Proving The Isometry Of Dual Bochner Spaces When S Is Atomic

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Hey guys! Today, we're diving deep into a fascinating topic in functional analysis: proving the isometry $(L^p(S;X))^*=L^q(S;X^*)$ when $S$ is an atomic measure space. This is a crucial result in the realm of Banach spaces, especially when dealing with Bochner spaces. We'll break down the concepts, theorems, and steps involved in this proof, making it super clear and easy to grasp. So, let's get started!

Understanding the Foundation

Before we jump into the proof, let's ensure we have a solid understanding of the fundamental concepts. This will make the entire process smoother and more intuitive. We'll start by defining the key terms and then move on to the theorems that play a crucial role in this proof.

Atomic Measure Space

First off, what exactly is an atomic measure space? Well, imagine a measure space $(S, \mathscr{A}, \mu)$ where the measure $\mu$ is concentrated on individual points. In simpler terms, an atomic measure space is one where the set $S$ can be broken down into a countable number of atoms. An atom, in this context, is a measurable set $A$ such that $\mu(A) > 0$, and for any measurable subset $B$ of $A$, either $\mu(B) = 0$ or $\mu(B) = \mu(A)$.

Think of it like this: you have a collection of points, and each point carries a certain "weight" or measure. The total measure is just the sum of these individual weights. This is in contrast to a continuous measure space, where the measure is spread out smoothly over the space. Atomic measure spaces are super common and pop up in various contexts, making their properties quite important.

Bochner Spaces

Next, let's talk about Bochner spaces. A Bochner space, denoted as $L^p(S; X)$, is essentially a generalization of the familiar $L^p$ spaces to functions taking values in a Banach space $X$. Here, $(S, \mathscr{A}, \mu)$ is our measure space, and $X$ is a Banach space (a complete normed vector space).

A function $f: S \rightarrow X$ is Bochner measurable if it can be approximated by simple functions. A simple function is just a finite linear combination of indicator functions, each multiplied by a vector in $X$. The Bochner integral extends the usual Lebesgue integral to these Banach space-valued functions. The norm in $L^p(S; X)$ is defined as:

$$||f||_{L^p(S;X)} = \left( \int_S ||f(s)||_X^p d\mu(s) \right)^{1/p},$$

for $1 \leq p < \infty$. For $p = \infty$, we have:

$$||f||_{L^\infty(S;X)} = \text{ess sup}_{s \in S} ||f(s)||_X.$$

Bochner spaces are fundamental in studying vector-valued functions and are widely used in areas like partial differential equations, stochastic analysis, and optimization.

Dual Spaces and Duality

Now, let's move on to dual spaces and duality. The dual space of a Banach space $X$, denoted as $X^*$, is the space of all bounded linear functionals from $X$ to the scalar field (usually the real numbers $\mathbb{R}$ or complex numbers $\mathbb{C}$). These bounded linear functionals are linear maps $f: X \rightarrow \mathbb{F}$ (where $\mathbb{F}$ is the scalar field) that satisfy the boundedness condition:

$$|f(x)| \leq M ||x||_X$$

for some constant $M$ and all $x \in X$. The norm of a bounded linear functional $f$ is defined as:

$$||f||_{X^*} = \sup_{||x||_X = 1} |f(x)|.$$

The concept of duality is all about understanding the relationship between a space and its dual. Duality theorems, such as the one we're aiming to prove, tell us how the dual of one space can be represented in terms of another related space. This is immensely useful because it allows us to transfer problems and solutions between different spaces, often simplifying the analysis.

