Question

Show that \(\ell_{4}\) is not isomorphic to a subspace of \(\ell_{2}\). Hint: Show that \(\ell_{4}\) is not of cotype 2 by considering the standard unit vectors.

Short answer

\(\ell_{4}\) is not isomorphic to any subspace of \(\ell_{2}\) because \(\ell_{4}\) is not of cotype 2, while \(\ell_{2}\) is.

Step-by-step solution

- Understand \ell_{2} and \ell_{4}

\(\ell_{2}\) is the space of all sequences \((x_n)\) such that \(\sum_{n=1}^{\infty} |x_n|^2 < \infty\). \(\ell_{4}\) is the space of all sequences \((y_n)\) such that \(\sum_{n=1}^{\infty} |y_n|^4 < \infty\).

- Standard Unit Vectors

For both \(\ell_{2}\) and \(\ell_{4}\), consider the standard unit vectors denoted by \(e_n\), where \(e_n\) is the sequence with 1 in the n-th position and 0 elsewhere.

- Cotype Definition

A Banach space \(X\) is of cotype 2 if there exists a constant \(C\) such that for any finite sequence \(\{x_i\}\) in \(X\), the following inequality holds: \[\left( \sum_i \|x_i\|^2 \right)^{1/2} \leq C \cdot \left( \mathbb E \left\| \sum_i r_i x_i \right\|^2 \right)^{1/2},\]where \(r_i\) are Rademacher functions.

- Sequence Norms in \ell_{2}

In the space \(\ell_{2}\), the norm for the sequence \((e_n)\) is: \[\|e_n\|_2 = \left( \sum_{i=1}^{\infty} |e_{n, i}|^2 \right)^{1/2} = 1.\]Thus, given any sequence of unit vectors in \(\ell_{2}\), the inequality for cotype 2 would hold with appropriate constants.

- Sequence Norms in \ell_{4}

In the space \(\ell_{4}\), the norm for the sequence \((e_n)\) is: \[\|e_n\|_4 = \left( \sum_{i=1}^{\infty} |e_{n, i}|^4 \right)^{1/4} = 1.\]However, the summation on the right side of the cotype inequality does not necessarily converge.

- Checking the Left Side of the Inequality

Checking the left side of the inequality leads to \[\left( \sum_{i=1}^m \|e_i\|_4^2 \right)^{1/2} = \left( \sum_{i=1}^m 1^2 \right)^{1/2} = \sqrt{m}\],which is different from the cotype inequality since in \(\ell_{2}\) spaces similarity would result in 1.

- Concluding Non-Isomorphism

Since \(\ell_{4}\) is not of cotype 2 due to the unbounded summation in the cotype inequality, it reveals that \(\ell_{4}\) cannot be isomorphic to any subspace of a Banach space of cotype 2, like \(\ell_{2}\). Therefore, \(\ell_{4}\) is not isomorphic to any subspace of \(\ell_{2}\).

Key concepts

Banach Spaces

In mathematics, a Banach space is a vector space with a norm, in which every Cauchy sequence converges within the space. Banach spaces are a cornerstone in functional analysis and are named after the Polish mathematician Stefan Banach.
Banach spaces help us generalize key properties of finite-dimensional vector spaces to infinite-dimensional contexts. This has pivotal applications in various mathematical fields, including differential equations and quantum mechanics.
Some common examples of Banach spaces include spaces of continuous functions, denoted as \(C([a, b])\), spaces of sequences like \details \ssubjects \donethe topic \at once... 😮 endofvideo is comfort... G'day friends 😃 We'll be talking today.

Cotype

Cotype is a measure of how closely a Banach space resembles a Hilbert space in terms of the distribution of its elements. Specifically, a Banach space \(X\) is said to have cotype 2 if there exists a constant \(C\) such that for any finite sequence \(\{x_i\}\) in \(X\), the following inequality holds: \[ \left( \sum_i \|x_i\|^2 \right)^{1/2} \leq C \cdot \left( \mathbb{E} \left\| \sum_i r_i x_i \right\|^2 \right)^{1/2}, \] where \( r_i \) are Rademacher functions.
Rademacher functions are a sequence of independent random variables taking the values \( \pm \). These functions are often employed in functional analysis and probability theory.
The importance of cotype lies in differentiating the structures of various Banach spaces. For instance, the \({l_2}\) space has cotype 2, \(while {\ell_4}\) does not, as demonstrated in our exercise.

Unit Vectors

Unit vectors, denoted as \({e_n}\), are vital in studying the structure of spaces like \({\backslash ell_4}\) and \({\backslash ell_2}\). A unit vector \({e_n}\) in sequence space is defined as a sequence with 1 in the nth position and 0 elsewhere.
For \({\backslash ell_2}\), the norm of the unit vector \(e_n\) is:\[ \| e_n \|_2 = \( \sum_{i=1}^{\backslash infty} |e_{n, i}|^2 \)^{1/2} = 1. \] This illustrates that any sequence of unit vectors in \({\backslash ell_2}\) will satisfy the cotype 2 inequality with appropriate constants.
However, in \({\backslash ell_4}\), the situation is different. The norm of the unit vector \(e_n\) is:\[ \| e_n \|_4 = \( \sum_{i=1}^{\backslash infty} |e_{n, i}|^4 \)^{1/4} = 1. \] Nonetheless, the summation on the right side of the cotype inequality may not converge, showing that \({\backslash ell_4}\) is not of cotype 2. This establishes significant differences between \({\backslash ell_4}\) and \({\backslash ell_2}\) spaces.