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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 be the inverse function of
. By changing variables, we have
(1)
where


(2)
Let


Define




Note that since we have

(2)
it follows by induction that



(3)
We show that is square-integrable.
Since it holds that and
, we have
(4)
for a constant


Then we obtain
(5)
Therefore


(3)
Suppose there exists a bounded inverse operator


Then we have for every

(6)
Hence it follows that





For , define
. Then it holds that
= 1.
Then we obtain
(7)
12
(1)
Since it holds that for every
, the function
is continuous. Hence from the intermediate value theorem, we obtain the desired result.
(2)
From the problem (1), we can inductively take a sequence in
such that
(8)
Define

![Rendered by QuickLaTeX.com B_1 = A\cap ([t_0,t_1)\cup [t,1])](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-87f23f906ab55e8d2142289c4c9b38e7_l3.png)



(3)
If






Assume for every measurable set
such that
.
If there is some such that
, then we can take some
such that
and
.
Then we have
(9)
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)
for every measurable set


Take such a measurable set

Since

(11)
From the problem (4), it follows that
(12)
and therefore from (11) we obtain
(13)
Then the proof is complete.
13
(1)
It is not difficult to check
(14)
For every

(15)
and therefore
(16)
Hence the probability density function


(17)
(2)
Since we have












(18)
Hence


(3)
From the independence of


(19)
Hence the sequence




(4)
We denote the support of


We show that a necessary and sufficient condition for

(20)
Note that from the same argument in the problem (2), the support



Suppose (20) holds. Then for every



(21)
where

![Rendered by QuickLaTeX.com \mathbb{P}[|S_N - a| < \epsilon / 2] > 0](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-cd47a5252a1e64b2e09661c38e778410_l3.png)


![Rendered by QuickLaTeX.com \mathbb{P}[ |\tilde{S}_N| < \epsilon / 2 ] > 0](http://paleperlite.com/wp-content/ql-cache/quicklatex.com-42f3f16392112d9bd487ab6a9fec70bb_l3.png)

From the Chebychev’s inequality, we have
(22)
Hence we obtain

Suppose (20) does not hold, that is, . Then for every
and
, we take
so that
, we have
(23)
Hence it follows that

In 9-(2), how does “f is square-integrable” show the desired result?
> YU
Thanks for your comment.
Because the function f satisfies (T – λI)f = 0.