Bases in Banach Spaces
[29/100%]Use Zorn's Lemma to show that if is a vector space then there exists a sequence that is a Hamel basis for .
Remark Although we focus in this volume on real and complex vector spaces, the argument of this exercise applies to any vector space over any field .
Let be an infinite-dimensional Banach space, and prove the following statements.
Any Hamel basis for must be uncountable.
Any infinite-dimensional subspace of that has a countable Hamel basis is a meager subset of , and cannot be a closed subspace of .
By Exercise 2.26, is a meager, dense subspace of . Show that does not have a countable Hamel basis.
Let be a Hamel basis for an infinite-dimensional Banach space . Then each can be written uniquely as where at most finitely many of the scalars are nonzero. Each is a linear functional on , and we call the sequence of coefficient functionals associated with the Hamel basis . This exercise addresses the question of whether these coefficient functionals can be continuous.
Show by example that it is possible for some particular functional to be continuous.
Show that for , where is the Kronecker delta.
Let . Show that .
Show that at most finitely many functionals can be continuous, i.e., is finite.
Give an example of an infinite-dimensional normed linear space that has a Hamel basis such that each of the associated coefficient functionals for is continuous.
Let be an unbounded linear functional on an infinite-dimensional Banach space (see Example 4.2). Then by Exercise 1.16, is a Banach space with respect to the norm . Set , and define .
Show that is a Banach space.
Show that even though , the normed space is not continuously embedded into , i.e., the mapping given by is not continuous.
Since the set of rational numbers is a field, we can consider the vector space over the field . By Exercise 4.1, there exists a Hamel basis for over . Hence every nonzero number can be written uniquely as for some and nonzero rational scalars . Use this to show that there exists a function that satisfies for all but does not satisfy for all . Thus is not linear, even though respects addition.
Let be a basis for a Banach space , and let be a sequence of nonzero scalars.
Show that is a basis for . In particular, is a normalized basis for .
If is an unconditional basis, will be an unconditional basis?
If is an absolutely convergent basis, will be an absolutely convergent basis?
Let be a separable Hilbert space.
Show that an orthonormal basis for is a normalized unconditional basis for .
Give an example of a basis for that is not an orthogonal sequence.
Give an example of a basis for that contains no orthogonal subsequences (equivalently, for all ).
Show that the standard basis is a normalized unconditional basis for for each , and is also a normalized unconditional basis for .
By Exercise 1.20, is a closed subspace of , and is a proper closed subspace of . Find a vector such that is a normalized unconditional basis for .
Show that is isometrically isomorphic to . Compare Exercise 1.75, which shows that we also have , and Exercise 4.22, which shows that and are topologically isomorphic but not isometrically isomorphic.
For each , define , where the 1 is repeated times. Show that is a normalized conditional basis for .
For each , define , where the 0 is repeated times. Show that is a normalized conditional basis for (this is called the summing basis for ).
Show that the standard basis is a monotone basis for for each index , and is also a monotone basis for . Is the basis for given in Exercise 4.8 monotone?
Show that any subsequence of a basis is a basic sequence.
Suppose that is a Schauder basis for a Hilbert space , and its biorthogonal system is . Show that if for every , then is an orthonormal basis for .
Show that if is an absolutely convergent basis for a Banach space , then
is a topological isomorphism of onto .
Conversely, show that every Banach space topologically isomorphic to has an absolutely convergent basis.
This exercise will give an alternative approach to proving Theorem 4.13. Let be a basis for a Banach space , and set , where the are the partial sum operators.
Show that is a norm on , and for all .
Suppose that is a Cauchy sequence in with respect to . With fixed, show that is Cauchy with respect to . Let be such that as , and observe that .
Show that for each we have , and use this to show that is Cauchy with respect to . Let be the element such that .
Show that , and use this to show that where is independent of .
Show that , and hence . Use this to show that , and conclude that is complete with respect to .
Show that and are equivalent norms on , and use this to show that . Conclude that is a Schauder basis for .
We say that a Banach space has the approximation property if the identity operator on can be uniformly approximated on every compact subset of by operators with finite rank. That is, given a compact set , there must exist continuous finite rank operators such that
Show that if has a basis, then has the approximation property.
Suppose that is a Banach space that has the approximation property and is an arbitrary Banach space. Show that if is a compact operator, then there exist continuous finite-rank operators such that as .
Remark Compact sets and compact operators are reviewed in Appendix B.
Let be a Banach space.
Show that if is a basis for , then the only possible vector such that is (weak convergence is defined in Section 2.10).
Use Theorem 2.39 to show that part (a) remains valid if we only assume that is a basic sequence in .
Give an example of a basis for a Banach space such that .
Give an example of a basis for a Banach space such that does not converge weakly to any vector in .
Prove the implication (a) (b) in Theorem 4.20.
Theorem 4.20: Let and be Banach spaces. If is a basis for and is a basis for , then the following two statements are equivalent.
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is equivalent to .
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converges in if and only if converges in . Definition 4.19: A basis for is equivalent to a basis for if there exists a topological isomorphism such that for all .
Suppose that is a basis for a Banach space is a sequence in a normed space , and there exists a topological isomorphism such that for every . Show that is a Banach space, is a basis for , and is equivalent to .
Suppose is a basis for a Banach space that is equivalent to a basis for a Banach space .
Show that is a bounded basis for if and only if is a bounded basis for .
Show that is an unconditional basis for if and only if is an unconditional basis for .
Show that is an absolutely convergent basis for if and only if is an absolutely convergent basis for .
Let be a basis for a Banach space and a basis for a Banach space . Show that is equivalent to if and only if there exist constants such that for all and we have
Let be the standard basis for . By Exercise 4.8, if we set , then is a basis for . Show that and are topologically isomorphic, and these two bases are equivalent.
Show that if and , then there exist with such that . Show that the analogous statement for can fail.
Show that is not isometrically isomorphic to (even so, note that their dual spaces and are each isometrically isomorphic to by Exercises 1.75 and 4.8).
Prove that the representation of functions given in equation (4.5) is unique.
Prove that is a Banach space with respect to the uniform norm.
Show that is a proper closed subspace of , and that is isometrically isomorphic to .
Show that is isometrically isomorphic to for each .
Show that if , then .
Prove Theorem 4.27.
Theorem 4.27: Let be a Banach space. If is a strong basis for , then is a weak basis for . Further, in this case is a weak Schauder basis for with coefficient functionals that are strongly continuous on .
Prove Theorem 4.28: let be a sequence in a Banach space , and let with norm .
is a Banach space.
If is a weak basis for , then is topologically isomorphic to via .
Finish the details of Example 4.31.
Example 4.31: Let , and note that is not separable. The sequence of standard basis vectors is contained in , although it does not form a strong basis for this space. We will show that is a weak* basis for . Choose any . Then for any we have
Hence in the weak* topology (even though this series need not converge strongly), and by Exercise 4.27 this representation is unique. Therefore is a weak* basis for .
Here a sequence in is a weak* basis for if for each there exist unique scalars such that , with convergence of this series in the weak* topology.
Prove the statements made in Example 4.32:
is a strong Schauder basis for whose biorthogonal system is .
The representation is the unique representation of with respect to weak* convergence, i.e., is a weak* basis for .
, so the functional determined by is not weak*-continuous on ; hence is not a weak* Schauder basis for .
Prove the statements made in Example 4.33:
is a strong Schauder basis for whose biorthogonal system is .
with weak* convergence, so is not weak* -independent; hence is not the unique weak*-convergent representation of , so is not a weak* basis for .