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.

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