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Use Zorn's lemma to prove the following: If \(E\) is an orthonormal set in a Hilbert space \(\mathscr{H}\), then \(\mathscr{H}\) has an (orthonormal) basis containing \(E\). In particular, every Hilbert space has an orthonormal basis. A similar Zorn's lemma argument shows that every vector space has a Hamel basis.

Short Answer

Expert verified
By Zorn's lemma, every orthonormal set is part of a maximal orthonormal set, i.e., a basis.

Step by step solution

01

Review the Definitions

Firstly, recall that an orthonormal set in a Hilbert space \( \mathscr{H} \) is a set of vectors that are all unit vectors and orthogonal to each other. A basis of \( \mathscr{H} \) is an orthonormal set such that every element in \( \mathscr{H} \) can be written as a possibly infinite linear combination of these vectors.
02

Define Partially Ordered Set

Consider the collection \( \mathcal{F} \) of all orthonormal sets in \( \mathscr{H} \) that contain \( E \). Set partial order \( \subseteq \) on \( \mathcal{F} \). This collection is non-empty because \( E \) itself is an element of \( \mathcal{F} \).
03

Check Chain Condition

Next, consider any chain \( \{ S_i \}_{i \in I} \) of orthonormal sets in \( \mathcal{F} \). Their union \( \bigcup_{i \in I} S_i \) is an upper bound of the chain, as the union of orthonormal sets is orthonormal. This satisfies the condition for Zorn's lemma.
04

Apply Zorn's Lemma

By Zorn's lemma, \( \mathcal{F} \) has a maximal element, say \( S^* \). This set \( S^* \) is maximal orthonormal with respect to inclusion of orthonormal sets containing \( E \).
05

Prove Maximiality Implies Basis

To show that \( S^* \) is an orthonormal basis, assume there is \( x \in \mathscr{H} \) orthogonal to all elements of \( S^* \). If \( x eq 0 \), then \( \{ x / \| x \| \} \cup S^* \) is orthonormal, contradicting the maximality of \( S^* \). Therefore, every vector in \( \mathscr{H} \) must be in the span of \( S^* \).
06

Conclusion

Thus, \( S^* \) is an orthonormal basis for \( \mathscr{H} \) that contains \( E \). This means every Hilbert space has an orthonormal basis, as \( E \) could have been chosen as any orthonormal set.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hilbert Space
A Hilbert space is a special kind of vector space that extends the methods of vector algebra and calculus. It is equipped with an inner product, which is a way to multiply two vectors to produce a scalar. This inner product allows us to discuss concepts like length (or norm) and orthogonality in a Hilbert space.

Hilbert spaces are often infinite-dimensional, which means they have infinitely many dimensions. This feature is critical because it helps in various fields such as quantum mechanics and functional analysis. For instance, the space of square-integrable functions, often used in quantum mechanics, is a Hilbert space.

Some key properties of Hilbert spaces include:
  • Completeness: Every Cauchy sequence in a Hilbert space converges to a limit within the space. This is analogous to how sequences of real numbers behave.
  • Orthogonality: Two vectors are orthogonal if their inner product is zero. This concept is vital for defining an orthonormal basis.
Orthonormal Basis
An orthonormal basis in a Hilbert space is a set of vectors that are both orthogonal to each other and each of unit length. This means that the vectors not only form a base for the space (allow every vector in the space to be expressed as a combination of them) but also all are of length one, and no two vectors are in any way dependent on each other from the orthogonal perspective.

The advantage of having an orthonormal basis is that it simplifies many mathematical computations. For example, the coefficients in a linear combination of basis vectors correspond directly to the inner product of the vector with the basis vectors.

Orthonormal bases are particularly useful because they simplify the procedure to find the projection of one vector onto another, leveraging the simplicity of their unit length and orthogonality. In Hilbert spaces, Zorn’s lemma is used to prove that such a basis always exists, providing a powerful way to analyze and work within these spaces.
Hamel Basis
A Hamel basis is another important concept within vector spaces, including Hilbert spaces. Unlike an orthonormal basis, a Hamel basis does not necessarily consist of orthogonal or normalized vectors.

Instead, a Hamel basis is a set of vectors in a vector space such that any vector in the space can be uniquely written as a finite linear combination of vectors in this set. Specifically, this means:
  • The vectors span the space: Any vector in the space can be expressed as a combination of these vectors.
  • The set is linearly independent: No vector in the basis can be expressed as a linear combination of others in the set.
This concept is crucial for understanding vector spaces because it allows the characterization of the space in terms of its dimension. However, constructing a Hamel basis can be a complex task, especially in infinite-dimensional spaces. Despite this complexity, every vector space has a Hamel basis, a fact often established using Zorn's lemma.
Partially Ordered Set
A partially ordered set, or poset, is a set equipped with a partial order relation. In a partial order, not every pair of elements need to be comparable. This is in contrast to a total order, where every pair of elements is comparable.

