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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


(3)
Then it holds that

(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

(4)
We decompose where
. Then it follows that
(8)
Then from (1), LHS of (8) converges to





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


Take

(10)
and hence


For every




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



It is enough to show that

Let


Hence

Let



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



(13)
Since it holds that


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)
(1)
This is no more than a simple calculation:
(15)
(2)
By (15) and Chebyshev’s inequality, we have for every ,
(16)
Hence


(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)
Fix


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(18)
Since



Again from (17), we have
(19)
and hence the law of



(4)
At first, we show the tightness of the sequence . Fix
. Then it holds that for every
and
(20)
From the problem 3, it follows that
(21)
Since


Suppose that the law of








(22)
by taking







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





