Question

Show that there is no \(T \in \mathcal{B}\left(\ell_{2}, \ell_{1}\right)\) such that \(T\) is an onto map. Hint: \(T^{*}\) would be an isomorphism of \(\ell_{\infty}\) into \(\ell_{2}\), which is impossible since \(\ell_{\infty}\) is nonseparable and \(\ell_{2}\) is separable.

Short answer

There is no onto map \(T \in \mathcal{B}(\ell_2, \ell_1)\) because the adjoint would map \(\ell_\infty\) onto \(\ell_2\), which is impossible.

Step-by-step solution

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Explain the step in detail

Title

Explain the step in good detail

- Understand the Given Spaces

The problem involves the Banach spaces \(\ell_2\) and \(\ell_1\). The space \(\ell_2\) consists of all square-summable sequences, while \(\ell_1\) consists of all absolutely summable sequences.

- Consider the Dual Spaces

Given the hint about \(T^*\), recall that the dual space of \(\ell_2\) is itself \(\ell_2\), and the dual space of \(\ell_1\) is \(\ell_\infty\), the space of bounded sequences.

- Analyze the Implication if \(T\) is Onto

If \(T \in \mathcal{B}(\ell_2, \ell_1)\) is an onto map, then for every element \(y \in \ell_1\), there exists an element \(x \in \ell_2\) such that \(T(x) = y\). This means that \(T^*\), the adjoint operator, maps \(\ell_\infty\) onto \(\ell_2\).

- Apply the Hint and Conclusion

Since \(T^*\) would be an isomorphism of \(\ell_\infty\) into \(\ell_2\) and \(\ell_\infty\) is nonseparable while \(\ell_2\) is separable, such an isomorphism is impossible. Hence, there can be no onto map \(T \in \mathcal{B}(\ell_2, \ell_1)\).

Key concepts

Banach Spaces

Banach spaces are a fundamental concept in functional analysis. A Banach space is a vector space complete with respect to a norm, meaning every Cauchy sequence in this space converges to a limit within the space itself. The spaces \( \ell_2 \) and \( \ell_1 \) involved in our exercise are both Banach spaces.

Specifically:

• \( \ell_2 \) consists of all sequences whose squares are summable, making it a space of square-summable sequences.


• \( \ell_1 \), on the other hand, consists of all sequences whose absolute values are summable, making it a space of absolutely summable sequences.

In the given problem, no operator \( T \) in \( \mathcal{B}(\ell_2, \ell_1) \) (which denotes bounded linear operators between \( \ell_2 \) and \( \ell_1 \)) can be “onto,” meaning it cannot map \( \ell_2 \) completely onto \( \ell_1 \). Understanding these spaces helps in grasping the essence of the problem.

Dual Spaces

Dual spaces play a crucial role in functional analysis, particularly when analyzing mappings and operators. A dual space of a given vector space is the set of all linear functionals on that space. Importantly:

• The dual space of \( \ell_2 \) is itself \( \ell_2 \). This implies any linear functional on \( \ell_2 \) can be represented as a sequence within \( \ell_2 \).


• The dual space of \( \ell_1 \) is \( \ell_\infty \), the space of all bounded sequences. This indicates every functional on \( \ell_1 \) can be identified with a bounded sequence.

In the context of our solution, the adjoint operator \( T^* \) maps from \( \ell_\infty \) to \( \ell_2 \). If \( T \) were onto, \( T^* \) would effectively be an isomorphism from \( \ell_\infty \) into \( \ell_2 \), which is deemed impossible due to the nature of the spaces involved.

Separability

Separability is a property of topological spaces, including Banach spaces, where a space is said to be separable if it contains a countable, dense subset. This concept is essential in understanding why \( T \) cannot be onto in our exercise.

Key points include:

• \( \ell_2 \) is separable. This means there exists a countable set such that every element in \( \ell_2 \) can be approximated arbitrarily well by elements from this countable set.


• \( \ell_\infty \), the dual space of \( \ell_1 \), is nonseparable, indicating it does not contain a countable dense subset. Thereby, it has a significantly different structure than \( \ell_2 \).

The hint in our problem highlights this contrast: if \( T \) were onto, \( T^* \) would map \( \ell_\infty \) onto \( \ell_2 \), meaning a nonseparable space would map onto a separable one—an impossibility given their structural differences.