Hölder's Inequality

One theorem that's crucial for our proof is Hölder's inequality. For scalar-valued functions, Hölder's inequality states that if $f \in L^p(S)$ and $g \in L^q(S)$, where $1 < p < \infty$, $1/p + 1/q = 1$, then $fg \in L^1(S)$ and

$$\int_S |f(s)g(s)| d\mu(s) \leq \left( \int_S |f(s)|^p d\mu(s) \right)^{1/p} \left( \int_S |g(s)|^q d\mu(s) \right)^{1/q}.$$

This inequality is also valid for Bochner spaces. If $f \in L^p(S; X)$ and $g \in L^q(S; X^*)$, where $X^*$ is the dual of $X$, then we have a similar inequality involving the duality pairing:

$$\int_S |\langle f(s), g(s) \rangle| d\mu(s) \leq \left( \int_S ||f(s)||_X^p d\mu(s) \right)^{1/p} \left( \int_S ||g(s)||_{X^*}^q d\mu(s) \right)^{1/q},$$

where $\langle f(s), g(s) \rangle$ denotes the application of the functional $g(s) \in X^*$ to the element $f(s) \in X$.

The Proposition: $(L^p(S;X))^*=L^q(S;X^*)$

Now that we've laid the groundwork, let's state the proposition we're aiming to prove. The proposition states that for $1 < p < \infty$ and $1/p + 1/q = 1$, and when $S$ is an atomic measure space, the dual space of $L^p(S; X)$, denoted as $(L^p(S; X))^*$, is isometric to $L^q(S; X^*)$. In mathematical notation:

$$(L^p(S;X))^* \cong L^q(S;X^*).$$

This result is powerful because it tells us that every bounded linear functional on $L^p(S; X)$ can be represented as integration against a function in $L^q(S; X^*)$, and vice versa. This representation simplifies many problems involving linear functionals on Bochner spaces.

Breaking Down the Proof

So, how do we prove this awesome proposition? The proof typically involves several key steps:

1. Defining a Natural Map: We start by defining a natural map $\Phi$ from $L^q(S; X^*)$ into $(L^p(S; X))^*$. This map essentially associates each function $g \in L^q(S; X^*)$ with a bounded linear functional on $L^p(S; X)$.
2. Showing $\Phi$ is a Bounded Linear Operator: We need to show that $\Phi$ is both linear and bounded. This ensures that the map behaves nicely and preserves the structure of the spaces involved.
3. Proving $\Phi$ is an Isometry: This is a crucial step where we demonstrate that $\Phi$ preserves the norm, i.e., $||\Phi(g)|| = ||g||$ for all $g \in L^q(S; X^*)$. This shows that the map is a faithful representation of $L^q(S; X^*)$ within $(L^p(S; X))^*$.
4. Showing $\Phi$ is Surjective: Finally, we need to prove that $\Phi$ maps onto the entire dual space $(L^p(S; X))^*$. This ensures that every bounded linear functional on $L^p(S; X)$ can be represented by a function in $L^q(S; X^*)$.

Let's dive into each of these steps in detail.

1. Defining the Natural Map $\Phi$

Given a function $g \in L^q(S; X^*)$, we define the map $\Phi: L^q(S; X^*) \rightarrow (L^p(S; X))^*$ as follows. For any $f \in L^p(S; X)$, we define $\Phi(g)$ as the linear functional that acts on $f$ by:

$$[\Phi(g)](f) = \int_S \langle f(s), g(s) \rangle d\mu(s),$$

where $\langle f(s), g(s) \rangle$ represents the duality pairing between $f(s) \in X$ and $g(s) \in X^*$. In simple terms, $\Phi(g)$ takes a function $f$ and returns the integral of the pointwise application of $g$ to $f$.

This definition is quite natural. It mirrors how we often represent linear functionals in $L^p$ spaces, using integration against a function in the dual space. The key here is to ensure that this map is well-defined and behaves as we expect.

2. Showing $\Phi$ is a Bounded Linear Operator

Next, we need to show that $\Phi$ is a bounded linear operator. This involves two parts:

• Linearity: We need to show that $\Phi$ preserves linear combinations. That is, for any $g_1, g_2 \in L^q(S; X^*)$ and scalars $a, b$, we need to show that $\Phi(ag_1 + bg_2) = a\Phi(g_1) + b\Phi(g_2)$.

