院試問題解答作成基本方針
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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
the following holds:
(1) 
(1)
By Hölder’s inequality, for every
s.t.
, we have
(2) 
Hence

. Define

where
(3) 
Then it holds that

and
(4) 
and therefore we obtain the result.
(2)
By Hölder’s inequality, we have for every
s.t.
,
(5) 
Then from (
1), it follows that
(6) 
Then from the problem 1, we obtain the result.
(3)
We denote the inner product of
as
. Note that
(7) 
From the assumption and (
1), we have

. Then from (
7), we obtain the result.
(4)
We decompose
where
. Then it follows that
(8) 
Then from (
1), LHS of (
8) converges to

as

, and RHS converges to

. Hence it holds that

and therefore

.
11
(1)
Assume
is an orthogonal projection. Since
, we have
(9) 
Then for every

, we have

.
Take

. Then it follows that
(10) 
and hence

. Therefore we obtain

.
For every

, it holds from (
9) that

. Hence

and from this,

easily follows.
Assume
. Then for every
, we have
and it easily follows that
.
(2)
Note that
(11) 
where

denotes

and

does the direct sum of Hilbert spaces.
It is enough to show that

.
Let

. Then we have

.
Hence

.
Let

. Then we have

, and therefore

.
(3)
Note that
for every
. Then from the problem 2, it is enough to show for every 
(12) 
Since it holds that

, we have for

and

,
(13) 
Since it holds that

for every

, 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
and
, there exists
and for every
, we have
(14) ![Rendered by QuickLaTeX.com \begin{align*}\sup_{x \in \mathbb{R}} | P[X_nS_k > x] - P[X_mS_k > x]| < \epsilon. \end{align*}](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-01e2fbe46f17165bef9abaf9bb011df7_l3.png)
(1)
This is no more than a simple calculation:
(15) ![Rendered by QuickLaTeX.com \begin{align*}P[S_k] = 1, \ \text{and} \ P[(S_k - P[S_k])^2] = 1/k. \end{align*}](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-e8690daf4f3f48eae077eed2eb81df0d_l3.png)
(2)
By (15) and Chebyshev’s inequality, we have for every
,
(16) ![Rendered by QuickLaTeX.com \begin{align*}P[|S_k - 1| > \epsilon] \leq 1/(\epsilon^2k) \xrightarrow{k \to \infty} 0.\end{align*}](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-6b7dedbdc450e1a4a6a5945e867769e1_l3.png)
Hence

converges to

in probability.
(3)
Fix
. From Helly’s selection theorem, there is a subsequence
the law of which converges vaguely to a finite measure
. It is enough to show that
is a probability measure and the law of
converges to
as
. By taking
to
along the subsequence
in (14), there exists an integer
and it holds for every continuity point
of
,
and
,
(17) ![Rendered by QuickLaTeX.com \begin{align*}\sup_{x \in \mathbb{R}}|\mu_k(x,\infty) - P[X_mS_k > x]| \leq \epsilon. \end{align*}](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-9943ada6cd70da8d6fdcb2b8e9b45b22_l3.png)
Fix

. By taking

so large that
![Rendered by QuickLaTeX.com P[X_NS_k > x] < \delta](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-aa23ed318ceac356129c33a114a7f99c_l3.png)
, we have from (
17)
(18) 
Since

and

can be taken arbitrary small, it follows that

.
Again from (
17), we have
(19) ![Rendered by QuickLaTeX.com \begin{align*}\lim_{m \to \infty}\sup_{x \in \mathbb{R}}|\mu_k(x,\infty) - P[X_mS_k > x]| = 0, \end{align*}](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-0b104b67730a92cf52f17fdd9c438150_l3.png)
and hence the law of

weakly converges to

as

.
(4)
At first, we show the tightness of the sequence
. Fix
. Then it holds that for every
and 
(20) ![Rendered by QuickLaTeX.com \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*}](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-5bff9b136d56048becd63139bdc70174_l3.png)
From the problem 3, it follows that
(21) ![Rendered by QuickLaTeX.com \begin{align*}\lim_{R \to \infty}\limsup_{n \to \infty}P[|X_n| > R] \leq P[|S_k - 1| \geq \epsilon]. \end{align*}](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-a6ff3d455dccdfd428f09700f9160378_l3.png)
Since

can be taken arbitrary large, we obtain the tightness of

from the problem 2.
Suppose that the law of

converges to the probability measures

and

along the subsequences

and

, respectively. Since it holds that for every

such that

and

,
(22) ![Rendered by QuickLaTeX.com \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*}](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-6b9fbc6b09e1637df1f05cda2e072796_l3.png)
by taking

and

to

along

and

, respectively and then taking

to

, we have
(23) 
Hence we have
(24) 
and similarly we can show

. Then since the point

such that

,

or

is at most countable, we obtain

.