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Chapter 7: Problem 18

$$ \text { Show that } \mathcal{K}\left(\ell_{2}\right) \text { contains an isometric copy of } \ell_{2} \text { . } $$

Short Answer

Expert verified
An isometric copy of \(\text{ℓ}_2\) can be embedded in \(\text{𝓚}(\text{ℓ}_2)\) via diagonal compact operators.

Step by step solution

01

- Understand the definitions

Define \(\text{𝓚}(\text{ℓ}_2)\) as the space of compact operators on \(\text{ℓ}_2\), and \(\text{ℓ}_2\) as the space of square-summable sequences.
02

- Identify target subspace

We need to show there is an isometric copy of \(\text{ℓ}_2\) inside \(\text{𝓚}(\text{ℓ}_2)\). This means finding a subspace in \(\text{𝓚}(\text{ℓ}_2)\) which is isometric to \(\text{ℓ}_2\).
03

- Construct the embedding

Consider the sequence of orthogonal projections \(P_n\) on \(\text{ℓ}_2\). Each \(P_n\) is a compact operator and \(P_nP_m = 0\) for \(n eq m\).
04

- Define the isometric embedding

Define the map \(T: \text{ℓ}_2 \rightarrow \text{𝓚}(\text{ℓ}_2)\) by \(T((a_n)) = \text{diag}(a_1, a_2, \text{...})\). Each \(T((a_n)) \) is compact as it is represented by a diagonal matrix with entries going to zero.
05

- Verify the isometric property

Verify that the map \(T\) preserves norms. For any sequence \((a_n) \in \text{ℓ}_2\), the operator norm of \(T((a_n))\) equals \(\text{sup}_n |a_n|\), which is the same as \(\text{ℓ}_2\) norm.
06

- Conclusion

Since \(T((a_n))\) maps \(\text{ℓ}_2\) into \(\text{𝓚}(\text{ℓ}_2)\) isometrically, \(\text{𝓚}(\text{ℓ}_2)\) contains an isometric copy of \(\text{ℓ}_2\).

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Key Concepts

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

Compact Operators
In the realm of functional analysis, a compact operator is immensely important. A compact operator on a Hilbert space, such as \(\text{ℓ}_2\), can be thought of similarly to a finite-dimensional matrix in infinite-dimensional space. Compact operators can turn a bounded sequence into a sequence that contains a convergent subsequence. This property is akin to what we see in finite-dimensional spaces and is crucial for understanding various function spaces and their properties.
Compact operators can be represented in an infinite matrix form, where the matrix elements approach zero as you move away from the diagonal. This means that they can approximate finite-rank operators very closely, making them a handy tool for analysis. In many ways, compact operators help us transfer our intuition from finite-dimensional vector spaces to infinite-dimensional ones. When dealing with compact operators, it's essential to remember that they exhibit properties that simplify a lot of complex analyses, thanks to their connection to finite-dimensional analogs. This simplification is precisely why we find compact operators playing a key role in proofs and solutions in functional spaces.
Space of Square-Summable Sequences
The space of square-summable sequences, denoted as \(\text{ℓ}_2\), is a key concept in functional analysis. Each element in \(\text{ℓ}_2\) is a sequence of numbers that, when squared and summed, converges to a finite value. Formally, a sequence \((a_n)\) belongs to \(\text{ℓ}_2\) if \[\sum_{n=1}^{\text{∞}} |a_n|^2 < \text{∞}.\]
Understanding \(\text{ℓ}_2\) is crucial for working with Hilbert spaces since \(\text{ℓ}_2\) itself is an example of a Hilbert space. Hilbert spaces have an inner product, which allows notions of angles and orthogonality to be extended from finite to infinite-dimensional spaces. This is why tools from basic geometry still apply within \(\text{ℓ}_2\).
What's remarkable about \(\text{ℓ}_2\) is its completeness. This means that any Cauchy sequence (a sequence where elements get arbitrarily close to each other) in \(\text{ℓ}_2\) will always converge to a limit within \(\text{ℓ}_2\). This property is vital for ensuring that limits and approximations within the space make sense and remain within the space.
Orthogonal Projections
Orthogonal projections are tools within vector spaces that help us break down complex problems into more manageable parts. An orthogonal projection in a Hilbert space like \(\text{ℓ}_2\) maps any vector in the space onto a subspace. The resulting vector is the closest vector within the subspace to the original vector, minimizing the distance between them.
Consider the sequence of orthogonal projections \(P_n\) on \(\text{ℓ}_2\). Each \(P_n\) is a compact operator, which introduces simplicity into complex realms. For indices \(n eq m\), \(P_n P_m = 0\), showing orthogonality between the projections.Understanding orthogonal projections is essential when dealing with isometric embeddings. When constructing an isometric copy of \(\text{ℓ}_2\) within \(\text{𝓚}(\text{ℓ}_2)\), one approach is to embed the sequences of numbers into a space of operators, while keeping properties such as orthogonality and norms intact. This translates to ensuring that the inner products and distances remain preserved in the embedding process.

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Most popular questions from this chapter

Let \(T\) be a diagonal operator on \(\ell_{2}\) associated with a bounded sequence of complex numbers \(\left\\{c_{i}\right\\} ;\) that is, \(T\left(\left(x_{i}\right)\right)=\left(c_{i} x_{i}\right) .\) Find the set of all eigenvalues of \(T\) and the spectrum of \(T\).

Show that there is a continuous map from \(B_{\ell_{2}}\) into \(B_{\ell_{2}}\) that has no fixed point. This means that the requirement of finite dimension cannot in general be dropped in Brouwer's theorem (see [Dug1]).

Let \(f(x, y)\) be a continuous bounded function on an open set \(\Omega\) in \(\mathbf{R}^{2}\) that is uniformly Lipschitz in \(y\) on \(\Omega\) (there is a constant \(K\) such that \(\left|f\left(x, y_{1}\right)-f\left(x, y_{2}\right)\right| \leq K\left|y_{1}-y_{2}\right|\) for every \(\left.\left(x, y_{1}\right),\left(x, y_{2}\right) \in \Omega\right)\) Let \(\left(x_{0}, y_{0}\right) \in \Omega\). Show that there is \(\delta>0\) such that on \(\left[x_{0}-\delta, x_{0}+\delta\right]\) there is a unique continuously differentiable solution to \(\frac{d y}{d x}=f(x, y)\) with the initial condition \(y\left(x_{0}\right)=y_{0}\)

Let \(K\) be a compact set in the scalar field. Show that there is an operator \(T \in \mathcal{B}\left(\ell_{2}\right)\) such that \(\sigma(T)=K\).

Let \(H\) be a separable Hilbert space. An operator \(T \in \mathcal{B}(H)\) is called a Hilbert-Schmidt operator if there is an orthonormal basis \(\left\\{e_{i}\right\\}\) of \(H\) such that \(\sum\left\|T\left(e_{i}\right)\right\|^{2}<\infty .\) Show that if \(\left\\{f_{i}\right\\}\) is another orthonormal basis of \(H\), then \(\sum\left\|T\left(f_{i}\right)\right\|^{2}=\sum\left\|T\left(e_{i}\right)\right\|^{2}\) The number \(\|T\|_{H S}=\left(\sum\left\|T\left(e_{i}\right)\right\|^{2}\right)^{\frac{1}{2}}\) is called the Hilbert-Schmidt norm of \(T .\) Show that \(\|T\|_{H S} \geq\|T\|\).
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