Proving Reflexivity: Is X×Y A Reflexive Space?

Hey everyone! Today, we're diving deep into the fascinating world of functional analysis, specifically exploring the concept of reflexive Banach spaces. Our main goal? To prove that the space $Z = X × Y$ is reflexive, given that $X$ and $Y$ are themselves reflexive Banach spaces. This is a crucial concept in understanding the structure and properties of these spaces, so let's break it down step by step.

Understanding Reflexive Banach Spaces

Before we jump into the proof, let's quickly recap what it means for a Banach space to be reflexive. Guys, this is super important! A Banach space is essentially a complete normed vector space – think of it as a vector space where we can measure distances and where sequences that "should" converge actually do converge. Reflexivity takes this a step further.

A Banach space $X$ is reflexive if the canonical embedding $J: X \rightarrow X^{**}$ is surjective. Okay, that's a mouthful! Let's unpack it. $X^{**}$ represents the double dual space of $X$. The dual space, $X^*$, consists of all bounded linear functionals on $X$ (linear transformations that map $X$ to the scalar field, usually real or complex numbers, and are "well-behaved" in terms of boundedness). The double dual, $X^{**}$, is then the dual of the dual space, meaning it's the space of bounded linear functionals on $X^*$.

Now, the canonical embedding $J$ is a natural map that takes an element $x$ from $X$ and maps it to a functional in $X^{**}$. This functional, denoted as $J(x)$, acts on elements $f$ in $X^*$ by simply evaluating $f$ at $x$, i.e., $J(x)(f) = f(x)$. Reflexivity demands that this map $J$ is not just injective (one-to-one) but also surjective (onto). Surjectivity means that every functional in $X^{**}$ can be obtained as the image of some element in $X$ under the map $J$. In simpler terms, a reflexive space is "big enough" to capture all the functionals on its dual space. So, to prove that $Z$ is reflexive, we'll need to show that this canonical embedding is surjective for $Z$.

Defining the Space Z = X × Y

Alright, let's define our space $Z$ more precisely. We're given that $Z = X × Y$, which is the Cartesian product of the Banach spaces $X$ and $Y$. This means that elements in $Z$ are ordered pairs of the form $z = (x, y)$, where $x$ belongs to $X$ and $y$ belongs to $Y$. But how do we make this into a Banach space itself? We need a norm!

The problem states that the norm on $Z$ is defined as: $|z| = (|x|^p + |y|^p){\frac{1}{p}}, \forall z = (x, y) \in X × Y$ here, $1 \leq p \leq \infty$. This is a generalization of the usual Euclidean norm. When $p = 2$, it's exactly the Euclidean norm, but we can consider other values of $p$ as well. The case when $p = \infty$ is interpreted as $\|z\| = max{\|x\|, \|y\|}$. Guys, remember that a norm needs to satisfy certain properties: it should be non-negative, it should be zero only for the zero vector, it should scale correctly with scalar multiplication, and it should satisfy the triangle inequality. It can be proven that this definition indeed gives a valid norm on $Z$.

Since $X$ and $Y$ are Banach spaces, and we've equipped $Z$ with a valid norm, we can also show that $Z$ is a Banach space. This is because completeness (the property that Cauchy sequences converge) is preserved under this construction. So, $Z$ is a Banach space, and now we're ready to tackle the main question: is it reflexive?

The Proof: Z = X × Y is Reflexive

Here's the heart of the matter! We want to prove that $Z$ is reflexive, given that $X$ and $Y$ are reflexive Banach spaces. Our strategy is to show that the canonical embedding $J_Z: Z \rightarrow Z^{**}$ is surjective. To do this, we'll leverage the reflexivity of $X$ and $Y$ and construct a suitable preimage in $Z$ for any functional in $Z^{**}$.

Let $F \in Z^{**}$ be an arbitrary element in the double dual of $Z$. This means $F$ is a bounded linear functional on $Z^*$. We need to find a $z = (x, y) \in Z$ such that $J_Z(z) = F$. In other words, we need to find $x \in X$ and $y \in Y$ such that $F(\phi) = \phi(z) = \phi(x, y)$ for all $\phi \in Z^*$.

