A recurring challenge in the analysis of partial differential equations and variational problems is the following: given a sequence of approximate solutions $\{u_n\}_{n=1}^\infty$ in a [Banach space](/page/Banach%20Space) $X$, bounded uniformly in the norm $\|u_n\|_X \le M$, can we extract a convergent subsequence? In finite dimensions, the Bolzano-Weierstrass theorem guarantees this. In infinite dimensions, it fails catastrophically — the closed unit ball of an infinite-dimensional normed space is never compact in the norm topology.

The standard remedy is to weaken the notion of convergence. Rather than demanding $\|u_{n_k} - u\|_X \to 0$, we ask only that $f(u_{n_k}) \to f(u)$ for every bounded linear functional $f \in X^*$ — this is weak convergence. But even this weaker notion does not automatically yield subsequential limits. Whether every bounded sequence admits a [weakly convergent](/page/Weak%20Convergence) subsequence depends on a structural property of the space $X$ itself: **reflexivity**.

[example: A Bounded Sequence in $L^1$ with No Weakly Convergent Subsequence] Let $X = L^1([0, 1])$ equipped with the norm $\|f\|_{L^1} = \int_0^1 |f(x)| \, dx$. Define the sequence of functions $u_n: [0, 1] \to \mathbb{R}$ by \begin{align*} u_n(x) := n \cdot \mathbb{1}_{[0, 1/n]}(x). \end{align*} Each $u_n$ is a "spike" of height $n$ concentrated on the interval $[0, 1/n]$. The sequence is bounded: \begin{align*} \|u_n\|_{L^1} = \int_0^1 n \cdot \mathbb{1}_{[0, 1/n]}(x) \, dx = n \cdot \frac{1}{n} = 1 \quad \text{for all } n \in \mathbb{N}. \end{align*} Suppose, for contradiction, that a subsequence $u_{n_k} \rightharpoonup u$ converges weakly in $L^1([0, 1])$. By the definition of weak convergence in $L^1$, this means \begin{align*} \int_0^1 u_{n_k}(x) g(x) \, dx \to \int_0^1 u(x) g(x) \, dx \quad \text{for every } g \in L^\infty([0, 1]). \end{align*} Taking $g = \mathbb{1}_{[a, b]}$ for any interval $[a, b] \subset (0, 1]$, we compute \begin{align*} \int_0^1 u_{n_k}(x) \mathbb{1}_{[a, b]}(x) \, dx = n_k \cdot \mathcal{L}^1([0, 1/n_k] \cap [a, b]) = 0 \end{align*} for all $n_k > 1/a$, since $[0, 1/n_k]$ and $[a, b]$ become disjoint. Therefore $\int_a^b u(x) \, dx = 0$ for every interval $[a, b] \subset (0, 1]$, which forces $u = 0$ a.e. on $(0, 1]$. But taking $g = \mathbb{1}_{[0, 1]}$ gives \begin{align*} \int_0^1 u_{n_k}(x) \, dx = 1 \to \int_0^1 u(x) \, dx = 0, \end{align*} a contradiction. Therefore no subsequence of $\{u_n\}$ converges weakly in $L^1([0, 1])$. The mass "concentrates" at the origin and escapes to a Dirac measure, which does not belong to $L^1$. [/example]

This failure is not an accident of a poorly chosen sequence. It reflects a fundamental structural deficiency of $L^1$: the space is **not reflexive**. As we shall see, reflexivity is equivalent to weak sequential [compactness](/page/Compact%20Space) of bounded sets, and the example above is the prototypical obstruction. The theory of reflexive spaces provides precise criteria for when the extraction of weakly convergent subsequences is possible — and these criteria underpin the entire existence theory for PDEs via the direct method of the [calculus of variations](/page/Calculus%20of%20Variations).

## Definition

To formulate reflexivity precisely, we must first understand how a [Banach space](/page/Banach%20Space) relates to its double dual. Every element $x$ of a normed space $X$ defines a bounded linear functional on the dual space $X^*$ by evaluation: $f \mapsto f(x)$. This construction yields a natural map from $X$ into $X^{**} = (X^*)^*$, the **bidual** of $X$.

