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 $(L^p(E))^*$. [quotetheorem:901] The surjectivity of $\Phi$ is the deep content — it asserts that *every* bounded linear functional on $L^p$ arises by integration against an $L^q$ function, with no "singular" functionals. This is what fails for $p = \infty$: $(L^\infty)^*$ is strictly larger than $L^1$, containing singular functionals that cannot be represented by integration. The restriction $p < \infty$ (equivalently, $q > 1$) is essential for surjectivity. When $1 < p < \infty$, both $p$ and its Hölder conjugate $q = p/(p-1)$ satisfy $1 < q < \infty$. We can therefore apply the Riesz representation theorem twice: \begin{align*} (L^p)^* &\cong L^q \quad \text{(first application, since } 1 < p < \infty\text{)}, \\ (L^q)^* &\cong L^p \quad \text{(second application, since } 1 < q < \infty\text{)}. \end{align*} Composing these identifications, the bidual satisfies $(L^p)^{**} = ((L^p)^*)^* \cong (L^q)^* \cong L^p$. A careful verification — tracing through the two isomorphisms — confirms that this composite identification coincides with the canonical embedding $J$. Therefore $L^p(E)$ is reflexive for every $1 < p < \infty$. [example: Tracing the Canonical Embedding for $L^p$] We verify explicitly that the canonical embedding $J: L^p \to (L^p)^{**}$ corresponds to the identity under the double Riesz identification. Let $u \in L^p(E)$, $g \in L^q(E)$, and $f \in L^p(E)$. Step 1. Under the first Riesz isomorphism $\Phi: L^q \to (L^p)^*$, the functional $\Phi_g \in (L^p)^*$ acts by $\Phi_g(f) = \int_E fg \, d\mu$. Step 2. The canonical embedding sends $u \in L^p$ to $J(u) \in (L^p)^{**}$, which acts on $(L^p)^*$ by evaluation: \begin{align*} J(u)(\Phi_g) = \Phi_g(u) = \int_E ug \, d\mu. \end{align*} Step 3. Under the second Riesz isomorphism $\Psi: L^p \to (L^q)^*$, the functional $\Psi_u \in (L^q)^*$ acts by $\Psi_u(g) = \int_E ug \, d\mu$. Comparing Steps 2 and 3: $J(u)(\Phi_g) = \Psi_u(g)$ for every $g \in L^q$. Since $\Phi$ is an isomorphism, every element of $(L^p)^*$ is of the form $\Phi_g$, and thus $J(u)$ corresponds to $\Psi_u$ under the identification $(L^p)^{**} \cong (L^q)^*$, which in turn corresponds to $u$ under $\Psi^{-1}$. The composite map is the identity on $L^p$, confirming that $J$ is surjective. [/example] ### The Failure of $L^1$ and $L^\infty$ The argument above breaks down at the endpoints $p = 1$ and $p = \infty$ for fundamentally different reasons. **The case $p = 1$.** The Riesz representation theorem gives $(L^1)^* \cong L^\infty$. However, $(L^\infty)^*$ is strictly larger than $L^1$. The dual of $L^\infty$ contains not only the functionals $g \mapsto \int fg \, d\mu$ for $f \in L^1$, but also singular functionals — elements of $(L^\infty)^*$ that cannot be represented by integration against any $L^1$ function. These singular functionals arise from finitely additive set functions that are absolutely [continuous](/page/Continuity) with respect to $\mu$ but not countably additive. Their existence is guaranteed by the [Hahn-Banach Theorem](/theorems/879), which extends functionals defined on subspaces of $L^\infty$ (such as the space of continuous functions) to all of $L^\infty$. Concretely, consider $L^1([0, 1])$. The functional $\Lambda: L^\infty([0, 1]) \to \mathbb{R}$ defined as a Banach limit (a translation-invariant extension of the ordinary limit to bounded sequences, which can be constructed via the Hahn-Banach theorem) or a functional arising from an ultrafilter on $\mathbb{N}$ cannot be written as $\Lambda(g) = \int_0^1 fg \, dx$ for any $f \in L^1$. Since $(L^1)^{**} \cong (L^\infty)^*$ contains such singular functionals, the canonical embedding $J: L^1 \to (L^1)^{**}$ is not surjective. The opening example of this article illustrates the same phenomenon from the sequential side: the bounded sequence $n \cdot \mathbb{1}_{[0, 1/n]}$ in $L^1$ has no [weakly convergent](/page/Weak%20Convergence) subsequence, which — as we shall prove — is impossible in a reflexive space. **The case $p = \infty$.