(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.
Hence . Define where
Then it holds that and
and therefore we obtain the result.
Then from (1), it follows that
Then from the problem 1, we obtain the result.
From the assumption and (1), we have . Then from (7), we obtain the result.
Then from (1), LHS of (8) converges to as , and RHS converges to . Hence it holds that and therefore .
Then for every , we have .
Take . Then it follows that
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 .
where denotes and does the direct sum of Hilbert spaces.
It is enough to show that .
Let . Then we have .
Let . Then we have , and therefore .
Since it holds that , we have for and ,
Since it holds that for every , we obtain the desired result.
(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.
By (15) and Chebyshev’s inequality, we have for every ,
Hence converges to in probability.
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 ,
Fix . By taking so large that , we have from (17)
Since and can be taken arbitrary small, it follows that .
Again from (17), we have
and hence the law of weakly converges to as .
From the problem 3, it follows that
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 ,
by taking and to along and , respectively and then taking to , we have
Hence we have
and similarly we can show . Then since the point such that , or is at most countable, we obtain .