To be more specific, a partial order is a binary relation \( \, \leq \, \) satisfying:
  • Reflexivity: Every element is related to itself. For any element \( a \), we have \( a \, \leq \, a \).
  • Antisymmetry: If two elements are related to each other, they are equal. If \( a \, \leq \, b \) and \( b \, \leq \, a \), then \( a = b \).
  • Transitivity: If an element is related to a second, which in turn is related to a third, then the first is related to the third. If \( a \, \leq \, b \) and \( b \, \leq \, c \), then \( a \, \leq \, c \).
Zorn's lemma is often used in the context of posets, particularly in proving the existence of maximal elements in certain sets, like the collection of orthonormal sets in a Hilbert space. In our original exercise, we define a poset to use Zorn's lemma, ensuring a maximal orthonormal basis for any orthonormal set.

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Most popular questions from this chapter

Suppose that \(X\) and \(Y\) are Banach spaces. (a) Show that \(X \times Y\) in the one-norm $$ \|(x, y)\| \equiv\|x\|_{X}+\|y\|_{Y} $$ is a Banach space. (b) Is \(X \times Y\) a Banach space in the norm $$ \|(x, y)\|_{\infty} \equiv \max \left(\|x\|_{X},\|y\|_{Y}\right) ? $$

A sequence \(\left\\{h_{n}\right\\}\) of vectors in a Hilbert space \(\mathscr{H}\) is said to be a Bessel sequence if $$ \sum_{n=1}^{\infty}\left|\left\langle h, h_{n}\right\rangle\right|^{2}<\infty $$ for every \(h \in \mathscr{H}\). A sequence \(\left\\{g_{n}\right\\}\) is said to be a Riesz-Fischer sequence if given any \(\left\\{c_{n}\right\\} \in \ell^{2}\) there exists (at least one) vector \(g \in \mathscr{H}\) such that $$ \left\langle g, g_{n}\right\rangle=c_{n} \text { for all } n $$ Note that an orthonormal basis is both a Bessel sequence and a Riesz-Fischer sequence. (a) Show that if \(\left\\{h_{n}\right\\}\) is a Bessel sequence, then there exists \(M<\infty\) so that $$ \sum_{n=1}^{\infty}\left|\left\langle h, h_{n}\right\rangle\right|^{2} \leq M\|h\|^{2} $$ for all \(h \in \mathscr{H}\). Hint: Apply the closed graph theorem to the map \(S: \mathscr{H} \rightarrow \ell^{2}\) defined by \(S h=\left\\{\left\langle h, h_{n}\right\rangle\right\\}\). (b) Show that if \(\left\\{g_{n}\right\\}\) is a Riesz-Fischer sequence, there exists \(m>0\) such that given \(\left\\{c_{n}\right\\} \in \ell^{2}\), the equations in (3.5) hold for at least one solution \(g\) satisfying $$ m\|g\|^{2} \leq \sum_{n=1}^{\infty}\left|c_{n}\right|^{2} . $$ Hint: The closed graph theorem again, applied to the appropriate map $$ T: \ell^{2} \rightarrow \mathscr{H} / N $$ where \(N\) is the orthogonal complement of the closed linear span of the vectors \(g_{n}\).

Let \(X\) be a compact Hausdorff space. A positive linear functional on \(C(X)\) is a (bounded) linear functional \(A\) with the additional property that \(A(f) \geq 0\) whenever \(f \geq 0\) on \(X\). Show that point evaluation at \(x_{0} \in X\) is a positive linear functional for each \(x_{0} \in X\). There is a representation theorem for the positive linear functionals on \(C(X)\) which says that for each positive linear functional \(\Lambda\) there is a unique positive, finite, regular Borel measure \(\mu\) on \(X\) with $$ \Lambda(f)=\int_{X} f d \mu . $$ (See Section A.5 in the Appendix for further discussion). If \(A\) is point evaluation at \(x_{0}\), what is the corresponding measure \(\mu\) ?

Show that if \(X\) is a Banach space that is not reflexive, then \(X^{*}\) is also not reflexive. Hint: Find a nonzero bounded linear functional on \(X^{* *}\) which is 0 on \(\left\\{x^{* *}: x \in X\right\\}\). The converse statement is also true; see p. 132 in [8].

Let \(X\) and \(Y\) be normed linear spaces and suppose \(T: X \rightarrow Y\) is linear. Show that \(T\) is continuous if \(\varphi \circ T\) is continuous for all \(\varphi\) in \(X^{*}\).

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