Question

Consider the right shift \(R\) on \(\ell_{2}\) and the diagonal operator \(D\) associated with \(d_{i}=2^{-i}\). Define a weighted shift operator \(T\) on the complex space \(\ell_{2}\) by \(T=R \circ D .\) Show that \(T\) is a compact operator with spectral radius 0 and \(T\) is one-to-one. Thus, \(T\) has no eigenvalues and \(\sigma(T)=\\{0\\}\).

Short answer

The operator \(T\) is compact with spectral radius 0, injective, has no eigenvalues, and its spectrum is \{0\}.

Step-by-step solution

- Define the Right Shift Operator

The right shift operator, denoted by \(R\), acts on a sequence \((x_1, x_2, x_3, ...)\) in \(\ell_2\) as follows: \(R(x_1, x_2, x_3, ...) = (0, x_1, x_2, x_3, ...)\).

- Define the Diagonal Operator

The diagonal operator \(D\) is associated with the sequence \(d_i = 2^{-i}\). Acting on a sequence \((x_1, x_2, x_3, ...)\), \(D\) operates by: \(D(x_1, x_2, x_3, ...) = (2^{-1}x_1, 2^{-2}x_2, 2^{-3}x_3, ...)\).

- Define the Weighted Shift Operator

The weighted shift operator \(T\) is defined by the composition of \(R\) and \(D\), so \(T = R \circ D\). For a sequence \((x_1, x_2, x_3, ...)\) in \(\ell_2\), \(T(x_1, x_2, x_3, ...) = R(D(x_1, x_2, x_3, ...)) = R(2^{-1}x_1, 2^{-2}x_2, 2^{-3}x_3, ...) = (0, 2^{-1}x_1, 2^{-2}x_2, 2^{-3}x_3, ...)\).

- Show T is Compact

To show \(T\) is compact, we can show that \(T\) maps bounded sequences to sequences with norms approaching 0. Given \(T(x_1, x_2, x_3, ...) = (0, 2^{-1}x_1, 2^{-2}x_2, 2^{-3}x_3, ...))\), for large \(n\), \(|2^{-n}x_n|\) approaches 0 since \(2^{-n}\) decays to 0 as \(n\) increases. Thus, \(T\) is compact.

- Compute the Spectral Radius

The spectral radius \(\rho(T)\) is the largest magnitude of the eigenvalues of \(T\). Since \(T\) is compact and its operator norm tends to zero, \(\rho(T) = 0\).

- Show T is One-to-One

To show \(T\) is one-to-one, assume \(T(x_1, x_2, x_3, ...) = (0, 2^{-1}x_1, 2^{-2}x_2, 2^{-3}x_3, ...) = 0\). This implies \(2^{-n}x_n = 0\) for all \(n\), which means \(x_n = 0\). Hence, \(T\) is injective.

- Determine the Spectrum

Since \(T\) is compact, has spectral radius 0, and is injective (hence no eigenvalues), the spectrum of \(T\) must be \({0}\). Hence, \(\sigma(T) = \{0\}\).

Key concepts

Right Shift Operator

The right shift operator, denoted as \( R \), is a fundamental concept in functional analysis. It acts on an infinite sequence of elements in the space \( \ell_2 \). Given a sequence \( (x_1, x_2, x_3, ...) \), the right shift operator transforms it by shifting each element one position to the right and inserting a zero at the beginning. For example, applying \( R \) to \( (x_1, x_2, x_3, ...) \) results in the sequence \( (0, x_1, x_2, x_3, ...) \). This operator is powerful when constructing other operators such as weighted shift operators by adjusting and combining these sequences efficiently.

Diagonal Operator

A diagonal operator, commonly used in operator theory, acts by scaling each element of a sequence by a specified factor. For the sequence \( (x_1, x_2, x_3, ...) \) in the space \( \ell_2 \), the diagonal operator \( D \) associated with the sequence \( d_i = 2^{-i} \) transforms it as follows: \( D(x_1, x_2, x_3, ...) = (2^{-1}x_1, 2^{-2}x_2, 2^{-3}x_3, ...) \). This means each element is multiplied by \( 2^{-i} \), where \( i \) is its position index. This operator can be particularly useful in controlling the decay rate of the elements.

Weighted Shift Operator

The weighted shift operator, denoted as \( T \), combines the effects of both the right shift operator \( R \) and the diagonal operator \( D \). For a given sequence \( (x_1, x_2, x_3, ...) \) in \( \ell_2 \), the weighted shift operator is defined as \( T = R \circ D \). This means performing the diagonal operation first, followed by the right shift operation:

Let's see this in action for the sequence \( (x_1, x_2, x_3, ...) \):
- Apply \( D \): \( D(x_1, x_2, x_3, ...) = (2^{-1}x_1, 2^{-2}x_2, 2^{-3}x_3, ...) \)
- Apply \( R \) to this result: \( R(2^{-1}x_1, 2^{-2}x_2, 2^{-3}x_3, ...) = (0, 2^{-1}x_1, 2^{-2}x_2, 2^{-3}x_3, ...) \)
This combined operation creates a sequence where each element is scaled and then shifted to the right by one position.

Spectral Radius

The spectral radius of an operator \( T \) is a crucial concept in understanding its properties. It is defined as the largest magnitude of the eigenvalues of \( T \), denoted as \( \rho(T) \). For compact operators like \( T \), whose norm approaches zero for large sequences, the spectral radius also tends to zero. This property can be shown mathematically by analyzing how the weighted coefficients decay. In our context, since \( T \) solely scales and shifts the sequence elements, resulting in negligible contribution from each element over multiple applications, hence \( \rho(T) = 0 \).

Eigenvalues

Eigenvalues are essential in understanding the behavior of an operator. An eigenvalue \( \lambda \) of an operator \( T \) is a scalar such that there exists a non-zero vector \( x \) where \( T(x) = \lambda x \). For the weighted shift operator \( T \), since it scales and then shifts the sequence, making all shifted sequences tend to zero, there cannot be a non-zero sequence remaining unchanged under \( T \). This means \( T \) is injective, leading to the conclusion that \( T \) has no eigenvalues.
As \( T \) is compact, injective, and has a spectral radius of zero, its spectrum consists solely of zero. Thus, \( \sigma(T) = \{ 0 \} \).