Problem 7 - An Introduction to Abstract Analysis

Question

Let \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) be two norms defined on a linear space \(X\). Suppose there exist constants \(a>0\) and \(b>0\) such that $$ a\|x\|_{1} \leq\|x\|_{2} \leq b\|x\|_{1}, \quad \text { for all } x \text { in } X $$ Show that \(\left(X,\|\cdot\|_{1}\right)\) is complete if and only if \(\left(X,\|\cdot\|_{2}\right)\) is complete.

Step-by-Step Solution

Answer

Question: Prove that if a linear space X is complete with respect to one norm, \(\|\cdot\|_{1}\), then it is complete with respect to another norm, \(\|\cdot\|_{2}\), provided the norms are related such that \(a\|x\|_{1}\leq\|x\|_{2}\leq b\|x\|_{1}\) for some \(a, b > 0\) and all \(x \in X\). Answer: We have shown that if a sequence is Cauchy with respect to one norm, it is also Cauchy with respect to the other norm. If X is complete with respect to \(\|\cdot\|_{1}\), then any Cauchy sequence with respect to \(\|\cdot\|_{2}\) will also be Cauchy with respect to \(\|\cdot\|_{1}\) and hence converge in X. Similarly, if X is complete with respect to \(\|\cdot\|_{2}\), then any Cauchy sequence with respect to \(\|\cdot\|_{1}\) will also be Cauchy with respect to \(\|\cdot\|_{2}\) and hence converge in X. Therefore, X is complete with respect to both norms.

Step 1: Sequential compactness for Cauchy Sequences

Let \(\{x_n\}_n\) be a Cauchy sequence in \(\left(X,\|\cdot\|_{1}\right)\). We want to show that it is also a Cauchy sequence in \(\left(X,\|\cdot\|_{2}\right)\).

Step 2: Show Cauchy condition under \(\|\cdot\|_{2}\)

Let \(\epsilon > 0\). Since \(\{x_n\}_n\) is Cauchy with respect to \(\|\cdot\|_{1}\), there exists an \(N \in \mathbb{N}\) such that for all \(m, n > N\): $$ \|x_m - x_n\|_{1} < \frac{\epsilon}{b} $$ Now, to show Cauchy property under \(\|\cdot\|_{2}\) for each \(m, n > N\): $$ \|x_m - x_n\|_{2} \leq b \|x_m - x_n\|_{1} < b\frac{\epsilon}{b} = \epsilon $$ Similarly, if \(\{y_n\}_n\) is a Cauchy sequence in \(\left(X,\|\cdot\|_{2}\right)\), we get:

Step 3: Show Cauchy condition under \(\|\cdot\|_{1}\)

Let \(\epsilon > 0\). Since \(\{y_n\}_n\) is Cauchy with respect to \(\|\cdot\|_{2}\), there exists an \(N \in \mathbb{N}\) such that for all \(m, n > N\): $$ \|y_m - y_n\|_{2} < \frac{\epsilon}{a} $$ Now, to show Cauchy property under \(\|\cdot\|_{1}\) for each \(m, n > N\): $$ \|y_m - y_n\|_{1} \geq \frac{1}{a} \|y_m - y_n\|_{2} > \frac{1}{a}\frac{\epsilon}{a} = \epsilon $$

Step 4: Conclude Completeness Equivalence

From Steps 2 and 3, we conclude that a sequence is Cauchy with respect to one norm if and only if it is Cauchy with respect to the other norm. Therefore, if \(\left(X,\|\cdot\|_{1}\right)\) is complete, then any Cauchy sequence in \(\left(X,\|\cdot\|_{2}\right)\) will also be Cauchy in \(\left(X,\|\cdot\|_{1}\right)\) and hence converge. Similarly, if \(\left(X,\|\cdot\|_{2}\right)\) is complete, then any Cauchy sequence in \(\left(X,\|\cdot\|_{1}\right)\) will also be Cauchy in \(\left(X,\|\cdot\|_{2}\right)\) and hence converge. And finally, we have that \(\left(X,\|\cdot\|_{1}\right)\) is complete if and only if \(\left(X,\|\cdot\|_{2}\right)\) is complete.

Key Concepts

Cauchy Sequence

A Cauchy sequence is a fundamental concept in the study of analysis within normed linear spaces. Crucial to understanding completeness, a Cauchy sequence is defined as a sequence \(\{x_n\}_n\) such that for every positive number \(\epsilon>0\), there exists an integer \(N\) where for all indices \(m, n>N\), the distance between \(x_m\) and \(x_n\) (under the given norm) is less than \(\epsilon\). In simpler terms, the elements of the sequence get arbitrarily close to each other as the sequence progresses.

This concept indicates that a sequence is ready to converge; however, convergence is not guaranteed unless the space is complete. In the scope of the exercise, this notion is used to establish a relationship between two norms and their associated completeness.

Norm Equivalence

The idea behind norm equivalence is crucial when discussing multiple norms on the same space. Two norms \(\|\cdot\|_{1}\) and \(\|\cdot\|_{2}\) on a linear space \(X\) are said to be equivalent if there exist constants \(a, b>0\) such that for all vectors \(x\) in \(X\), \(a\|x\|_{1} \leq\|x\|_{2} \leq b\|x\|_{1}\). Norm equivalence implies that a sequence that is Cauchy with respect to one norm is also Cauchy with respect to the other. In our exercise, this concept showed us that the completeness of the space is preserved across equivalent norms, emphasizing the flexibility and interconnectedness of different norm structures in giving us the same 'big picture' about sequence behaviors in the space.

Sequential Compactness

The term sequential compactness defines a property of a space that translates to every sequence in that space having a convergent subsequence, whose limit is also within the space. It's a form of 'containment' for sequences under a given norm. In normed linear spaces, sequential compactness often coincides with 'completeness' plus 'total boundedness' (every sequence has a Cauchy subsequence), but this relationship can depend on specific properties of the space. Our exercise used the sequential compactness implicitly by dealing with Cauchy sequences under two norms, emphasizing the importance of the space's structure in determining the behavior of sequences.

Convergence of Sequences

Understanding convergence of sequences is critical when studying the analysis in normed spaces. A sequence \(\{x_n\}_n\) is said to converge to a limit \(x\) in a normed space \((X, \|\cdot\|)\) if, for every \(\epsilon>0\), there exists a positive integer \(N\) such that for all \(n>N\), we have \(\|x_n - x\| < \epsilon\). The limit point \(x\) is the 'destination' the sequence is approaching. In terms of the exercise, this pertains directly to the concept of completeness, which guarantees that every Cauchy sequence in the space will indeed converge—that is, it will have that 'destination' within the space itself, an assurance that is critical for various analytical tasks and proofs involving sequences.