Problem 10
Question
If \(S\) is any subset of \(\mathcal{H}\), and \(V\) the closed subspace generated by \(S, V=\overline{L(S)}\), show that \(S^{1}=\left\\{u \in \mathcal{H} \mid\\{u|x\rangle=0\right.\) for all \(x \in S\\}=V^{1}\)
Step-by-Step Solution
Verified Answer
The orthocomplement of \(S\), denoted by \(S^{1}\), is in fact equal to the orthocomplement of \(V\), \(V^{1}\), where \(V\) is the closed subspace generated by \(S\). This is shown by first proving that \(S^{1} \subseteq V^{1}\) and then that \(V^{1} \subseteq S^{1}\), which can then be used to conclude that \(S^{1} = V^{1}\).
1Step 1: Show that \(S^{1} \subseteq V^{1}\)
Start by assuming that \(u \in S^{1}\). This means that for every \(x \in S\), \(u, x = 0\), where ',' represents the inner product. Hence, \langle u, x \rangle = 0 for all \(x \in S\). Because every \(x \in S\) is also in \(V\) (since \(V\) is generated by \(S\)), then \langle u, x \rangle = 0 holds for all \(x \in V\). Therefore, \(u \in V^{1}\), effectively showing that \(S^{1} \subseteq V^{1}\).
2Step 2: Show that \(V^{1} \subseteq S^{1}\)
Again, start by assuming that \(u \in V^{1}\). This means that \langle u, x \rangle = 0 for all \(x \in V\). Now take a sequence \(x_n \in L(S)\) such that \(x_n\) converges to \(x\) in \(V\). Since \(x_n \rightarrow x\), and \(u \in V^{1}\) (meaning that \langle u, x_n \rangle -> 0), it must also be the case that \langle u, x \rangle = 0 for all \(x \in V\), meaning that \(u \in S^{1}\). This shows that \(V^{1} \subseteq S^{1}\).
3Step 3: Conclude that \(S^{1} = V^{1}\)
Having shown in steps 1 and 2 that \(S^{1} \subseteq V^{1}\) and \(V^{1} \subseteq S^{1}\), it can now be concluded that \(S^{1} = V^{1}\), which is the outcome the exercise requested.
Key Concepts
SubspaceInner ProductOrthogonalityConvergence
Subspace
In the context of Hilbert spaces, a **subspace** is a subset that retains the structure of the larger space. For any subset \( S \) in a Hilbert space \( \mathcal{H} \), the subspace \( V \) generated by \( S \) is crucial. This means \( V \) includes all possible linear combinations of the elements in \( S \), and if \( V \) is closed, it also contains all limits of convergent sequences in \( S \).
For example:
For example:
- The set of all vectors in \( \mathcal{H} \) that can be formed as \( a_1 x_1 + a_2 x_2 + \ldots + a_n x_n \) where \( x_i \in S \) is part of the subspace.
- If \( V \) is closed, then even if a sequence of vectors from \( S \) converges to another vector, that vector is also in \( V \).
Inner Product
The **inner product** is a special operation that takes two vectors from a Hilbert space and returns a scalar. It helps to define geometric concepts like angle and length within abstract spaces.
Consider two vectors \( u \) and \( v \) in a Hilbert space \( \mathcal{H} \). The inner product is denoted as \( \langle u, v \rangle \) and must satisfy several properties:
Consider two vectors \( u \) and \( v \) in a Hilbert space \( \mathcal{H} \). The inner product is denoted as \( \langle u, v \rangle \) and must satisfy several properties:
- Linearity: \( \langle au + bw, v \rangle = a \langle u, v \rangle + b \langle w, v \rangle \)
- Symmetry: \( \langle u, v \rangle = \overline{\langle v, u \rangle} \)
- Positivity: \( \langle u, u \rangle \ge 0 \) and equals zero only if \( u \) is the zero vector.
Orthogonality
**Orthogonality** refers to the concept of vectors being perpendicular to each other, which is central in Hilbert space theory. Two vectors \( u \) and \( v \) are orthogonal if their inner product is zero, i.e., \( \langle u, v \rangle = 0 \).
In subspaces:
In subspaces:
- If every vector \( u \) in subspace \( V^{\perp} \) (the orthogonal complement) is orthogonal to every vector in \( V \), then \( V^{\perp} \) contains all vectors in \( \mathcal{H} \) orthogonal to \( V \).
- This concept helps decompose spaces into simpler, independent elements, enhancing the ability to analyze complex structures within Hilbert spaces.
Convergence
**Convergence** in a Hilbert space involves a sequence of vectors approaching a specific vector in terms of distance. This concept is vital for understanding limits and continuity within these spaces.
A sequence \( \{ x_n \} \) is said to converge to \( x \) if for any small positive value \( \epsilon \), there exists an integer \( N \) such that for all \( n > N \), the distance \( \| x_n - x \| < \epsilon \).
This means:
A sequence \( \{ x_n \} \) is said to converge to \( x \) if for any small positive value \( \epsilon \), there exists an integer \( N \) such that for all \( n > N \), the distance \( \| x_n - x \| < \epsilon \).
This means:
- As \( n \) becomes larger, \( x_n \) gets arbitrarily close to \( x \).
- Convergence ensures stability and predictability of sequences, allowing us to effectively work within subspaces like \( V \).
Other exercises in this chapter
Problem 5
Let \(\ell_{0}\) be the subset of \(\ell^{2}\) consisting of sequences with only finutely many terms different from zero. Show that \(\ell_{0}\) is a vector sub
View solution Problem 7
In the Hulbert space \(L^{2}([-1,1])\) let \(\left.\mid f_{n}(x)\right\\}\) be the sequence of functions \(1, x, x^{2}, \ldots, f_{n}(x)=x^{n} \ldots .\) (a) Ap
View solution Problem 11
Which of the following is a vector subspace of \(\ell^{2}\), and which are closed? In each case find the space of vectors orthogonal to the set. (a) \(V_{N}=\le
View solution Problem 13
If \(A: \mathcal{H} \rightarrow \mathcal{H}\) is an operator such that \(A u \perp u\) for all \(u \in \mathcal{H}\), show that \(A=0\).
View solution