a) What is Is the representation unique for
b) What is (n terms) Do we have uniqueness here?
c) If is any normed linear space of dimension greater than prove that
We claim that and that the representation is unique. Clearly Let . Then where For
any we have the representation If with and real then and It follows that Thus and Thus the representation is unique. Uniqueness of representation for follows from this.
The answer to (b) is
and the representation with for all is unique if and only if Clearly any is a sum of complex numbers with modulus at most and hence it can be written as a sum of elements of Now fix and suppose with for all Then and if then with for all . But for all so for all and Thus the representation is unique when We claim that it is not unique if There are infinitely many complex numbers such that and In fact one such number is where is the polar form of By continuity the inequality holds for any sufficiently close to . Any representation of as a sum of elements of gives a similar representation for Hence the representation is not unique.
c) In the case of complex scalars we can restrict to a one-dimensional subspace and in the case of real scalars we can look at a two dimensional subspace containing a given vector with .
Note that uniqueness fails: let and in . Then .
2. If are independent identically distributed positive random variables does it follow that assuming that all the products and expectations exist.
No! Let take values and with probabilty each. Then so . We claim that . Since it follows that is a non-negative martingale. Hence it converges a.s. We shall show that the limit is a.s. thereby completing the proof. Let . Then . By SLLN applied to we see that . Hence a.s. because .
3. Consider the inequalities and . Are these true when and are open sets in ? Are these true when and are compact sets in ?
Suppose and are compact. Translate by . Then the inequality does not change and the proof is therefore reduced to the case say. In this case ( because and (because and are disjoint) so . The reverse inequality fails when the Cantor set. It is not true that for all open sets: there is an open set such that and . Since we would have and choose compact sets such that and . Then .