \leq 2^{-t} t^{x-1}.   \end{align*}


Since the function t \mapsto 2^{-t} t^{x-1} is integrable on (0,\infty) for every x > 0, we obtain from the dominated convergence theorem \varphi(x) =  \int_{0}^{\infty}\mathrm{e}^{-t}t^{x-1}dt.

(2)
Define g(t,x) = \mathrm{e}^{-t}t^{x-1}. Fix x_0 > 0 and take C,\epsilon so that \log t \leq Ct^{\epsilon} \ (t > 0). Since we have \frac {\partial}{\partial x}g(t,x) = g(t,x) \log t,
it holds that

(2)   \begin{align*} \left| \frac{\partial}{\partial x}g(t,x) \right| \leq C g(t,x + \epsilon).  \end{align*}


Hence for every \delta > 0 such that x_0 - \delta > 0, the function u(t) := \frac{\partial}{\partial x}g(t,x) for x \in [x_0 - \delta, x_0 + \delta] is dominated by the integrable function Cg(t,x_0 + \epsilon + \delta). Then we can interchange the differentiation and integration of \varphi and \varphi is differentiable.

7

(1)はアスコリ-アルツェラの定理を使ってコンパクト性を示します. 定番の方法といえるでしょう.
(2)は具体的に共役作用素が求まることがポイントです. 最終的には常微分方程式の境界値問題になります.

(1)
Let \{ u_n \}_n be a sequence in the unit ball of L^2([0,1]).
From Arzela-Ascoli theorem, it is enough to show that the sequence \{Tu_n\}_n is uniformly bounded and equicontinuous in C([0,1]).
From Hölder’s inequality, we have for every 0 \leq s < t \leq 1,

(3)   \begin{align*}     |Tu_n(t) - Tu_n(s)| \leq \sqrt{t - s}||u_n||_2 \leq \sqrt{t - s},  \end{align*}


where \| \cdot \| denotes the norm of L^2([0,1]).
From this, the equicontinuity and uniform boundedness is obvious.

(2)
From an elementary calculation, we can check that

(4)   \begin{align*}     T^{\ast}u(t) = \int_{t}^{1}u(s)ds.  \end{align*}


Since the operator T^{\ast}T is self-adjoint, its eigenvalues must be real.
Take \lambda \in \mathbb{R} and suppose T^{\ast}Tu = \lambda u and u \neq 0. Then it follows that

(5)   \begin{align*}     \int_{t}^{1}ds\int_{0}^{s}u(r)dr = \lambda u(t) \ (t \in [0,1]).  \end{align*}


Since the LHS of (5) is continuous, the function u is continuous. Then it follows that LHS of (5) is C^2 function and therefore u is so.
Differentiating both sides twice in (5), we obtain

(6)   \begin{align*}     -u(t) = \lambda u''(t).  \end{align*}


Since u \neq 0, it follows that \lambda \neq 0. Then we have

(7)   \begin{align*}     u(1) = 0, \ u'(0) = 0.  \end{align*}

If \lambda > 0, from (6), it holds that

(8)   \begin{align*}     u(t) = A \cos \frac{t}{\sqrt{\lambda}} + B \sin \frac{t}{\sqrt{\lambda}}  \end{align*}


for constants A,B \in \mathbb{R}. Then from (7), we have A \neq 0, B = 0 and \lambda = \frac{1}{(1/2 + n)^2\pi^2} \ (\exists n \in \mathbb{N}). Hence u must be of the form

(9)   \begin{align*}      u(t) = A \cos \left(\frac{1}{2} + n\right)\pi t,  \end{align*}


and we can easily check that this function satisfies (5).

If \lambda < 0, it follows from (6) that

(10)   \begin{align*}     u(t) = A\mathrm{e}^{\sqrt{-\lambda}t} + B\mathrm{e}^{-\sqrt{-\lambda}t}  \end{align*}


for constants A,B \in \mathbb{R}. From (7), we have A = B = 0 and this contradicts to the assumption u \neq 0.

