# 数学代写|泛函分析作业代写Functional Analysis代考|Bidual, Reflexive Spaces

#### Doug I. Jones

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## 数学代写|泛函分析作业代写Functional Analysis代考|Bidual, Reflexive Spaces

The Bidual Space. Let $U$ be a normed space and $U^{\prime}$ its topological dual. Then $U^{\prime}$ equipped with the dual norm is itself a normed space (always complete, even if $U$ is not) and it also makes sense to speak of the space of all continuous linear functionals on $U^{\prime}$. The dual of the dual of a normed space $U$ is again a Banach space, denoted $U^{\prime \prime}$ and called the bidual of $U$
$$U^{\prime \prime} \stackrel{\text { def }}{=}\left(U^{\prime}\right)^{\prime}$$
It turns out that any normed space $U$ is isomorphic, in a natural way, to a subspace of its bidual $U^{\prime \prime}$. To see this, let $u \in U$ and $f \in U^{\prime}$. Then $\langle f, u\rangle=f(u)$ is a linear functional on $U$ (by the choice of $f$ ). However, for each fixed $u,\langle f, u\rangle$ is also a linear functional on $U^{\prime}$ (by definition of vector space operations in $U^{\prime}$ ). More precisely, for each $u \in U$, we define a corresponding linear functional $F_u$ on $U^{\prime}$, called the evaluation at $u$ and defined as
$$U^{\prime} \ni f \longrightarrow F_u(f) \stackrel{\text { def }}{=}\langle f, u\rangle \in \mathbb{R}(\boldsymbol{C})$$
From the inequality
$$\left|F_u(f)\right|=|\langle f, u\rangle| \leq|u|_U|f|_{U^{\prime}}$$
follows that
$$\left|F_u\right|_{U^{\prime \prime}} \leq|u|_U$$

## 数学代写|泛函分析作业代写Functional Analysis代考|Weak Topologies, Weak Sequential Compactness

The topological properties of normed linear spaces are complicated by the fact that topologies can be induced on such spaces in more than one way. This leads to alternative notions of continuity, compactness, and convergence for a normed space $U$.

Weak Topology. Let $U$ be a normed space and let $U^{\prime}$ denote its dual. For each continuous, linear functional $f \in U^{\prime}$ we introduce a corresponding seminorm $p_f$ on $U$ defined as
$$p_f(\boldsymbol{u}) \stackrel{\text { def }}{=}|f(\boldsymbol{u})|=|\langle f, \boldsymbol{u}\rangle|, f \in U^{\prime}$$

By Corollary 5.5.2, for each $\boldsymbol{u} \neq \mathbf{0}$, there exists a functional $f \in U^{\prime}$ taking a non-zero value at $\boldsymbol{u}$, which implies that the family of seminorms
$$p_f: U \rightarrow[0, \infty), f \in U^{\prime}$$
satisfies the axiom of separation (see Section 5.5). Consequently, the $p_f$ seminorms can be used to construct a locally convex topology on $U$. We refer to it as the weak topology in contrast to the topology induced by the norm and called the strong topology.

Indeed, it follows immediately from the definition of locally convex spaces that the weak topology is weaker than the one induced by the norm. To see this, consider an arbitrary element from the base of neighborhoods for the zero vector in $U$
$$B\left(I_0, \varepsilon\right)=\left{\boldsymbol{u} \in U:|f(\boldsymbol{u})| \leq \varepsilon, f \in I_0\right}$$
where $I_0 \subset U^{\prime}$ is finite.

# 泛函分析代写

## 数学代写|泛函分析作业代写Functional Analysis代考|Bidual, Reflexive Spaces

$$U^{\prime \prime} \stackrel{\text { def }}{=}\left(U^{\prime}\right)^{\prime}$$

$$U^{\prime} \ni f \longrightarrow F_u(f) \stackrel{\text { def }}{=}\langle f, u\rangle \in \mathbb{R}(\boldsymbol{C})$$

$$\left|F_u(f)\right|=|\langle f, u\rangle| \leq|u|U|f|{U^{\prime}}$$

$$\left|F_u\right|_{U^{\prime \prime}} \leq|u|_U$$

## 数学代写|泛函分析作业代写Functional Analysis代考|Weak Topologies, Weak Sequential Compactness

$$p_f(\boldsymbol{u}) \stackrel{\text { def }}{=}|f(\boldsymbol{u})|=|\langle f, \boldsymbol{u}\rangle|, f \in U^{\prime}$$

$$p_f: U \rightarrow[0, \infty), f \in U^{\prime}$$

$$B\left(I_0, \varepsilon\right)=\left{\boldsymbol{u} \in U:|f(\boldsymbol{u})| \leq \varepsilon, f \in I_0\right}$$

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## MATLAB代写

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