[definition: Canonical Embedding] Let $X$ be a [normed vector space](/page/Normed%20Vector%20Space) over $\mathbb{R}$ with topological dual%20Dual) $X^*$. The **canonical embedding** is the map \begin{align*} J: X &\to X^{**} \\ x &\mapsto J(x), \end{align*} where $J(x): X^* \to \mathbb{R}$ is the evaluation functional defined by \begin{align*} J(x)(f) := f(x) \quad \text{for all } f \in X^*. \end{align*} [/definition]

The canonical embedding is not an arbitrary construction — it is the unique natural way to place $X$ inside $X^{**}$. A fundamental result confirms that this map preserves all metric information.

[quotetheorem:875]

The inequality $\|J(x)\|_{X^{**}} \le \|x\|_X$ is immediate from the definition: for any $f \in X^*$ with $\|f\|_{X^*} \le 1$, we have $|J(x)(f)| = |f(x)| \le \|f\|_{X^*}\|x\|_X \le \|x\|_X$. The reverse inequality $\|J(x)\|_{X^{**}} \ge \|x\|_X$ is the substantial content, and it depends on the existence of a norming functional — a consequence of the Hahn-Banach Theorem. For any nonzero $x \in X$, the Hahn-Banach theorem produces $f_0 \in X^*$ with $\|f_0\|_{X^*} = 1$ and $f_0(x) = \|x\|_X$, so $\|J(x)\|_{X^{**}} \ge |J(x)(f_0)| = |f_0(x)| = \|x\|_X$.

The fact that $J$ is an isometry has an important consequence: $J(X)$ is a closed subspace of $X^{**}$ whenever $X$ is complete (since isometric images of complete spaces are complete). The question of reflexivity is whether this subspace exhausts the entire bidual.

[definition: Reflexive Space] A Banach space $X$ is called **reflexive** if the canonical embedding $J: X \to X^{**}$ is surjective, that is, if $J(X) = X^{**}$. [/definition]

Two points of caution are essential here.

First, reflexivity requires surjectivity of the **canonical** embedding $J$, not merely the existence of *some* isometric isomorphism between $X$ and $X^{**}$. R. C. James constructed a remarkable Banach space $\mathcal{J}$ that is isometrically isomorphic to its bidual $\mathcal{J}^{**}$ via a non-canonical map, yet the canonical embedding $J: \mathcal{J} \to \mathcal{J}^{**}$ has codimension one, so $\mathcal{J}$ is **not** reflexive. The definition of reflexivity is therefore intrinsically tied to the specific map $J$, not just to the abstract isomorphism class of $X$.

Second, reflexivity is a property of the Banach space as a whole, not of individual elements. A space is either reflexive or it is not; there is no useful notion of a "reflexive element."

[remark: Reflexivity and Completeness] The definition of reflexivity presupposes that $X$ is a Banach space (complete normed space). For incomplete normed spaces, the canonical embedding is still well-defined and still an isometry, but the image $J(X)$ may not be closed in $X^{**}$. Some authors define "reflexive" for general normed spaces, but this leads to pathologies — for instance, an incomplete normed space can have $J(X) = X^{**}$ only if $X$ is already complete. In this article, $X$ is always a Banach space. [/remark]

## Reflexive and Non-Reflexive Spaces

The definition of reflexivity is abstract — it asks whether every element of the bidual $X^{**}$ arises as evaluation at some point of $X$. To develop intuition for which spaces are reflexive, we examine the most important concrete function spaces, beginning with the [$L^p$ spaces](/page/%24L%5Ep%24%20Spaces) that are the workhorses of PDE theory.

### The $L^p$ Spaces for $1 < p < \infty$

The Riesz representation theorem for $L^p$ spaces provides a concrete identification of the dual. For a $\sigma$-finite measure space $(E, \mathcal{E}, \mu)$ and $1 \le p < \infty$ with Hölder conjugate $q = p/(p-1)$, the map $g \mapsto \Phi_g$ defined by $\Phi_g(f) = \int_E fg \, d\mu$ is an isometric isomorphism from $L^q(E)$ onto $(L^p(E))^*$.