Therefore the eigenvalues of T^{\ast}T are \left\{\frac{1}{(1/2 + n)^2\pi^2} ,  n \in \mathbb{N}\right\}.

院試問題 東大 複雑理工学専攻 平成31年 専門基礎科目

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複雑理工学専攻

1

単純な計算問題なので省略.

2

線形代数の基本的な問題です.

(1)
This is obvious.

(1)   \begin{align*} A = \left( \begin{matrix} 0 & 1 \\ 1 & 1 \\ \end{matrix} \right).  \end{align*}

(2) For \lambda \in \mathbb{C}, we have

(2)   \begin{align*} \det (\lambda - A) = \lambda^2 - \lambda - 1.  \end{align*}

Therefore \det (\lambda - A) = 0 \Leftrightarrow \lambda = (1 \pm \sqrt{5})/2, and we obtain

(3)   \begin{align*} \lambda_\pm = \frac{1 \pm \sqrt{5}}{2}.  \end{align*}

(3) Since it holds that

(4)   \begin{align*} (\lambda_\pm - A) = \left( \begin{matrix} \lambda_\pm & -1 \\ -1 & - \lambda_\mp \\ \end{matrix} \right),  \end{align*}

an eigenvector u_\pm for the eigenvalue \lambda_\pm is

(5)   \begin{align*} u_\pm = \left( \begin{matrix} 1 \\ \lambda_\pm \\ \end{matrix} \right). \end{align*}

Hence we obtain

(6)   \begin{align*} P^{-1}AP = \left( \begin{matrix} \lambda_+ & 0 \\ 0 & \lambda_- \\ \end{matrix} \right), \end{align*}

where P = (u_+ ,u_-). Then it follows that

(7)   \begin{align*} P^{-1}A^nP = \left( \begin{matrix} \lambda_+^n & 0 \\ 0 & \lambda_-^n \\ \end{matrix} \right), \end{align*}

and we obtain the desired result. (4) Since we have

(8)   \begin{align*} \left( \begin{matrix} x_n \\ x_{n+1} \\ \end{matrix} \right) = A^{n-1} \left( \begin{matrix} x_1 \\ x_2 \\ \end{matrix} \right), \end{align*}

we can easily check that

(9)   \begin{align*} x_n = \frac{1}{\sqrt{5}}\left( \frac{1}{\lambda_+^2} + \frac{1}{\lambda_+} \right)\lambda_+^n + \frac{1}{\sqrt{5}}\left( \frac{1}{\lambda_-^2} + \frac{1}{\lambda_-} \right)\lambda_-^n = \frac{1}{\sqrt{5}}(\lambda_+^n - \lambda_-^n).  \end{align*}

Here we used the fact \lambda_\pm^2 - \lambda_\pm -1 = 0. (5) From the problem (3) and (4), we have

(10)   \begin{align*} A^n     = \left( \begin{matrix} x_{n-1} & x_n \\ x_n & x_{n+1} \\ \end{matrix} \right). \end{align*}

Hence we obtain

(11)   \begin{align*} y_n = a x_{n-2} + b x_{n-1}. \end{align*}

(6) It is obvious that D_2 = 2. By considering the cofactor expansion of D_n with respect to the first column, we have

(12)   \begin{align*} D_n &= D_{n-1} + \det \left( \begin{matrix} 1 & 0 & \ldots & 0 \\ -1 &&& \\ 0 &&& \\ \vdots &&\text{\Huge $D_{n-2}$}& \\ 0 &&&  \end{matrix} \right) \\ &= D_{n-1} + D_{n-2}. \end{align*}

Hence from the problem (5), we obtain

(13)   \begin{align*} D_n = x_{n-2} + 2x_{n-1}. \end{align*}

3

フーリエ変換の問題です. (4)以外は単純な計算問題です. (4)はs(\omega/2)^3の逆フーリエ変換を直接計算できるかと思いましたが, 計算がうまくいかなかったため, L^2におけるフーリエ変換のユニタリ性(特に単射性)を用いました. 少し大げさな解き方かもしれません.

(1)
We denote the Fourier transform of g as G.