** The space $L^\infty$ is not reflexive either. One way to see this: the subspace $c_0$ of sequences converging to zero (viewed inside $\ell^\infty$) is a closed, [separable](/page/Separable%20Space)) subspace of $\ell^\infty$, but $\ell^\infty$ is not separable. If $\ell^\infty$ were reflexive, then $c_0$ would also be reflexive (since closed subspaces of reflexive spaces are reflexive, as we prove below), which would make $c_0^*$ separable. But $c_0^* \cong \ell^1$, and $\ell^1$ is separable — so this test does not immediately yield a contradiction. The actual failure is that $c_0$ is not reflexive: $c_0^{**} \cong (\ell^1)^* \cong \ell^\infty$, and the canonical embedding $J: c_0 \to \ell^\infty$ has image $c_0 \subsetneq \ell^\infty$. Since $c_0$ is a closed non-reflexive subspace of $\ell^\infty$, the space $\ell^\infty$ cannot be reflexive. ### Hilbert Spaces Every [Hilbert space](/page/Hilbert%20Space) $H$ is reflexive. The [Riesz representation theorem](/theorems/221) provides an isometric (conjugate-linear in the complex case, isometric in the real case) identification $H^* \cong H$. Applying this twice gives $H^{**} \cong H^* \cong H$, and tracing through the maps confirms that the composite agrees with the canonical embedding $J$. In particular, the [Sobolev spaces](/page/Sobolev%20Space) $H^k(U) = W^{k,2}(U)$ are reflexive for every $k \ge 0$. More generally, $W^{k,p}(U)$ is reflexive for $1 < p < \infty$, since $W^{k,p}(U)$ embeds isomorphically as a closed subspace of a finite product of $L^p$ spaces (via the map $u \mapsto (D^\alpha u)_{|\alpha| \le k}$), and finite products of reflexive spaces are reflexive. ### Summary Table The following table collects the reflexivity status of the most common function spaces. In each case, $(E, \mathcal{E}, \mu)$ is a $\sigma$-finite measure space with infinitely many atoms or non-atomic. | Space | Reflexive? | Reason | |---|---|---| | $L^p(E)$, $1 < p < \infty$ | Yes | Double Riesz representation | | $L^1(E)$ | No | $(L^\infty)^* \supsetneq L^1$ | | $L^\infty(E)$ | No | Contains non-reflexive closed subspace $c_0$ | | $\ell^p$, $1 < p < \infty$ | Yes | Special case of $L^p$ | | $\ell^1$ | No | Special case of $L^1$ | | $\ell^\infty$ | No | $c_0^{**} \cong \ell^\infty \supsetneq c_0$ | | $c_0$ | No | $c_0^{**} \cong \ell^\infty$ | | $H^k(U) = W^{k,2}(U)$ | Yes | Hilbert space | | $W^{k,p}(U)$, $1 < p < \infty$ | Yes | Closed subspace of products of $L^p$ | | $C([0,1])$ | No | $C([0,1])^* \cong \mathcal{M}([0,1])$, not reflexive | ## Weak Compactness and the Unit Ball The most important consequence of reflexivity is that it restores a form of [compactness](/page/Compact%20Space) to the unit ball — not in the norm topology (which is impossible in infinite dimensions), but in the [weak topology](/page/Weak%20Topology). This section develops the precise relationship between reflexivity and weak compactness, beginning with the foundational result that motivated much of the theory. ### The Banach-Alaoglu Theorem and Its Limitations The starting point is the Banach-Alaoglu theorem, which guarantees compactness in the [dual space](/page/Dual%20Space), but only in the [weak-$*$ topology](/page/Weak*%20Topology). [quotetheorem:212] This theorem applies to **every** normed space $X$, reflexive or not. The closed unit ball $B_{X^*}$ is always weak-$*$ compact. However, weak-$*$ compactness is a property of the dual $X^*$, not of $X$ itself. For the original space $X$, Banach-Alaoglu alone tells us nothing about weak compactness of $B_X$. The weak-$*$ topology $\sigma(X^*, X)$ is generally coarser than the weak topology $\sigma(X^*, X^{**})$. A net in $X^*$ converges weak-$*$ if it converges when tested against elements of $X$; it converges weakly if it converges when tested against all elements of $X^{**}$. Since $X \subset X^{**}$ (via the canonical embedding), weak convergence is a stronger requirement. Consequently, weak-$*$ compactness does not imply weak compactness unless $X = X^{**}$. To transfer compactness from $X^{**}$ back to $X$, we need the canonical embedding $J$ to be surjective. If $X$ is reflexive, then $J: X \to X^{**}$ is an isometric isomorphism, and the weak topology $\sigma(X, X^*)$ on $X$ corresponds exactly to the weak-$*$ topology $\sigma(X^{**}, X^*)$ on $X^{**}$ under this identification. In this case, Banach-Alaoglu applied to $X^{**}$ (with $X^*$ playing the role of the predual) yields weak-$*$ compactness of $B_{X^{**}}$, which — via $J^{-1}$ — translates to weak compactness of $B_X$. When $X$ is **not** reflexive, the image $J(B_X)$ is a proper subset of $B_{X^{**}}$. Although $B_{X^{**}}$ is weak-$*$ compact, the subset $J(B_X)$ is not closed in this topology (by Goldstine's Lemma, it is merely dense). So the compactness does not transfer. ### Goldstine's Lemma The gap between $J(B_X)$ and $B_{X^{**}}$ is precisely quantified by the following density result. [quotetheorem:898] Goldstine's Lemma says that the unit ball of $X$ is always "almost" as large as the unit ball of $X^{**}$ — its image is dense. The difference between reflexive and non-reflexive spaces is whether this dense set is actually the entire ball: - If $X$ is reflexive: $J(B_X) = B_{X^{**}}$ (closed and dense, hence equal). - If $X$ is not reflexive: $J(B_X) \subsetneq B_{X^{**}}$ (dense but not closed). This dichotomy illustrates a recurring theme in functional analysis: many structural questions reduce to whether a dense subspace is actually the whole space. Goldstine's Lemma ensures that the canonical image is always dense; reflexivity is the condition that promotes density to equality. ### Kakutani's Characterisation The interplay between reflexivity and weak compactness is captured precisely by Kakutani's theorem, which gives a clean topological characterisation of reflexivity. [quotetheorem:897] This result transforms reflexivity from an algebraic condition (surjectivity of $J$) into a topological one (weak compactness of the unit ball). The forward direction follows from the Banach-Alaoglu argument sketched above. The converse is more subtle: if $B_X$ is weakly compact, then $J(B_X)$ is a weak-$*$ compact (hence weak-$*$ closed) subset of $B_{X^{**}}$, and Goldstine's Lemma ensures it is also weak-$*$ dense. A subset that is both dense and closed in a topological space must be the entire space, so $J(B_X) = B_{X^{**}}$, which gives surjectivity of $J$. Note what Kakutani's theorem does *not* say: it does not assert that every weakly closed bounded set is weakly compact. It says only that the **closed unit ball** is weakly compact, and the equivalence with reflexivity means this is an all-or-nothing property of the space. However, once reflexivity is established, it follows that *every* bounded and weakly closed subset of $X$ is weakly compact (since any such set is a weakly closed subset of a scalar multiple of $B_X$). This applies, in particular, to closed bounded convex sets — a fact that is central to the direct method of the [calculus of variations](/page/Calculus%20of%20Variations). ### The Eberlein-Smulian Theorem In infinite-dimensional spaces, compactness and sequential compactness are generally distinct properties. A topological space can be compact without being sequentially compact, and vice versa. However, for the weak topology on a Banach space, these notions coincide — a remarkable fact due to Eberlein and Smulian. [quotetheorem:987] The equivalence $(1) \Leftrightarrow (2)$ is the most frequently used part. Combined with Kakutani's theorem, it yields the result that is most directly applied in PDE theory. [quotetheorem:214] This is the theorem that makes reflexivity indispensable for existence theory. In a reflexive Banach space, every bounded sequence has a [weakly convergent](/page/Weak%20Convergence) subsequence — the infinite-dimensional analogue of Bolzano-Weierstrass, with weak convergence replacing strong convergence. Without reflexivity, this fails, as the opening example in $L^1$ demonstrates. The converse direction is important: if a Banach space enjoys the property that every bounded sequence has a weakly convergent subsequence, then it must be reflexive. Weak sequential compactness of bounded sets is not a bonus feature that some reflexive spaces happen to have — it is a *characterisation* of reflexivity (for [Banach spaces](/page/Banach%20Space)). The logical chain is worth recording explicitly: 1. **Banach-Alaoglu** (all [normed spaces](/page/Normed%20Vector%20Space)): $B_{X^*}$ is weak-$*$ compact. 2. **Kakutani** (characterisation): $X$ reflexive $\Leftrightarrow$ $B_X$ is weakly compact. 3. **Eberlein-Smulian** (Banach spaces): weak compactness $\Leftrightarrow$ weak sequential compactness. 