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\}.

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