(14)   \begin{align*}     G(\omega) &= \int_{-\infty}^{\infty}g(t)\mathrm{e}^{-\i \omega t}dt = 2\int_{0}^{1/2} \cos \omega t dt = \frac{2 \sin (\omega /2)}{\omega} = s(\omega/2).  \end{align*}



(2)

(15)   \begin{align*}     h(t) &= \int_{-\infty}^{\infty}g(\tau)g(t - \tau)d\tau  \\     &= \int_{-1/2}^{1/2}g(t - \tau)d\tau  \\     &=\int_{t - 1/2}^{t + 1/2}g(\tau)d\tau  \\     &= (1 - |t|)1\{|t| \leq 1\}.     \end{align*}



(3)
From the problem (1) and (2). we have from Fubini’s theorem

(16)   \begin{align*}     \int_{-\infty}^{\infty}k(t)\mathrm{e}^{-\i \omega t}dt      &= \int_{-\infty}^{\infty}(g\ast g)(t)\mathrm{e}^{-\i \omega t}dt \\     &= \left(\int_{-\infty}^{\infty}g(t)\mathrm{e}^{-\i \omega t}dt\right)^2 \\     &= s(\omega/2)^2. \end{align*}



(4)
It is not difficult to check that the Fourier transform of g\ast g \ast g is s(\omega/2)^3.
Since the function s(\omega/2)^2 is square-integrable on \mathbb{R}, we have from the unitarity of the Fourier transform l = g \ast g \ast g.

4

(2)の冒頭で示す離散の場合の部分積分公式を使うと計算が楽です.

(1)

(17)   \begin{align*}     \mathrm{Pr}[X \geq a] = \sum_{i = a}^{4}2^{-i} + 2^{-4} = 2^{-a+1}(1 - 2^{-5+a}) + 2^{-4} = 2^{-a + 1}. \end{align*}



(2)
Note that for every function f: \{1,2,\cdots,k\} \rightarrow \mathbb{R} and random variable W, we have

(18)   \begin{align*}        \sum_{i=1}^{k}f(i)\mathrm{Pr}[W = i] &= \sum_{i=1}^{k}f(i)(\mathrm{Pr}[W \geq i] - \mathrm{Pr}[W \geq i + 1])  \label{} \\     &= \sum_{i=1}^{k}f(i)\mathrm{Pr}[W \geq i] - \sum_{i=1}^{k-1}f(i)\mathrm{Pr}[W \geq i + 1] \label{} \\     &= \sum_{i=1}^{k}f(i)\mathrm{Pr}[W \geq i] - \sum_{i=2}^{k}f(i-1)\mathrm{Pr}[W \geq i] \label{} \\     &= f(1) + \sum_{i=2}^{k}(f(i) - f(i-1))\mathrm{Pr}[W \geq i].    \end{align*}


Hence we obtain

(19)   \begin{align*}     E[Z] &= \sum_{i=1}^{5}2^{i}\mathrm{Pr}[X = i] = 2 + \sum_{i = 2}^{5}2^{i-1}\mathrm{Pr}[X \geq i] = 2 + \sum_{i = 2}^{5}1 = 6,  \\     E[Z^2] &= 4 + \sum_{i = 2}^{5}3 \cdot 4^{i-1}\cdot 2^{-i + 1} = 4 + 3 \sum_{i = 2}^{5}2^{i-1} = 94,  \\     \mathrm{Var}(Z) &= 94 - 36 = 58.  \end{align*}



(3)
From the independence of {X_n}_n, we have

(20)   \begin{align*}     \mathrm{Pr}[Y_n \geq a] = \mathrm{Pr}[X_i \geq a \ \text{for all}  1 \leq i \leq n] = \mathrm{Pr}[X \geq a]^n = 2^{n(-a + 1)}. \end{align*}



(4)
From the formula (18), we obtain

(21)   \begin{align*}     E[Y_n] = 1 + \sum_{i = 2}^{k} \mathrm{Pr}[Y_n \geq i] = \frac{2^n - 2^{-n(k-1)}}{2^n -1}.  \end{align*}