4. **Conclusion**: $X$ reflexive $\Leftrightarrow$ every bounded sequence in $X$ has a weakly convergent subsequence. ### Weak Lower Semicontinuity of the Norm Weak convergence is weaker than norm convergence — the norm can drop in the limit. This asymmetry is captured by the following result. [quotetheorem:215] The proof is a direct application of the definition of weak convergence and the Hahn-Banach theorem. For any $\varepsilon > 0$, choose $f \in X^*$ with $\|f\|_{X^*} = 1$ and $f(x) \ge \|x\|_X - \varepsilon$ (such $f$ exists by the Hahn-Banach theorem). Then $\|x\|_X - \varepsilon \le f(x) = \lim_n f(x_n) \le \liminf_n \|x_n\|_X$. Since $\varepsilon > 0$ is arbitrary, the inequality follows. The hypothesis that $X$ is a Banach space can be weakened to $X$ being a normed space; the result depends only on the Hahn-Banach theorem. The inequality can be strict. For instance, in $L^2([0, 2\pi])$, the sequence $u_n(x) = \sin(nx)$ converges weakly to $0$ (by the Riemann-Lebesgue lemma), yet $\|u_n\|_{L^2} = \sqrt{\pi}$ for every $n$, so $\|0\|_{L^2} = 0 < \sqrt{\pi} = \lim_n \|u_n\|_{L^2}$. The norm "leaks" through oscillation. This lower semicontinuity property is what allows the direct method to work: we minimise a functional $I(u)$, extract a weakly convergent subsequence $u_{n_k} \rightharpoonup u$, and then show $I(u) \le \liminf I(u_{n_k})$ — that is, the functional is weakly lower semicontinuous. The norm inequality is the simplest instance of this pattern. ## Sufficient Conditions for Reflexivity Verifying reflexivity directly from the definition — showing that every element of $X^{**}$ is in the range of $J$ — is rarely practical. Instead, one typically deduces reflexivity from more concrete geometric or structural properties of the space. This section collects the main sufficient conditions. ### Uniform Convexity and the Milman-Pettis Theorem A particularly clean sufficient condition for reflexivity comes from the geometry of the unit ball. A Banach space is **uniformly convex** if midpoints of line segments between points on the unit sphere are bounded away from the sphere — the unit ball has no "flat edges." [definition: Uniformly Convex Space] A Banach space $X$ is **uniformly convex** if for every $\varepsilon > 0$ there exists $\delta > 0$ such that for all $x, y \in X$ with $\|x\| \le 1$, $\|y\| \le 1$, and $\|x - y\| \ge \varepsilon$, we have \begin{align*} \left\| \frac{x + y}{2} \right\| \le 1 - \delta. \end{align*} [/definition] The function $\delta_X(\varepsilon) := \inf\{1 - \|(x+y)/2\| : \|x\| = \|y\| = 1, \|x - y\| \ge \varepsilon\}$ is the **modulus of convexity** of $X$. Uniform convexity asserts that $\delta_X(\varepsilon) > 0$ for every $\varepsilon > 0$. [example: Uniform Convexity of $L^p$ for $1 < p < \infty$] The spaces $L^p(E)$ for $1 < p < \infty$ are uniformly convex. This is the content of the **Clarkson inequalities**. For $2 \le p < \infty$ and $f, g \in L^p(E)$: \begin{align*} \left\| \frac{f + g}{2} \right\|_{L^p}^p + \left\| \frac{f - g}{2} \right\|_{L^p}^p \le \frac{1}{2}\left(\|f\|_{L^p}^p + \|g\|_{L^p}^p\right). \end{align*} If $\|f\|_{L^p} \le 1$, $\|g\|_{L^p} \le 1$, and $\|f - g\|_{L^p} \ge \varepsilon$, then \begin{align*} \left\| \frac{f + g}{2} \right\|_{L^p}^p \le 1 - \left(\frac{\varepsilon}{2}\right)^p, \end{align*} so $\|(f+g)/2\|_{L^p} \le (1 - (\varepsilon/2)^p)^{1/p} < 1$, confirming uniform convexity with an explicit modulus. The spaces $L^1$ and $L^\infty$ are **not** uniformly convex. In $L^1([0, 1])$, consider $f = 2 \cdot \mathbb{1}_{[0, 1/2]}$ and $g = 2 \cdot \mathbb{1}_{[1/2, 1]}$, both of which lie on the unit sphere: $\|f\|_{L^1} = \|g\|_{L^1} = 1$. Their difference satisfies $\|f - g\|_{L^1} = \|2\mathbb{1}_{[0,1/2]} - 2\mathbb{1}_{[1/2,1]}\|_{L^1} = 2$, so $f$ and $g$ are maximally separated. Yet the midpoint $(f + g)/2 = \mathbb{1}_{[0,1/2]} + \mathbb{1}_{[1/2,1]} = \mathbb{1}_{[0,1]}$ has $\|(f+g)/2\|_{L^1} = 1$. The midpoint lies on the unit sphere itself, so $\delta_{L^1}(2) = 0$, violating uniform convexity. Geometrically, the unit ball of $L^1$ has a "flat face" connecting $f$ and $g$. [/example] The Milman-Pettis theorem connects this geometric property to reflexivity. [quotetheorem:899] The converse is false: there exist reflexive [Banach spaces](/page/Banach%20Space) that are not uniformly convex. For example, the space $\ell^p \oplus_1 \ell^q$ (direct sum with the $\ell^1$ norm) for distinct $1 < p, q < \infty$ is reflexive (since both summands are reflexive) but not uniformly convex (the unit ball has "corners" coming from the $\ell^1$ direct sum structure). The Milman-Pettis theorem is therefore a strict implication, not a characterisation. The theorem provides the most common route to proving reflexivity of [$L^p$ spaces](/page/%24L%5Ep%24%20Spaces): rather than tracing through the Riesz representation theorem twice, one can simply invoke the Clarkson inequalities to establish uniform convexity of $L^p$ for $1 < p < \infty$, and then apply Milman-Pettis. ### Hereditary Properties When building new Banach spaces from existing ones — by taking subspaces, quotients, or products — a natural question arises: if the ingredient spaces are reflexive, is the result? And conversely, if a subspace or quotient fails to be reflexive, what does that say about the ambient space? These permanence questions are essential in practice because most function spaces encountered in PDE theory are constructed as subspaces or products of $L^p$ spaces, and one needs to know that reflexivity propagates through these constructions. [quotetheorem:996] This follows from the fact that the canonical embedding $J_Y: Y \to Y^{**}$ can be related to the restriction of $J_X: X \to X^{**}$, combined with the Hahn-Banach theorem to identify $Y^{**}$ with a quotient of $X^{**}$. The converse question — when does reflexivity of a subspace imply reflexivity of the ambient space — requires additional hypotheses. Reflexivity of a single subspace says nothing about the ambient space (every Banach space contains the reflexive subspace $\{0\}$). However, reflexivity is well-behaved with respect to quotients. [quotetheorem:997] The reflexivity of quotients can be understood through the dual characterisation: $(X/Y)^* \cong Y^\perp$ (the annihilator of $Y$ in $X^*$), and $(Y^\perp)^* \cong X^{**}/Y^{\perp\perp}$. When $X$ is reflexive, $X^{**} = J(X)$, and these identifications simplify to show that the canonical embedding of $X/Y$ is surjective. Together with the subspace result, this gives a complete picture of how reflexivity interacts with short exact sequences. [quotetheorem:998] Together, these three results show that reflexivity behaves well with respect to the short exact sequence $0 \to Y \to X \to X/Y \to 0$: $X$ is reflexive if and only if both $Y$ and $X/Y$ are reflexive. This is the **three-space property** of reflexivity. Additional permanence properties include: - **Finite products.** If $X_1, \ldots, X_n$ are reflexive Banach spaces, then $X_1 \times \cdots \times X_n$ (with any of the standard product norms) is reflexive. - **Isomorphic invariance.** If $X$ is reflexive and $Y$ is isomorphic to $X$ (as a Banach space), then $Y$ is reflexive. - **Duality.** $X$ is reflexive if and only if $X^*$ is reflexive. For the forward direction, if $J_X: X \to X^{**}$ is surjective, then a direct computation shows that $J_{X^*}: X^* \to X^{***}$ is surjective as well: given $\Phi \in X^{***}$, the composition $\Phi \circ J_X^{-*}$ defines an element of $X^{**}$ whose pre-image under $J_X$ provides the representing element. For the converse, if $X^*$ is reflexive, then $X^{**}$ is reflexive (by the forward direction), and since $J_X(X)$ is a closed subspace of the reflexive space $X^{**}$, the subspace $J_X(X)$ is reflexive. But $J_X$ is an isometric isomorphism from $X$ onto $J_X(X)$, so $X$ is reflexive. ## James' Characterisation The theorems of Kakutani and Eberlein-Smulian characterise reflexivity through [compactness](/page/Compact%20Space) properties of the unit ball. James' theorem offers a strikingly different perspective, linking reflexivity to the **norm-attaining** behaviour of bounded linear functionals. For a bounded linear functional $f \in X^*$ on a [Banach space](/page/Banach%20Space) $X$, the operator norm is defined as \begin{align*} \|f\|_{X^*} = \sup\{f(x) : x \in X, \, \|x\|_X \le 1\}. \end{align*} The supremum is always finite (by definition of boundedness), but it need not be attained — there may be no $x_0 \in X$ with $\|x_0\| \le 1$ and $f(x_0) = \|f\|_{X^*}$. James' theorem asserts that reflexivity is precisely the condition under which every functional achieves its supremum. [quotetheorem:999] The forward direction (reflexive implies norm-attaining) is a direct consequence of weak compactness. If $X$ is reflexive, the closed unit ball $B_X$ is weakly compact by Kakutani's theorem. The functional $f$ is weakly [continuous](/page/Continuity) (by definition of the [weak topology](/page/Weak%20Topology)). A continuous real-valued function on a compact set attains its supremum, so there exists $x_0 \in B_X$ with $f(x_0) = \sup_{B_X} f = \|f\|_{X^*}$. The converse direction is much harder and represents one of the deeper results in Banach space theory. The proof, due to R. C. James (1964), uses a combinatorial argument with basic sequences and is considerably more technical than the other characterisations of reflexivity. [example: A Non-Attaining Functional on $c_0$] We exhibit a bounded linear functional on $c_0$ that does not attain its norm, confirming that $c_0$ is not reflexive (by the contrapositive of James' theorem). Recall that $c_0 = \{x = (x_1, x_2, x_3, \ldots) \in \mathbb{R}^{\mathbb{N}} : \lim_{n \to \infty} x_n = 0\}$ with the supremum norm $\|x\|_\infty = \sup_n |x_n|$, and $c_0^* \cong \ell^1$ via the pairing $f_a(x) = \sum_{n=1}^\infty a_n x_n$ for $a = (a_1, a_2, \ldots) \in \ell^1$. Define $a \in \ell^1$ by $a_n = 2^{-n}$ for $n \in \mathbb{N}$, and let $f_a \in c_0^*$ be the corresponding functional: \begin{align*} f_a(x) = \sum_{n=1}^\infty \frac{x_n}{2^n}. \end{align*} The norm of $f_a$ is $\|f_a\|_{c_0^*} = \|a\|_{\ell^1} = \sum_{n=1}^\infty 2^{-n} = 1$. We claim that no $x \in c_0$ with $\|x\|_\infty \le 1$ achieves $f_a(x) = 1$. Suppose $x \in c_0$ satisfies $\|x\|_\infty \le 1$ and $f_a(x) = 1$. Since $|x_n| \le 1$ and $a_n > 0$, we have $a_n x_n \le a_n$ with equality if and only if $x_n = 1$. For $\sum a_n x_n = \sum a_n = 1$, we need $x_n = 1$ for all $n$. But the constant sequence $(1, 1, 1, \ldots)$ does not belong to $c_0$ (it does not converge to $0$). Therefore $f_a(x) < 1$ for every $x \in c_0$ with $\|x\|_\infty \le 1$. [/example] This example is instructive: the supremum of $f_a$ over $B_{c_0}$ is approached by the sequence $x_m = (\underbrace{1, \ldots, 1}_{m}, 0, 0, \ldots)$ with $f_a(x_m) = \sum_{n=1}^m 2^{-n} = 1 - 2^{-m} \to 1$, but the limit of this sequence in $\ell^\infty$ is $(1, 1, 1, \ldots)$, which lies in $\ell^\infty \setminus c_0$. The "missing" supremum escapes to the bidual. ## The Direct Method of the Calculus of Variations The primary application of reflexivity in PDE theory is through the **direct method** of the [calculus of variations](/page/Calculus%20of%20Variations) — a general strategy for proving existence of minimisers for functionals. Reflexivity provides the compactness step that makes the method work. The direct method proceeds as follows. Suppose we wish to minimise a functional $I: X \to \mathbb{R} \cup \{+\infty\}$ over a [Banach space](/page/Banach%20Space) $X$ (or over a closed convex subset $K \subset X$). The strategy has three steps: **Step 1: Boundedness.** Show that a minimising sequence $\{u_n\}_{n=1}^\infty$ (i.e., $I(u_n) \to \inf_X I$) is bounded in $X$. This typically requires **coercivity**: $I(u) \to +\infty$ as $\|u\|_X \to \infty$. **Step 2: Weak [compactness](/page/Compact%20Space).** Extract a [weakly convergent](/page/Weak%20Convergence) subsequence $u_{n_k} \rightharpoonup u$ in $X$. This step requires $X$ to be reflexive (or, more precisely, that the minimising sequence lies in a weakly compact set). **Step 3: Weak lower semicontinuity.** Show that $I(u) \le \liminf_{k \to \infty} I(u_{n_k})$. Combined with the definition of a minimising sequence, this gives $I(u) = \inf_X I$, so $u$ is a minimiser. [example: Existence of a Minimiser for a Model Variational Problem] Let $U \subset \mathbb{R}^n$ be a bounded [open set](/page/Open%20Set) with $C^1$ boundary, let $g \in L^2(U)$, and consider the functional $I: H^1_0(U) \to \mathbb{R}$ defined by \begin{align*} I(u) := \frac{1}{2} \int_U |\nabla u|^2 \, d\mathcal{L}^n - \int_U gu \, d\mathcal{L}^n. \end{align*} This is the energy functional associated with the Poisson equation $-\Delta u = g$ in $U$ with Dirichlet boundary conditions $u = 0$ on $\partial U$. We prove that $I$ attains its minimum on $H^1_0(U)$. **Step 1: Coercivity.