  Let's verify this. For any $f \in L^p(S; X)$, we have:

  $$\begin{aligned} [\Phi(ag_1 + bg_2)](f) &= \int_S \langle f(s), ag_1(s) + bg_2(s) \rangle d\mu(s) \\ &= \int_S [a \langle f(s), g_1(s) \rangle + b \langle f(s), g_2(s) \rangle] d\mu(s) \\ &= a \int_S \langle f(s), g_1(s) \rangle d\mu(s) + b \int_S \langle f(s), g_2(s) \rangle d\mu(s) \\ &= a [\Phi(g_1)](f) + b [\Phi(g_2)](f). \end{aligned}$$

  Thus, $\Phi(ag_1 + bg_2) = a\Phi(g_1) + b\Phi(g_2)$, so $\Phi$ is linear. Awesome!

• Boundedness: We need to show that there exists a constant $M$ such that $||\,\Phi(g)\,||_{(L^p(S; X))^*} \leq M ||g||_{L^q(S; X^*)}$ for all $g \in L^q(S; X^*)$. In other words, we need to control the norm of the linear functional $\Phi(g)$ by the norm of the function $g$.

  To do this, we use Hölder's inequality. Recall that the norm of a linear functional $T \in (L^p(S; X))^*$ is given by:

  $$||T||_{(L^p(S; X))^*} = \sup_{||f||_{L^p(S; X)} = 1} |T(f)|.$$

  So, we want to estimate $|[\Phi(g)](f)|$ for $f \in L^p(S; X)$ with $||f||_{L^p(S; X)} = 1$. Using Hölder's inequality for Bochner spaces, we have:

  $$\begin{aligned} |[\Phi(g)](f)| &= \left| \int_S \langle f(s), g(s) \rangle d\mu(s) \right| \\ &\leq \int_S |\langle f(s), g(s) \rangle| d\mu(s) \\ &\leq \left( \int_S ||f(s)||_X^p d\mu(s) \right)^{1/p} \left( \int_S ||g(s)||_{X^*}^q d\mu(s) \right)^{1/q} \\ &= ||f||_{L^p(S; X)} ||g||_{L^q(S; X^*)}. \end{aligned}$$

  Since we're considering $f$ with $||f||_{L^p(S; X)} = 1$, we get:

  $$|[\Phi(g)](f)| \leq ||g||_{L^q(S; X^*)}.$$

  Taking the supremum over all such $f$, we find:

  $$||\Phi(g)||_{(L^p(S; X))^*} = \sup_{||f||_{L^p(S; X)} = 1} |[\Phi(g)](f)| \leq ||g||_{L^q(S; X^*)}.$$

  Thus, $\Phi$ is bounded with a bound of $M = 1$. Fantastic!

3. Proving $\Phi$ is an Isometry

Now, let's show that $\Phi$ is an isometry. This means we need to prove that $||\,\Phi(g)\,||_{(L^p(S; X))^*} = ||g||_{L^q(S; X^*)}$ for all $g \in L^q(S; X^*)$. We've already shown that $||\,\Phi(g)\,||_{(L^p(S; X))^*} \leq ||g||_{L^q(S; X^*)}$; now, we need to show the reverse inequality.

Since $S$ is an atomic measure space, we can write $S = \{s_i\}_{i \in I}$, where $I$ is a countable index set. Let $\mu_i = \mu(\{s_i\})$ be the measure of the atom $\{s_i\}$. Then, for any $f \in L^p(S; X)$ and $g \in L^q(S; X^*)$, the integrals become sums:

$$\int_S ||f(s)||_X^p d\mu(s) = \sum_{i \in I} ||f(s_i)||_X^p \mu_i,$$

and

$$\int_S ||g(s)||_{X^*}^q d\mu(s) = \sum_{i \in I} ||g(s_i)||_{X^*}^q \mu_i.$$

For each $s_i$, we can use the definition of the norm in $X^*$ to find a vector $x_i \in X$ with $||x_i||_X = 1$ such that

$$|\langle x_i, g(s_i) \rangle| \geq (1 - \epsilon) ||g(s_i)||_{X^*},$$

for any $\epsilon > 0$. Now, let's define a function $f \in L^p(S; X)$ by

$$f(s_i) = \mu_i^{q/p} ||g(s_i)||_{X^*}^{q/p - 1} \text{sgn}(\langle x_i, g(s_i) \rangle) x_i,$$

where $\text{sgn}(z) = z/|z|$ if $z \neq 0$ and $0$ if $z = 0$. Then,

$$||f(s_i)||_X = \mu_i^{q/p} ||g(s_i)||_{X^*}^{q/p - 1} ||x_i||_X = \mu_i^{q/p} ||g(s_i)||_{X^*}^{q/p - 1}.$$

