Dual Space and Adjoint Operator

1 Dual Space

1.1 Dual Space

Definition 1.1 (Dual space) Let $X$ be a normed space. $X^{\ast}=\mathcal{L}(X,K)$ is called the dual space of $X$. Since $K$ is complete, $X^{\ast}$ is also a Banach space.

Definition 1.2 (Bidual space) We can define another intermediate mapping $x^{\ast\ast}: X^{\ast}\rightarrow R$ by $x^{\ast\ast}(f)=f(x)$. The space $X^{\ast\ast}=\mathcal{L}(X^{\ast},K)$ is called Bidual space, which is also a Banach space.

Definition 1.3 (Canonical mapping/embedding) We can define a mapping $J:X\rightarrow X^{\ast\ast}$ where $Jx(f)=f(x)$. $J$ is a linear isometry.

Proof. Since $x^{\ast\ast}(f)=|f(x)|\leq||x||||f||$, $||x^{\ast\ast}||\leq ||x||$. Due to Hahn-Banach theorem, there exists $f_0\in E^{\ast}$ such that $\Vert f_0\Vert=1$ and $|f_0(x)|=||x||$, and thus $||x^{\ast\ast}||\geq||x||$. Therefore, $||x^{\ast\ast}||=||x||$.

Definition 1.4 (Reflexive) In the sense of isometry, $X$ is a subspace of $X^{\ast\ast}$. $X$ is said to be reflexive if the canonical embedding $J$ into its bidual $X^{\ast\ast}$ is surjective.

1.2 Dual Space of Hilbert Space

Theorem 1.1 (Projection theorem) Let $H$ be a Hilbert space and $H_0 \subset H$ a closed subspace. Then for every $x \in H$ there exist uniquely $y \in H_0$ and $z \in H_0^{\perp}$ s.t. $x = y + z$.

Theorem 1.2 (Riesz’s representation theorem) Let $H$ be a Hilbert space $f$ be a bounded linear functional on $H$, then there exists one and only one $h\in H$ such that

$$f(x)=(x,h), ||f||_{H^\ast}=||h||_{H}$$

The Riesz representation theorem suggests that the mapping $H\rightarrow H^{\ast}$ is surjective, and hence an isometric isomorphism. It is not hard to see that $H^{\ast}$ is also a Hilbert space. Repeat the argument for $H^*$ and mapping $H^{\ast}\rightarrow H^{\ast\ast}$. We can prove that a Hilbert space is reflexive.

Corollary 1.3 (Lax-Milgram) Let $V$ be a Hilbert space, $a : H \times H → \mathbb R$ a bilinear, bounded and coercive form on $H$ . Then, for any $f \in H’$ the abstract problem

$$\text{find } u \in H : a(u, v) = f(v), \forall v \in H$$

has a unique solution $u \in V$ such that $\Vert u\Vert_H ≤\Vert f\Vert_{H’}/\alpha$ where $\alpha$ is the coercivity constant of the bilinear form.

1.3 Examples of Dual Spaces

Distribution.

2 Adjoint Operators

Definition 2.1 (Adjoint Operator) Let $X$ and $Y$ be normed linear spaces and $T\in \mathcal{L}(X,Y)$. The adjoint operator of $T$, denoted by $T^{\ast}$ , is the operator $T^{\ast}:Y^{\ast}\rightarrow X^{\ast}$ defined by

$$(T^{\ast}y^{\ast})(x)=y^{\ast}(Tx) \text{ for all } y^{\ast}\in Y^{\ast}, x\in X^{\ast}$$

Theorem 2.1 Let $X$ and $Y$ be normed linear spaces and $T\in \mathcal{L}(X,Y)$.

(a) $T^{\ast}$ is bounded linear operator on $Y^{\ast}$.

(b) $||T||=||T^{\ast}||$.

3 Weak Topology

3.1 Type of Convergence

Definition 3.1 (Types of convergence for sequence)

Let $X$ be normed linear space, $x,\{x_n\}\in X$.

(1) Strong Convergence: $||x_n-x||\rightarrow 0$.

(2) Weak convergence: If $\forall$ $f\in X^{\ast}$, $f(x_n)\rightarrow f(x)$, then we call $x_n\rightharpoonup x$ in weak topology.

(3) Weak-$\ast$ convergence [Concepts for dual space of X]. Let $X$ be a normed space and $X^{\ast}$ be its dual space. We define $f_n\mathop{\rightarrow}\limits_{}^{w^{\ast}} f$ if $f_n(x)\rightarrow f(x)$, for all $x\in X$.

Note: We can define similar convergences for functionals [strong convergence and weak convergence, weak-$\ast$ convergence is originally defined on $X^{\ast}$]. It is not the case for ‘convergence for operators’. For example, when we define weak convergence for $X^{\ast}$: we say $f_n\rightharpoonup f$, if $X^{\ast\ast}(f_n)\rightarrow X^{\ast\ast}(f)$ for all $x^{\ast \ast}\in X^{**}$. Proposition 3.4 and its proof are a very good explanation for this.