** By the [Poincaré inequality](/theorems/76) for $W^{1,2}_0(U)$, there exists $C_P > 0$ such that $\|u\|_{L^2(U)} \le C_P \|\nabla u\|_{L^2(U)}$ for all $u \in H^1_0(U)$. Using the Cauchy-Schwarz inequality and then Young's inequality: \begin{align*} I(u) &= \frac{1}{2}\|\nabla u\|_{L^2}^2 - \int_U gu \, d\mathcal{L}^n \\ &\ge \frac{1}{2}\|\nabla u\|_{L^2}^2 - \|g\|_{L^2}\|u\|_{L^2} \\ &\ge \frac{1}{2}\|\nabla u\|_{L^2}^2 - C_P\|g\|_{L^2}\|\nabla u\|_{L^2} \\ &\ge \frac{1}{4}\|\nabla u\|_{L^2}^2 - C_P^2\|g\|_{L^2}^2, \end{align*} where the last step applies $ab \le \frac{1}{4}a^2 + b^2$ with $a = \|\nabla u\|_{L^2}$ and $b = C_P\|g\|_{L^2}$. Since $\|\nabla u\|_{L^2}$ is an equivalent norm on $H^1_0(U)$ (by the Poincaré inequality), $I(u) \to +\infty$ as $\|u\|_{H^1_0} \to \infty$. In particular, any minimising sequence $\{u_n\}$ with $I(u_n) \to \inf_{H^1_0} I > -\infty$ must satisfy $\sup_n \|\nabla u_n\|_{L^2} < \infty$. **Step 2: Weak compactness.** The space $H^1_0(U) = W^{1,2}_0(U)$ is a [Hilbert space](/page/Hilbert%20Space), hence reflexive. The bounded sequence $\{u_n\}$ therefore has a subsequence $u_{n_k} \rightharpoonup u$ in $H^1_0(U)$ by the Weak Sequential Compactness theorem stated above. **Step 3: Weak lower semicontinuity.** We must show $I(u) \le \liminf_k I(u_{n_k})$. We treat the two terms separately. For the linear term: since $u_{n_k} \rightharpoonup u$ in $H^1_0(U) \subset L^2(U)$ and $g \in L^2(U)$, the functional $v \mapsto \int_U gv \, d\mathcal{L}^n$ is a bounded linear functional on $H^1_0(U)$, so $\int_U gu_{n_k} \, d\mathcal{L}^n \to \int_U gu \, d\mathcal{L}^n$. For the quadratic term: by the weak lower semicontinuity of the norm applied to the sequence $\nabla u_{n_k} \rightharpoonup \nabla u$ in $L^2(U; \mathbb{R}^n)$: \begin{align*} \|\nabla u\|_{L^2}^2 \le \liminf_{k \to \infty} \|\nabla u_{n_k}\|_{L^2}^2. \end{align*} Combining: \begin{align*} I(u) = \frac{1}{2}\|\nabla u\|_{L^2}^2 - \int_U gu \, d\mathcal{L}^n \le \liminf_{k \to \infty} \left(\frac{1}{2}\|\nabla u_{n_k}\|_{L^2}^2 - \int_U gu_{n_k} \, d\mathcal{L}^n\right) = \liminf_{k \to \infty} I(u_{n_k}) = \inf_{H^1_0} I. \end{align*} Since $u \in H^1_0(U)$ and $I(u) \le \inf I$, we conclude $I(u) = \inf I$, so $u$ is a minimiser. The Euler-Lagrange equation for this minimiser recovers the weak formulation of $-\Delta u = g$: for any $v \in H^1_0(U)$, computing $\frac{d}{dt}I(u + tv)\big|_{t=0} = 0$ gives $\int_U \nabla u \cdot \nabla v \, d\mathcal{L}^n = \int_U gv \, d\mathcal{L}^n$. [/example] The structure of this argument — coercivity for boundedness, reflexivity for weak compactness, convexity or lower semicontinuity for passing to the limit — is the template for virtually all existence proofs in elliptic PDE theory and the calculus of variations. Without reflexivity, Step 2 fails, and the entire approach collapses. This is precisely why the spaces $L^p$ for $1 < p < \infty$ and the [Sobolev spaces](/page/Sobolev%20Space) $W^{k,p}$ for $1 < p < \infty$ are the natural setting for variational problems, while $L^1$ and $L^\infty$ require fundamentally different techniques (such as concentration-compactness or relaxation methods). ## Standard Arguments Involving Reflexivity This section collects the most common techniques that use reflexivity in practice. These patterns recur throughout functional analysis and PDE theory. ### Extracting Weak Limits from Bounded Sequences The most frequent use of reflexivity is the extraction of [weakly convergent](/page/Weak%20Convergence) subsequences. The argument pattern is: 1. Establish a uniform bound $\|u_n\|_X \le M$ (usually from an energy estimate or an a priori bound). 2. Invoke reflexivity of $X$ and the Weak Sequential [Compactness](/page/Compact%20Space) theorem to extract $u_{n_k} \rightharpoonup u$. 