Now, let's compute the $L^p$ norm of $f$:

$$\begin{aligned} ||f||_{L^p(S; X)}^p &= \sum_{i \in I} ||f(s_i)||_X^p \mu_i \\ &= \sum_{i \in I} (\mu_i^{q/p} ||g(s_i)||_{X^*}^{q/p - 1})^p \mu_i \\ &= \sum_{i \in I} \mu_i^q ||g(s_i)||_{X^*}^{q - p} \mu_i \\ &= \sum_{i \in I} \mu_i ||g(s_i)||_{X^*}^q \\ &= ||g||_{L^q(S; X^*)}^q. \end{aligned}$$

So, $||f||_{L^p(S; X)} = ||g||_{L^q(S; X^*)}^{q/p}$. Now, let's normalize $f$ by defining $f_0 = f / ||f||_{L^p(S; X)}$, so $||f_0||_{L^p(S; X)} = 1$. Then,

$$\begin{aligned} [\Phi(g)](f_0) &= \int_S \langle f_0(s), g(s) \rangle d\mu(s) \\ &= \frac{1}{||f||_{L^p(S; X)}} \sum_{i \in I} \langle f(s_i), g(s_i) \rangle \mu_i \\ &= \frac{1}{||g||_{L^q(S; X^*)}^{q/p}} \sum_{i \in I} \mu_i^{q/p} ||g(s_i)||_{X^*}^{q/p - 1} \text{sgn}(\langle x_i, g(s_i) \rangle) \langle x_i, g(s_i) \rangle \mu_i \\ &= \frac{1}{||g||_{L^q(S; X^*)}^{q/p}} \sum_{i \in I} \mu_i^{1 + q/p} ||g(s_i)||_{X^*}^{q/p - 1} |\langle x_i, g(s_i) \rangle| \\ &\geq \frac{1}{||g||_{L^q(S; X^*)}^{q/p}} \sum_{i \in I} \mu_i^{1 + q/p} ||g(s_i)||_{X^*}^{q/p - 1} (1 - \epsilon) ||g(s_i)||_{X^*} \\ &= \frac{1 - \epsilon}{||g||_{L^q(S; X^*)}^{q/p}} \sum_{i \in I} \mu_i^{1 + q/p} ||g(s_i)||_{X^*}^{q/p} \\ &= (1 - \epsilon) ||g||_{L^q(S; X^*)}. \end{aligned}$$

Taking the supremum over all $f_0$ with norm 1, we get:

$$||\Phi(g)||_{(L^p(S; X))^*} \geq (1 - \epsilon) ||g||_{L^q(S; X^*)}.$$

Since $\epsilon > 0$ is arbitrary, we have:

$$||\Phi(g)||_{(L^p(S; X))^*} \geq ||g||_{L^q(S; X^*)}.$$

Combining this with the earlier inequality, we conclude that

$$||\Phi(g)||_{(L^p(S; X))^*} = ||g||_{L^q(S; X^*)}.$$

So, $\Phi$ is an isometry! This is a huge step forward.

4. Showing $\Phi$ is Surjective

Finally, we need to show that $\Phi$ is surjective, meaning that for every bounded linear functional $T \in (L^p(S; X))^*$, there exists a function $g \in L^q(S; X^*)$ such that $\Phi(g) = T$.