Proposition 3.1 (Sufficient and necessary condition for weak convergence) $x_n\rightharpoonup x$ iff

(1) $\{||x_n||\}$ is bounded.

(2) $\exists$ a dense subset $M\subseteq X^{\ast}$ such that for any $f\in M$, $f(x_n)\rightarrow f(x)$.

Proposition 3.2

If $X$ is finite dimensional, then $\text{strong convergence}\iff \text{weak convergence}$.

Proposition 3.3 (Weak-$\ast$ convergence) Let $X$ be Banach space and $f,\{f_n\}\in X^{\ast}$. Then $f_n\mathop{\rightarrow}\limits_{}^{w^{\ast}} f$ iff

(1) $\{||f_n||\}$ is bounded.

(2) $\exists$ a dense subset $M\subseteq X$ such that for any $x\in M$, $f(x_n)\rightarrow f(x)$.

Definition 3.2 (Types of convergence for operators)

Let $X,Y$ be normed linear spaces, $T,\{T_n\}\in \mathcal L(X,Y)$.

(1) Uniform Convergence.

(2) Strong Convergence: If $\forall$ $x\in X$, $T_nx$ convergences strongly to $Tx$, then we call $T_n\rightarrow T$.

(3) Weak Convergence: If $\forall$ $x\in X$, $T_nx$ convergences weakly to $Tx$ in $Y$, then we call $T_n\rightharpoonup T$.

Let $X,Y$ be normed spaces and $T\in \mathcal L(X,Y)$. The mapping $GT=T^{\ast}$ is a distance-preseved mapping and $Im(G)$ is isometric isomorphism wrt $\mathcal L(X,Y)$.

Proposition 3.4 (Proposition 5.13 in [2], Weak Convergence on $X^{}$ and weak-$$ convergence) Let $X$ be normed spaces and $X^{\ast}$ be the dual space, then

$$\text{weak convergence} \Rightarrow \text{weak-}{\ast} \text{convergence}$$

in $X^{\ast}$. If $X$ is reflexive,

$$\text{weak convergence} \ on \ X^* \iff\text{weak-}{\ast} \text{convergence}$$

3.2 Separability

Theorem 3.5 (Banach-Alaoglu Theorem, Theorem 5.5.7 in [1]). Let $X$ be a normed linear space. Then the closed unit ball in $X$ is weak-$*$ compact.

Proof. We can turn to [1] and link here.

Proposition 3.6 (Sequential Banach-Alaoglu Theorem). If $X$ is a separable normed space, then any bounded sequence $\{f_n\}\in X^{\ast}$ admits a weakly-* convergent subsequence. This is equivalent to Helly selection Theorem. (We cannot treat it as weaker Version of Banach-Alaoglu Theorem.)

Proposition 3.7. Let $X$ be a normed space, then $X^{\ast}$ is separable $\Rightarrow$ $X$ is separable.

3.3 Weak Convergence in $L^2$

Proposition 3.8 (Weak Convergence on $L^p$, Example 4 in Section 5.3 [2]). In $L^p[a,b]$, $x_n\rightharpoonup x_0$ iff:

(1) For every $t\in[a,b]$, we have

$$\int_{a}^t x_n(s)ds\rightarrow \int_a^t x_0(s)ds$$

(2) $(\Vert x_n\Vert )$ is bounded.

Proof. It relies on Proposition 3.1 in this note.

Proposition 3.9 (Weak Convergence on $L^2$ or generally a hilbert space $H$) This link provieds a theorem: Every bounded sequence in Hilbert space admits a weakly convergent sequence.

4 Finite Dimentional Normed Space

A small supplement.

Definition 4.1 (Linear isometry and homeomorphism in normed spaces) In mathematics, an isometry is a distance-preserving transformation between metric spaces, usually assumed to be bijective. Linear isometry is often respect with normed vector spaces. Homeomorphism represents the topological isomorphism.

Proposition 4.1 (Several Properties) I met them in my “Smooth Manifolds” course.

• All norms on finite dimensional normed space $X$ are equivalent. [1, Theorem 2.6.1]
• Every finite dimensional normed space is linear isomorphic (线性同构，一定意义的代数上) and homeomorphic (同胚，拓扑上) to $\mathbb{R}^n$. [2, Theorem 2.1]

Proposition 4.2 (More properties about the finite dim normed spaces)

• Every finite dimensional normed space is complete. Heine-Borel Theorem could be generalized to finite dimensional normed space.
• A normed space $X$ is finite dimensional iff $X$ is locally compact, i.e. every point $x$ in $X$ has a compact neighborhood (through Riesz’s Lemma). [2, Theorem 2.2]
• A normed space $X$ is finite-dimensional iff its closed unit ball is compact. [1, Theorem 2.6.4]

References

[1] Functional Analysis Notes, by Mr. Andrew Pinchuck, https://uomustansiriyah.edu.iq/media/lectures/9/9_2017_09_30!12_03_54_PM.pdf

[2] A Sketch of Real-Value Function and Functional Analysis, by Wang Shengwang, Zheng Weixing.