The key here is to understand the structure of the dual space $Z^*$. It can be proven that $Z^*$ is isometrically isomorphic to $X^* × Y^*$. This means there's a norm-preserving linear bijection between $Z^*$ and $X^* × Y^*$. Intuitively, this means that any bounded linear functional on $Z$ can be represented as a combination of bounded linear functionals on $X$ and $Y$. More precisely, for every $\phi \in Z^*$, there exist unique $f \in X^*$ and $g \in Y^*$ such that $\phi(x, y) = f(x) + g(y)$ for all $(x, y) \in Z$.

Now, let's use this representation to define functionals $F_1$ on $X^*$ and $F_2$ on $Y^*$. For any $f \in X^*$, define $F_1(f) = F(\phi)$, where $\phi \in Z^*$ is given by $\phi(x, y) = f(x)$. Similarly, for any $g \in Y^*$, define $F_2(g) = F(\phi)$, where $\phi \in Z^*$ is given by $\phi(x, y) = g(y)$. It can be proven that $F_1$ and $F_2$ are bounded linear functionals, so $F_1 \in X^{**}$ and $F_2 \in Y^{**}$.

This is where the reflexivity of $X$ and $Y$ comes into play! Since $X$ is reflexive, the canonical embedding $J_X: X \rightarrow X^{**}$ is surjective. This means there exists an $x \in X$ such that $J_X(x) = F_1$. Similarly, since $Y$ is reflexive, there exists a $y \in Y$ such that $J_Y(y) = F_2$. In other words, we have: $F_1(f) = J_X(x)(f) = f(x), \forall f \in X^$ $F_2(g) = J_Y(y)(g) = g(y), \forall g \in Y^$

Now, let's consider the element $z = (x, y) \in Z$. We claim that $J_Z(z) = F$. To prove this, we need to show that $J_Z(z)(\phi) = F(\phi)$ for all $\phi \in Z^*$. Let $\phi \in Z^*$ be arbitrary. As we discussed earlier, there exist unique $f \in X^*$ and $g \in Y^*$ such that $\phi(x, y) = f(x) + g(y)$. Then:

$J_Z(z)(\phi) = \phi(z) = \phi(x, y) = f(x) + g(y)$

On the other hand:

$F(\phi) = F(\phi)$, where $\phi(x, y) = f(x) + g(y)$.

Recall how we defined $F_1$ and $F_2$: $F_1(f) = F(\phi)$ where $\phi(x, y) = f(x)$, and $F_2(g) = F(\phi)$ where $\phi(x, y) = g(y)$. Therefore:

$F(\phi) = F_1(f) + F_2(g)$

Now, using the reflexivity of $X$ and $Y$:

$F_1(f) + F_2(g) = J_X(x)(f) + J_Y(y)(g) = f(x) + g(y)$

Combining these results, we have:

$J_Z(z)(\phi) = f(x) + g(y) = F(\phi)$

This holds for all $\phi \in Z^*$, so we've proven that $J_Z(z) = F$. Since $F \in Z^{**}$ was arbitrary, this shows that the canonical embedding $J_Z$ is surjective. Therefore, $Z = X × Y$ is reflexive.

Conclusion

So, there you have it, guys! We've successfully proven that if $X$ and $Y$ are reflexive Banach spaces, then their product space $Z = X × Y$, equipped with the norm $\|z\| = (\|x\|^p + \|y\|^p)^{\frac{1}{p}}$, is also a reflexive Banach space. This is a powerful result that highlights how reflexivity is preserved under certain operations. Understanding reflexivity is crucial for many advanced topics in functional analysis, so hopefully this detailed explanation has been helpful!

This proof relies heavily on the properties of reflexive spaces and the structure of dual spaces. We used the fact that $Z^*$ is isomorphic to $X^* × Y^*$ and leveraged the surjectivity of the canonical embeddings for $X$ and $Y$. Remember, the key idea is to find a pre-image in $Z$ for any functional in $Z^{**}$, and we accomplished this by carefully constructing the elements $x$ and $y$ using the reflexivity of $X$ and $Y$. Keep practicing these concepts, and you'll become a functional analysis pro in no time!