Let $T \in (L^p(S; X))^*$ be a bounded linear functional. Since $S$ is atomic, we have $S = \{s_i\}_{i \in I}$. For each $i \in I$, define a map $T_i: X \rightarrow \mathbb{F}$ by

$$T_i(x) = T(\delta_{s_i} x),$$

where $\delta_{s_i}$ is the indicator function of the singleton $\{s_i\}$, i.e., $\delta_{s_i}(s) = 1$ if $s = s_i$ and $0$ otherwise. The function $\delta_{s_i} x$ is the function that takes the value $x$ at $s_i$ and $0$ elsewhere.

Each $T_i$ is a bounded linear functional on $X$, so $T_i \in X^*$. We define the function $g: S \rightarrow X^*$ by $g(s_i) = T_i$ for each $i \in I$. Now, we need to show that $g \in L^q(S; X^*)$ and that $\Phi(g) = T$.

First, let's show that $g \in L^q(S; X^*)$. We need to show that $\sum_{i \in I} ||g(s_i)||_{X^*}^q \mu_i < \infty$. For any $x \in X$ with $||x||_X \leq 1$, we have

$$|T_i(x)| = |T(\delta_{s_i} x)| \leq ||T||_{(L^p(S; X))^*} ||\delta_{s_i} x||_{L^p(S; X)} = ||T||_{(L^p(S; X))^*} ||x||_X \mu_i^{1/p} \leq ||T||_{(L^p(S; X))^*} \mu_i^{1/p}.$$

Thus, $||T_i||_{X^*} \leq ||T||_{(L^p(S; X))^*} \mu_i^{1/p}$, so $||g(s_i)||_{X^*} \leq ||T||_{(L^p(S; X))^*} \mu_i^{1/p}$. Now,

$$\begin{aligned} \sum_{i \in I} ||g(s_i)||_{X^*}^q \mu_i &\leq \sum_{i \in I} (||T||_{(L^p(S; X))^*} \mu_i^{1/p})^q \mu_i \\ &= ||T||_{(L^p(S; X))^*}^{q} \sum_{i \in I} \mu_i^{q/p} \mu_i \\ &= ||T||_{(L^p(S; X))^*}^{q} \sum_{i \in I} \mu_i^{q/p + 1} \\ &= ||T||_{(L^p(S; X))^*}^{q} \sum_{i \in I} \mu_i^q < \infty. \end{aligned}$$

So, $g \in L^q(S; X^*)$. Finally, we need to show that $\Phi(g) = T$. For any $f \in L^p(S; X)$, we have

$$\begin{aligned} [\Phi(g)](f) &= \int_S \langle f(s), g(s) \rangle d\mu(s) \\ &= \sum_{i \in I} \langle f(s_i), g(s_i) \rangle \mu_i \\ &= \sum_{i \in I} \langle f(s_i), T_i \rangle \mu_i \\ &= \sum_{i \in I} T_i(f(s_i)) \mu_i \\ &= \sum_{i \in I} T(\delta_{s_i} f(s_i)) \mu_i \\ &= T(\sum_{i \in I} \delta_{s_i} f(s_i) \mu_i) \\ &= T(f). \end{aligned}$$

Thus, $\Phi(g) = T$, and $\Phi$ is surjective! We've done it!

Conclusion

Alright guys, that was quite a journey! We've successfully proven that when $S$ is an atomic measure space, the dual space of $L^p(S; X)$ is isometric to $L^q(S; X^*)$. This result is a cornerstone in functional analysis and is super useful in various applications. We started by understanding the fundamental concepts, broke down the proof into manageable steps, and conquered each step with careful reasoning. Great job sticking with it! Understanding these kinds of theorems not only deepens our knowledge but also equips us with powerful tools for tackling complex problems in analysis. Keep exploring, and happy analyzing!

• atomic measure space: What is an atomic measure space, and how does it differ from a continuous measure space?
• Bochner spaces: Explain Bochner spaces and their significance in functional analysis.
• dual spaces: What are dual spaces and duality theorems, and why are they important?
• Hölder's inequality: How does Hölder's inequality apply to Bochner spaces?
• isometry $(L^p(S;X))^*=L^q(S;X^*)$: Prove the isometry $(L^p(S;X))^*=L^q(S;X^*)$ when $S$ is an atomic measure space.