Problem 31
Question
Let \(V\) be a Banach space and \(W\) te a vector subspace of \(V .\) Define its closture \(\bar{W}\) to be the union of \(W\) and all hmuts of Cauchy sequences of elements of \(W\). Show that \(\bar{W}\) is a closed vector subspace of \(V\) in the sense that the limit points of all Cauchy scquences in \(\bar{W}\) lie in \(\bar{W}\) (note that the Cauchy sequences may include the newly added limit points of \(W\) ).
Step-by-Step Solution
Verified Answer
The closure of the vector subspace \(W\) of the Banach space \(V\), denoted as \(\bar{W}\), is a closed vector subspace of \(V\). This follows from the fact that \(\bar{W}\) satisfies the properties of a subspace, and it contains all its limit points, thus making it closed.
1Step 1: Proving Subspace
To show that \(\bar{W}\) is a subspace, confirm the three properties of a vector space. 1) The set is not empty - the subspace includes at least the zero vector. 2) The sum of any two vectors in the set remains within the set. 3) The scalar multiple of any vector in the set remains within the set.
2Step 2: Dealing with Scalability
Let \(x\) be in \(\bar{W}\) and \(c\) be a scalar. There exists a sequence in \(W\) that converges to \(x\) (we can consider the sequence itself if \(x\) is in \(W\))). Then \(c*x_n\) is in \(W\), and as \(W\) is a subspace, \(c*x_n\) is Cauchy in \(W\) (because \(x_n\) converges and hence is Cauchy), and \(c*x_n\) converges to \(c*x\) in \(V\), because multiplication by a scalar is continuous in a normed vector space. So \(c*x\) is also in \(\bar{W}\).
3Step 3: Dealing with Vector Addition
Let \(x\) and \(y\) be in \(\bar{W}\). From the definition of closure, there exist sequences \(x_n\) in \(W\) and \(y_n\) in \(W\) such that \(x_n\) converges to \(x\) and \(y_n\) converges to \(y\). Then \(x_n + y_n\) is in \(W\) (as \(W\) is a subspace), and is Cauchy (as the sum of two Cauchy sequences is Cauchy), and converges to \(x+y\) (as vector addition is continuous in a normed vector space). Hence, \(x+y\) is in \(\bar{W}\).
4Step 4: Showing Closedness
To show \(\bar{W}\) is closed, consider a Cauchy sequence \((z_n)\) in \(\bar{W}\). Each \(z_n\) is the limit of a Cauchy sequence in \(W\), so we can create a new sequence from \(W\) which is still Cauchy and has the same limit. Since \(W\) is complete (as it is a subspace of a Banach space), the limit is contained in \(W\). Therefore, \(\bar{W}\) contains all its limit points and is closed.
Key Concepts
Vector SubspaceCauchy SequenceClosureNormed Vector Space
Vector Subspace
A vector subspace, often simply called a subspace, is a subset of a vector space. This subset still follows the rules of vector addition and scalar multiplication, maintaining its structure as a vector space. To confirm a set is a subspace, it must:
- Include the zero vector; meaning it cannot be empty and the vector space's origin must exist within it.
- Be closed under addition; if you take any two elements (vectors) from this subspace, their sum should also be in the subspace.
- Be closed under scalar multiplication; this means multiplying any vector in the subspace by a scalar should still result in a vector that is in the subspace.
Cauchy Sequence
A Cauchy sequence is a concept rooted in analysis, particularly useful in understanding convergence in metric and normed spaces. A sequence is termed "Cauchy" if, as you progress along the sequence, the terms get arbitrarily close to each other. More formally, for every positive number \( \epsilon \), no matter how small, there exists an index \( N \) such that for all indices \( m, n \geq N \), the distance \( |a_m - a_n| \) is less than \( \epsilon \).
This essentially means the terms of the sequence bunch up together as you move towards infinity, suggesting that they approach some limit. However, the limit itself need not be within the space in which the sequence lies, unless the space is complete. In a complete space, such as a Banach space, every Cauchy sequence has a limit within the space.
This essentially means the terms of the sequence bunch up together as you move towards infinity, suggesting that they approach some limit. However, the limit itself need not be within the space in which the sequence lies, unless the space is complete. In a complete space, such as a Banach space, every Cauchy sequence has a limit within the space.
Closure
The concept of closure extends a subset of a vector space by incorporating all limit points of sequences. For a vector subspace \( W \), its closure \( \bar{W} \) includes all elements of \( W \) as well as the limits of all Cauchy sequences from \( W \).
Here's how it works:
Here's how it works:
- Every element of \( W \) is naturally in \( \bar{W} \).
- Any limit point, meaning a point approached by terms of a Cauchy sequence from \( W \), is also included in \( \bar{W} \).
Normed Vector Space
A normed vector space is essentially a vector space equipped with a function called a 'norm'. This norm assigns a non-negative length or size to each vector in the space, providing a formal way to measure 'distance' and 'size' in the context of abstract vector spaces. The properties defining a norm include:
- Positivity: For any vector \( x \), the norm \( \|x\| \geq 0 \) and \( \|x\| = 0 \) if and only if \( x \) is the zero vector.
- Scalar Multiplication: For any scalar \( c \) and vector \( x \), \( \| c \cdot x \| = |c| \cdot \|x\| \).
- Triangle Inequality: For any two vectors \( x \) and \( y \), \( \| x + y \| \leq \|x\| + \|y\| \).
Other exercises in this chapter
Problem 29
Show that if \(x_{n} \rightarrow x\) in a normed vector space then $$ \frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \rightarrow x $$
View solution Problem 30
Show that if \(x_{n}\) is a sequence in anormed vector space \(V\) such that every subsequence has a subsequence comergent to \(x\), then \(x_{n} \rightarrow x\
View solution Problem 32
Show that cvery space \(F(S)\) is complete with respect to the supremum norm of Example 10.26. Hence show that the vector space \(\ell_{\infty}\) of bounded inf
View solution Problem 33
Show that the set \(V^{\prime}\) consisting of bounded linear functionals on a Banach space \(V\) is a normed vector space with respect to the norm $$ \|\varphi
View solution