Question

Show that (a) the unit operator is not compact, and that (b) the inverse of a compact operator cannot be bounded.

Short answer

(a) The unit operator is not compact because it cannot map a sequence with no convergent subsequence (e.g., a sequence of orthogonal vectors) to a sequence with a convergent subsequence. (b) The inverse of a compact operator cannot be bounded because it would contradict the definition of boundedness; for any sequence in the domain of the inverse operator, there are more than a finite number of elements in the sequence within the unit ball in Hilbert space, contradicting the assumption of boundedness.

Step-by-step solution

Understanding compact operators

Before proceeding with the proofs, it's crucial to understand what is meant by a compact operator. In functional analysis, an operator \(T\) is called compact if it maps any bounded sequence in its domain to a sequence in its range that has a convergent subsequence. In other words, for any sequence \(\{x_n\}\) in the domain of \(T\) such that \(\|x_n\|\leq M\) for all \(n\), \(T({x_n})\) has a convergent subsequence.

Proving that the unit operator is not compact

Consider the unit operator \(I\) in infinite-dimensional Hilbert space \(\mathcal{H}\). The unit operator is defined as \(I(x) = x\), for all \(x \in \mathcal{H}\). Take a sequence \(\{x_n\}\) consisting of the standard basis vectors in \(\mathcal{H}\). As each vector in the sequence is orthogonal to all the others, the sequence does not have any convergent subsequence. However, \(I(x_n) = x_n\), which means that the image of the sequence under the unit operator also does not have a convergent subsequence. Therefore, the unit operator is not a compact operator.

Understanding bounded operators

A bounded operator is an operator \(T\) for which there exists a constant \(C\) such that \(\|T(x)\|\leq C\|x\|\) for every \(x\) in the domain of \(T\). When \(C\) does not exist, the operator is said to be unbounded.

Proving that the inverse of a compact operator cannot be bounded

Suppose \(T: \mathcal{H} \rightarrow \mathcal{H}\) is a compact operator and \(T\) is also invertible. This means that for any \(y \in \mathcal{H}\), there exists \(x \in \mathcal{H}\) such that \(y=T(x)\). Now, let's assume that the inverse \(T^{-1}: \mathcal{H} \rightarrow \mathcal{H}\) is bounded. Then \(\|T^{-1}(y)\| \leq C\|y\|\) for some constant \(C\). Now take \(y_n=T(e_n)\), where \(\{e_n\}\) is any sequence in \(\mathcal{H}\). There are only finitely many \(y_n\) in the unit ball of \(\mathcal{H}\), contradicting the assumption that \(T^{-1}\) is bounded. Therefore, the inverse of a compact operator cannot be bounded.

Key concepts

Unit Operator

The unit operator, often symbolized as 'I', plays a foundational role in the study of linear operators on a Hilbert space. It acts as the identity map, taking each vector to itself. In simple terms, if you imagine a function that just hands back whatever you give to it, that's exactly what a unit operator does in a mathematical space.

Think of it as the equivalent of multiplying a number by one; no matter the number, the result is the same. So, when you apply the unit operator to any vector in the Hilbert space, it doesn't change. It might sound simple, but its presence is critical when talking about transformations and operations that can be done within this abstract mathematical universe.

In the context of our exercise, one key point is that in an infinite-dimensional Hilbert space, the unit operator is not compact. It means that even if you have a bunch of vectors that don't stray too far from the origin, when you apply the unit operator, you won't necessarily find a bunch that point in the same general direction - a property that would hint at compactness.

Bounded Operator

A bounded operator is a type of linear operator that ensures things don't get out of hand - mathematically speaking! When you apply a bounded operator to a vector, it guarantees that the length of the resulting vector isn't disproportionately larger than the original. There's a limit, a boundary, to how much it can stretch a vector. Another way to understand a bounded operator is by thinking about it as a function that doesn't allow outputs to go haywire as inputs are changed.

Imagine stretching a rubber band; there's only so far it can stretch before it snaps. A 'bounded' rubber band would be one that you could only stretch up to a certain length, no matter how much force you apply. In our solution, we discussed that a key aspect of bounded operators is the existence of a magic number 'C' (our snapping point for the rubber band) such that no matter the vector you apply the operator to, the length of the result won't exceed 'C' times the length of the original.

Hilbert Space

A Hilbert space is a concept in mathematics that's as beautiful as it is complex. It's a stage where vectors live and interact, following the rules of geometry and algebra, but these vectors aren't your everyday arrows in space - they can represent an infinite number of dimensions! It's a space where you can do all the things you're used to doing with vectors in two or three dimensions, like adding them together or finding a vector perpendicular to others, but these actions have far-reaching consequences in the realms of quantum mechanics and signal processing, to name a few.

In a Hilbert space, we can talk about infinite series of vectors - a whole symphony of components that can converge to form new vectors, stretching out in all sorts of directions but still dwelling within the confines of this meticulously orderly mathematical universe. Our exercise delves into the peculiar behavior of operators acting within a Hilbert space and reveals that their nature — being compact or not, bounded or unbounded — heavily influences the kinds of sequences of vectors we can expect to encounter.