# 院試問題　東大　数理科学研究科　平成31年　専門科目B

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## 9

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

(1)

where . Hence .

(2)
Let and .
Define , and for .
Note that since we have and

(2)

it follows by induction that . Hence the sequence is strictly increasing and diverges to .

Define .
Then we have

(3)

We show that is square-integrable.
Since it holds that and , we have

(4)

for a constant and every .
Then we obtain

(5)

Therefore is square-integrable and is not injective.

(3)
Suppose there exists a bounded inverse operator of .
Then we have for every that

(6)

Hence it follows that . Then it is enough to show that there exists a sequence in such that and .

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 , and for . Then it is obvious that satisfies the desired properties.

(3)
If -a.e. , it is clear that for every measurable set such that .

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 such that .
Take such a measurable set .
Since is an orthonormal basis, we have

(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 , we have

(15)

and therefore

(16)

Hence the probability density function of is

(17)

(2)
Since we have a.s., the support of is contained in . For and , there exists a number such that . Since the support of is , we have from the independence of and ,

(18)

Hence and it follows that .

(3)
From the independence of , we have for

(19)

Hence the sequence is Cauchy in and converges to some random variable in .

(4)
We denote the support of as .
We show that a necessary and sufficient condition for is

(20)

Note that from the same argument in the problem (2), the support of is .
Suppose (20) holds. Then for every , and , we have

(21)

where . Since for every large , we obtain if we can show for large .
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 .

# 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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## 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 .
From an elementary calculation, we have for . Therefore it follows that for ,

(1)

Since the function is integrable on for every , we obtain from the dominated convergence theorem .

(2)
Define . Fix and take so that . Since we have ,
it holds that

(2)

Hence for every such that , the function for is dominated by the integrable function . Then we can interchange the differentiation and integration of and 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 be a sequence in the unit ball of .
From Arzela-Ascoli theorem, it is enough to show that the sequence is uniformly bounded and equicontinuous in .
From Hölder’s inequality, we have for every ,

(3)

where denotes the norm of .
From this, the equicontinuity and uniform boundedness is obvious.

(2)
From an elementary calculation, we can check that

(4)

Since the operator is self-adjoint, its eigenvalues must be real.
Take and suppose and . Then it follows that

(5)

Since the LHS of (5) is continuous, the function is continuous. Then it follows that LHS of (5) is function and therefore is so.
Differentiating both sides twice in (5), we obtain

(6)

Since , it follows that . Then we have

(7)

If , from (6), it holds that

(8)

for constants . Then from (7), we have and . Hence must be of the form

(9)

and we can easily check that this function satisfies (5).

If , it follows from (6) that

(10)

for constants . From (7), we have and this contradicts to the assumption .

Therefore the eigenvalues of are .

# 院試問題　東大　複雑理工学専攻　平成31年　専門基礎科目

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## 2

(1)
This is obvious.
(1)

(2)
For , we have
(2)
Therefore , and we obtain
(3)

(3)
Since it holds that
(4)
an eigenvector  for the eigenvalue  is
(5)

Hence we obtain
(6)
where .
Then it follows that
(7)
and we obtain the desired result.

(4)
Since we have
(8)
we can easily check that
(9)
Here we used the fact .

(5)
From the problem (3) and (4), we have
(10)
Hence we obtain
(11)

(6)
It is obvious that .
By considering the cofactor expansion of  with respect to the first column, we have
(12)
Hence from the problem (5), we obtain
(13)    

## 3

フーリエ変換の問題です. (4)以外は単純な計算問題です. (4)はの逆フーリエ変換を直接計算できるかと思いましたが, 計算がうまくいかなかったため, におけるフーリエ変換のユニタリ性(特に単射性)を用いました. 少し大げさな解き方かもしれません.

(1)
We denote the Fourier transform of as .

(14)

(2)

(15)

(3)
From the problem (1) and (2). we have from Fubini’s theorem

(16)

(4)
It is not difficult to check that the Fourier transform of is .
Since the function is square-integrable on , we have from the unitarity of the Fourier transform .

## 4

(2)の冒頭で示す離散の場合の部分積分公式を使うと計算が楽です.

(1)

(17)

(2)
Note that for every function and random variable , we have

(18)

Hence we obtain

(19)

(3)
From the independence of , we have

(20)

(4)
From the formula (18), we obtain

(21)