Question

Let \(X\) be a normed space with two norms \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) such that \(X\) in both of them is a complete space. Assume that \(\|\cdot\|_{1}\) is not equivalent to \(\|\cdot\|_{2}\). Let \(I_{1}\) be the identity map from \(\left(X,\|\cdot\|_{1}\right)\) onto \(\left(X,\|\cdot\|_{2}\right)\) and \(I_{2}\) be the identity map from \(\left(X,\|\cdot\|_{2}\right)\) onto \(\left(X,\|\cdot\|_{1}\right) .\) Show that neither \(I_{1}\) nor \(I_{2}\) are continuous. Hint: The Banach open mapping theorem.

Short answer

Neither \(I_{1}\) nor \(I_{2}\) are continuous because the norms are not equivalent.

Step-by-step solution

- Understand the given norms and identity maps

We have a normed space \(X\) with two norms: \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\). It is given that \(X\) is complete under both norms, which means it is a Banach space in both norms. The norms are not equivalent, implying that there is no constant \(C > 0\) such that \(C^{-1}\|x\|_{1} \leq \|x\|_{2} \leq C\|x\|_{1}\) for all \(x \in X\).

- Define identity maps

Define \(I_{1}\) as the identity map from \((X, \|\cdot\|_1)\) to \((X, \|\cdot\|_2)\), i.e., \(I_{1}(x) = x\) for all \(x \in X\). Similarly, \(I_{2}\) is the identity map from \((X, \|\cdot\|_2)\) to \((X, \|\cdot\|_1)\), i.e., \(I_{2}(x) = x\) for all \(x \in X\).

- Recall the Open Mapping Theorem

The Banach Open Mapping Theorem states that if \(T\) is a bounded (continuous) linear operator between Banach spaces and \(T\) is surjective, then \(T\) is an open map. This means that if \(T\) maps open sets to open sets, then \(T\) must be continuous.

- Apply the theorem to our identity maps

If \(I_{1}\) were continuous, then by the Banach Open Mapping Theorem, \(I_{1}\) would map open sets in \((X, \|\cdot\|_1)\) to open sets in \((X, \|\cdot\|_2)\). Since the norms are not equivalent, there exist sequences in \(X\) that converge under one norm but not under the other. This violates the open mapping condition.

- Conclude non-continuity for \(I_{1}\)

Since the norms are not equivalent, \(I_{1}\) cannot satisfy the conditions of the Open Mapping Theorem and, therefore, cannot be continuous.

- Repeat the reasoning for \(I_{2}\)

Similarly, applying the Open Mapping Theorem to \(I_{2}\) shows that if it were continuous, it would map open sets to open sets. Again, due to the non-equivalence of the norms, \(I_{2}\) fails to satisfy this condition. Thus, \(I_{2}\) is also not continuous.

- Final conclusion

Neither \(I_{1}\) nor \(I_{2}\) are continuous maps because the norms \(\|\cdot\|_1\) and \(\|\cdot\|_2\) are not equivalent, violating the necessary conditions presented by the Open Mapping Theorem.

Key concepts

Normed Spaces

In mathematical analysis, a normed space is a vector space furnished with a function called a norm. This norm assigns a strictly positive length or size to each vector in the space, except for the zero vector, which is given a norm of zero. Normed spaces are essential because they offer a structure that can measure the size of elements in a vector space and their distances.

A few key points about normed spaces:

• A normed space is typically denoted by \( (X, \|\bullet\|) \), where \( X \) is the vector space and \( \|\bullet\| \) is the norm.
• The norm must satisfy certain properties: positivity, scalar multiplication, the triangle inequality, and definiteness (meaning if \( \|x\| = 0 \, then \ x = 0 \)).
• When a normed space is complete with respect to its norm (all Cauchy sequences converge), it is called a Banach space.

Understanding normed spaces forms the foundation for more advanced topics such as the Open Mapping Theorem and concepts of continuity in functional analysis.

Open Mapping Theorem

The Open Mapping Theorem is a fundamental result in functional analysis. It emphasizes the relationship between continuous linear operators and open sets. Specifically, the theorem states that: \( If T is a bounded (continuous) linear operator between two Banach spaces and T is surjective (onto), then T is an open mapping. \) This means that T maps open sets in the domain to open sets in the codomain.

Here are some points to elaborate on:

• For the operator to be open, for any open set in the input space, the image under T must be an open set in the output space.
• The importance of surjectivity: An operator must cover the entire codomain space for the theorem to apply.
• The implication for continuity: If the theorem’s conditions are met, then T is continuous. Conversely, if T is not continuous, it can't be open under these conditions.

In the provided exercise, understanding this theorem helps deduce why the identity maps \(I_1\) and \(I_2\) are not continuous between spaces with non-equivalent norms.

Continuity

Continuity in the context of normed spaces is concerned with whether small changes in the input of a map lead to small changes in the output. A map or function between normed spaces is said to be continuous if, roughly speaking, whenever a sequence converges, its image under the function also converges.

A few key notes on continuity:

• A function \( f \: X \to Y \) between normed spaces is continuous if for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \| x \ - \ y \| < \ \delta \), we have \( \| f(x) - f(y) \| < \epsilon \).
• The concept of a continuous map plays a crucial role in functional analysis since it ensures the robustness of mathematical structures and transformations in various applications.
• In our particular problem, the identity maps \(I_1\) and \(I_2\) between different normed spaces are analyzed for continuity. Due to the non-equivalence of \( \| \bullet \|_1 \) and \( \| \bullet \|_2 \, these maps fail to be continuous.

Understanding continuity within normed spaces provides a key conceptual foundation for exploring more complex mathematical relationships and transformations.