院試問題 京大 数学・数理解析専攻 平成29年 専門科目

Pocket
LINEで送る

院試問題解答作成基本方針

問題は↓から見られます.
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.

コメントを残す

メールアドレスが公開されることはありません。

このサイトはスパムを低減するために Akismet を使っています。コメントデータの処理方法の詳細はこちらをご覧ください