where . Hence .
Let and .
Define , and for .
Note that since we have and
it follows by induction that . Hence the sequence is strictly increasing and diverges to .
for a constant and every .
Then we obtain
Therefore is square-integrable and is not injective.
Suppose there exists a bounded inverse operator of .
Then we have for every that
Hence it follows that . Then it is enough to show that there exists a sequence in such that and .
Since it holds that for every , the function is continuous. Hence from the intermediate value theorem, we obtain the desired result.
From the problem (1), we can inductively take a sequence in such that
Define , and for . Then it is obvious that satisfies the desired properties.
If -a.e. , it is clear that for every measurable set such that .
and this contradicts to the assumption.
This is obvious from Schwarz’s inequality.
From the problem (3), it is enough to show that
for every measurable set such that .
Take such a measurable set .
Since is an orthonormal basis, we have
From the problem (4), it follows that
and therefore from (11) we obtain
Then the proof is complete.
For every , we have
Hence the probability density function of is
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 ,
Hence and it follows that .
From the independence of , we have for
Hence the sequence is Cauchy in and converges to some random variable in .
We denote the support of as .
We show that a necessary and sufficient condition for is
Note that from the same argument in the problem (2), the support of is .
Suppose (20) holds. Then for every , and , we have
where . Since for every large , we obtain if we can show for large .
From the Chebychev’s inequality, we have
Hence we obtain .
Suppose (20) does not hold, that is, . Then for every and , we take so that , we have
Hence it follows that .