Question

Show that if \(T\) is a one-to-one operator from \(\ell_{1}\) into \(\ell_{2}\), then \(T\left(\ell_{1}\right)\) is not closed in \(\ell_{2}\)

Short answer

The image of \(\ell_1\) under a one-to-one operator T into \(\ell_2\) is not closed due to the discrepancy between the density properties of \(\ell_1\) and the closure properties in \(\ell_2\).

Step-by-step solution

- Understand the Scenario

We need to show that if there is a linear operator T that is one-to-one from the space \(\ell_1\) into the space \(\ell_2\), then the image of \(\ell_1\) under T is not closed in \(\ell_2\).

- Recall Definitions

Recall that \(\ell_1\) is the space of absolutely summable sequences, and \(\ell_2\) is the space of square summable sequences. A set in a metric space is closed if it contains all its limit points.

- Consider T as an Operator

Since T is a linear one-to-one operator, it maps \(\ell_1\) sequences to \(\ell_2\) sequences. Each sequence in \(\ell_1\) will have a unique image in \(\ell_2\).

- Analyze Closure in \(\ell_2\)

We need to show that there exists a sequence in \(\ell_2\) that is the limit of a sequence of images of \(\ell_1\) sequences, but this limit is not itself an image of a sequence in \(\ell_1\).

- Observe Density

It is known that \(\ell_1\) is dense in \(\text{c}_0\), the space of sequences converging to 0. Since \(\text{c}_0\) is a strict subset of \(\ell_2\), we use this density property.

- Construct a Contradiction

Assume \(T(\ell_1)\) is closed. Then, the image of \(\ell_1\) under T would form a closed subspace in \(\ell_2\). Considering the density argument, there would be sequences in \(\text{c}_0\) that are limits of \(\ell_1\) images. However, these limits do not necessarily lie in \(\ell_1\).

- Conclusion

Thus, the presumed closure means there’s an inherent contradiction due to the density property and the nature of \(\ell_2\). Hence, \(T(\ell_1)\) cannot be closed in \(\ell_2\).

Key concepts

Linear Operators

Linear operators are fundamental in understanding transformations between different types of spaces. A linear operator, such as our operator \( T \), satisfies the properties of additivity and homogeneity. This means for any sequences \( x, y \in \ell_{1} \) and scalar \( a \), the operator satisfies \( T(x + y) = T(x) + T(y) \) and \( T(ax) = aT(x) \).
In our problem, \(T\) is one-to-one, which ensures that different sequences in \(\ell_{1}\) map to different sequences in \(\ell_{2}\). This injectiveness is crucial in arguing that we cannot have certain limits of sequences in \(\ell_{2}\) being part of the image of \(\ell_{1}\) under \(T\).
Understanding this property helps us later when proving that \(T(\ell_{1})\) is not closed in \(\ell_{2}\).

Closed Sets

A set is closed if it contains all its limit points. In simple terms, if you have a sequence of points within this set that converges, the limit of that sequence must also be within the set.
In our exercise, we need to prove that \(T(\ell_{1})\), the image of \(\ell_{1}\) under \(T\), does not form a closed set in \(\ell_{2}\).
This involves demonstrating the existence of a sequence in \(\ell_{2}\) that is the limit of a sequence of images in \(\ell_{1}\) but the limit itself is not an image of any sequence in \(\ell_{1}\). This contradicts the closure property.

Space of Sequences

The spaces \(\ell_{1}\) and \(\ell_{2}\) are examples of sequence spaces.
\(\ell_{1}\) consists of sequences whose absolute values' series is summable, making it the space of absolutely summable sequences.
\(\ell_{2}\) consists of sequences whose squares' series is summable, i.e., the space of square summable sequences.
These sequence spaces are fundamentally important as they provide contexts where different types of convergence and limits behave differently. In particular, the relationship between these spaces (how one maps to the other through an operator) lays the groundwork for understanding higher mathematical constructs such as closure and density.

Metric Spaces

At its core, a metric space is a set where a notion of distance between elements is defined. Both \(\ell_{1}\) and \(\ell_{2}\) are metric spaces with specific metrics:
For \(\ell_{1}\), the metric is \(d(x, y) = \sum_{i=1}^{\infty} |x_{i} - y_{i}|\).
For \(\ell_{2}\), the metric is \(d(x, y) = \left( \sum_{i=1}^{\infty} |x_{i} - y_{i}|^{2} \right)^{1/2}\).
These metrics define the topology of the spaces, guiding how convergence and limits work. In our example, understanding these distances aids in comprehending why the image of \(\ell_{1}\) sequences under \(T\) does not cover all possible sequences in \(\ell_{2}\), particularly in terms of closure.

Density

Density is a concept describing how a subset fills up a space. A subset is dense in a space if every point in the space is a limit of some points from the subset.
In the given exercise, \(\ell_{1}\) is dense in \(c_{0}\), the space of sequences converging to zero. Since \(c_{0}\) is a strict subset of \(\ell_{2}\), any sequence in \(\ell_{2}\) can be approximated arbitrarily closely by sequences from \(\ell_{1}\).
This density property helps illustrate why the image of \(\ell_{1}\) under \(T\) cannot be closed in \(\ell_{2}\). If it were closed, the density argument shows that the sequences in \(T(\ell_{1})\) would absurdly cover more than just \(\ell_{1}\), leading to a contradiction.