3. Identify the limit $u$ by passing to the limit in a weak formulation (using continuity of linear operators and lower semicontinuity of convex functionals). The identification step (3) often requires compact embeddings (such as [Rellich-Kondrachov](/theorems/64)) to upgrade weak convergence to strong convergence in a lower-order space. For instance, if $u_{n_k} \rightharpoonup u$ in $W^{1,p}(U)$ and $U$ is bounded with $C^1$ boundary, then $u_{n_k} \to u$ strongly in $L^q(U)$ for $q < p^*$, which allows us to pass to the limit in nonlinear terms. ### Weak Convergence and Nonlinear Terms A major difficulty in PDE theory is that weak convergence does not preserve nonlinear operations. If $u_n \rightharpoonup u$ in $L^p$, it does **not** follow that $u_n^2 \rightharpoonup u^2$ or that $f(u_n) \rightharpoonup f(u)$ for a nonlinear $f$. The sequence $u_n(x) = \sin(nx)$ in $L^2([0, 2\pi])$ converges weakly to $0$, but $u_n^2(x) = \sin^2(nx)$ converges weakly to $1/2$ (the average value), not to $0^2 = 0$. To handle nonlinearities, one combines reflexivity with: - **Compact embeddings** — to get strong convergence in a lower-order space, which is enough to pass to the limit in nonlinear terms that involve lower-order derivatives. - **Convexity arguments** — convex functionals are weakly lower semicontinuous, so $\int F(\nabla u) \, d\mathcal{L}^n \le \liminf \int F(\nabla u_n) \, d\mathcal{L}^n$ when $F$ is convex. - **Monotonicity methods** — for monotone operators $A: X \to X^*$, the Minty-Browder technique uses the inequality $\langle A(u_n) - A(v), u_n - v \rangle \ge 0$ to identify weak limits without strong convergence. ### Testing Reflexivity via Separability of the Dual A useful negative test for reflexivity exploits the following fact: if $X$ is reflexive and [separable](/page/Separable%20Space), then $X^*$ is separable. Equivalently, if $X$ is separable but $X^*$ is not, then $X$ is not reflexive. This provides a quick way to rule out reflexivity without computing the bidual. [example: $\ell^1$ is Not Reflexive via Separability] The space $\ell^1$ is separable — the [countable set](/page/Countable%20Set) of sequences with finitely many rational entries is dense. However, $(\ell^1)^* \cong \ell^\infty$, and $\ell^\infty$ is not separable: the uncountable collection $\{e_S : S \subset \mathbb{N}\}$ where $(e_S)_n = \mathbb{1}_S(n)$ satisfies $\|e_S - e_T\|_\infty = 1$ for $S \neq T$, so no countable set can be dense. Since $\ell^1$ is separable but $(\ell^1)^*$ is not, the space $\ell^1$ cannot be reflexive. Note the logical structure: reflexivity of $X$ implies that $X^*$ is reflexive, which (together with separability of $X^*$) would imply separability of $X^{**}$. But the key implication used here is simpler: if $X$ is reflexive, then $X$ is isomorphic to $X^{**}$, and if $X^*$ separates points of $X$ (which it always does), then separability of $X$ implies separability of $X^*$. For a proof, one uses the fact that the unit ball of $X^*$ in the [weak-$*$ topology](/page/Weak*%20Topology) is metrisable when $X$ is separable, and compact metrisable spaces are separable. [/example] ### Upgrading Weak-$*$ to Weak Convergence In a reflexive space, the weak and weak-$*$ topologies on $X^*$ coincide, because $X^{**} = X$ under the canonical embedding, so $\sigma(X^*, X^{**}) = \sigma(X^*, X)$. This simplification means that the [Sequential Banach-Alaoglu Theorem](/theorems/496), which provides weak-$*$ convergent subsequences in the dual of a separable space, automatically yields **weak** convergent subsequences when the predual is reflexive. This is useful in regularity theory: if we have a bounded sequence of derivatives $\{\nabla u_n\}$ in $L^p(U; \mathbb{R}^n)$ with $1 < p < \infty$, we can extract a weakly convergent subsequence $\nabla u_{n_k} \rightharpoonup v$ in $L^p$. The limit $v$ is then identified as $\nabla u$ (the [weak derivative](/page/Weak%20Derivative) of the limit of $u_n$) through the closedness of the gradient operator. ## References - H. Brezis, *Functional Analysis, [Sobolev Spaces](/page/Sobolev%20Space) and Partial Differential Equations*, Springer (2011). - J. Conway, *A Course in Functional Analysis*, 2nd edition, Springer (1990). - L. C. Evans, *Partial Differential Equations*, 2nd edition, AMS (2010). - R. C. James, "Characterizations of Reflexivity," *Studia Mathematica* **23** (1964), 205–216. - R. E. Megginson, *An Introduction to [Banach Space](/page/Banach%20Space) Theory*, Springer (1998). - W. Rudin, *Functional Analysis*, 2nd edition, McGraw-Hill (1991).