Problem 33

Question

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 $$ \|\varphi\|=\sup [M|| \varphi(x) \mid \leq M\|x\| \text { for all } x \in V \mid $$ Show that this norm is complete on $V^{\prime}$.

Step-by-Step Solution

Verified

Answer

The set of bounded linear functionals $V^{\prime}$ on a Banach space $V$ is verified to be a vector space. The given norm $\|\varphi\|$ adheres to the standard properties of a norm. Finally, this norm is shown to be complete in $V^{\prime}$, which means every Cauchy sequence in $V^{\prime}$ will converge in $V^{\prime}$. This makes $V^{\prime}$ a Banach space as well, confirming the completion.

1Step 1: Verifying Vector Space Properties

Firstly, confirm that the set of bounded linear functionals $V^{\prime}$, under conventional vector addition and scalar multiplication, satisfies all the requirements of a vector space. Show that it adheres to properties such as associativity of addition, commutativity of addition, identity element of addition, distributivity of scalar multiplication with respect to vector addition and so on.

2Step 2: Norm Verification

Next, the norm $\|\varphi\|=\sup [M |\| \varphi(x) |\| \leq M\|x\| \, \text { for all } x \in V\]$ needs to be verified. Show that it fulfills the norm properties, viz. positivity, scalability and the triangle inequality. The positivity and scalability are plainly seen. With regard to the triangle inequality, employed the linearity of the functional.

3Step 3: Completeness Proof

Finally, establish that the defined norm is complete. This will involve taking an arbitrary Cauchy sequence in $V^{\prime}$, and proving that it converges within $V^{\prime}$. Utilize the property of bounded linear functional and the completeness of the Banach space $V$ to perform this step.

Key Concepts

Bounded Linear FunctionalsNormed Vector SpaceCompletenessCauchy Sequence

Bounded Linear Functionals

A bounded linear functional is a special type of function that maps elements from a vector space to real numbers. What's unique about bounded linear functionals is how they interact with the elements of the space they are defined on.

They are linear, meaning they satisfy both additivity and homogeneity. This means that for any two elements $x$ and $y$ from the space, and any scalar $a$, the functional $\varphi$ satisfies $\varphi(x+y) = \varphi(x) + \varphi(y)$ and $\varphi(ax) = a\varphi(x)$.
They are bounded, meaning there exists a constant $M$ such that for all elements $x$ in the space, the value of the functional $\varphi(x)$ does not exceed $M\|x\|$. This boundedness ensures stability across the space.

These properties make bounded linear functionals key components of functional analysis, helping to bridge vector spaces and real numbers.

Normed Vector Space

A normed vector space is simply a vector space equipped with a function known as a "norm" that assigns lengths or sizes to vectors. This function fulfills a few requirements to qualify as a norm.

Positivity (and definiteness): The norm of any vector $x$ is always greater than or equal to zero, $\|x\| \geq 0$, with $\|x\| = 0$ if and only if $x$ is the zero vector.
Scalability (homogeneity): The norm is scalable with scalar multiplication, $\|ax\| = |a|\|x\|$ for any scalar $a$.
Triangle inequality: The norm satisfies the triangle inequality, $\|x + y\| \leq \|x\| + \|y\|$ for any vectors $x$ and $y$.

When each vector in a vector space has these properties, the space allows for a more structured method of analyzing vectors and their relationships, forming the baseline for more advanced structures like Banach spaces.

Completeness

Completeness is a crucial property for a space, especially in analysis. A space is considered complete if every Cauchy sequence within it converges to a point that is also in the space. In our case, completeness is shown for the space of bounded linear functionals, implying that if you take any sequence of functionals where the values get arbitrarily close as the sequence progresses, there exists a limit within the same space. This concept ensures that no limits "escape" the space they are part of. Completeness in a normed space turns it into a Banach space, enabling full-fledged analysis using both algebraic and topological concepts.

Cauchy Sequence

A Cauchy sequence is a sequence of elements from a space where the elements "stick closer together" as you move along the sequence. Formally, for any small positive $\epsilon$, there exists a natural number $N$ such that for all $m, n \geq N$, the distance between the elements $x_m$ and $x_n$ is less than $\epsilon$.

Cauchy sequences focus on the elements themselves rather than their limit, unlike convergence which considers the "destination" of the sequence.
Not all Cauchy sequences converge in every space; only those in complete spaces have this property.

In functional analysis, particularly in Banach spaces, Cauchy sequences are used to determine completeness and to evaluate convergence without explicitly knowing the limits of sequences.

Problem 32

Problem 34

Other exercises in this chapter

Problem 31

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

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 34

We say two norms $\|u\|_{1}$ and $\|u\|_{2}$ on a vector space $V$ are equivalent if there exist constants $A$ and $B$ such that $$ \|u\|_{1} \leq A\|

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