Every vector space $X$ has a set of linear functionals $f: X \to \mathbb{C}$ — the *algebraic dual* $X^\#$. But the algebraic dual is too large to be useful: it contains pathological functionals constructed by the axiom of choice that ignore the topology of $X$ entirely. In analysis, one works instead with the **topological dual** $X'$ (or $X^*$): the subspace of linear functionals that are *continuous* with respect to the topology on $X$. The passage from $X^\#$ to $X'$ is where the topology of the underlying space enters, and different topologies on $X$ produce different duals — this is the mechanism behind the chain of inclusions $\mathcal{S}'(\mathbb{R}^n) \hookrightarrow \mathcal{D}'(\mathbb{R}^n)$ in distribution theory. Having formed the dual space $X'$, one can equip it with several natural topologies — the norm topology (for [Banach spaces](/page/Banach%20Space)), the [weak-* topology](/page/Weak*%20Topology), and the strong dual topology — each giving different convergence properties and compactness theorems. The choice of topology on the dual is just as important as the choice of topology on $X$ itself, and the interplay between them governs existence arguments in PDE theory, the Fourier analysis of distributions, and the duality theory of locally convex spaces. [motivation] ## Motivation ### The Algebraic Dual Is Too Large Given a vector space $X$ over $\mathbb{C}$, the **algebraic dual** $X^\#$ is the set of all [linear maps](/page/Linear%20Map) $f: X \to \mathbb{C}$. For finite-dimensional spaces, $X^\#$ has the same dimension as $X$ and every linear functional is automatically continuous in any topology compatible with the vector space structure. But for infinite-dimensional spaces, $X^\#$ is enormous — it contains linear functionals that are wildly discontinuous. For instance, let $X = L^2([0,1], \mathcal{L}^1)$ and let $\{e_\alpha\}_{\alpha \in A}$ be a Hamel basis (an algebraic basis, whose existence requires the axiom of choice). Define $f: X \to \mathbb{C}$ by $f(e_\alpha) = \alpha$ for some indexing where the values are unbounded, then extend by linearity. The resulting functional is linear but discontinuous — it sends a bounded sequence to an unbounded sequence of numbers. Such functionals are useless in analysis because they do not respect [limits](/page/Limit): if $x_n \to x$ in $X$, there is no guarantee that $f(x_n) \to f(x)$. What we need is a notion of dual space that restricts attention to those functionals that are compatible with the topology of $X$ — so that evaluating a functional commutes with taking limits. ### The Topology of $X$ Determines Which Functionals Are "Good" A linear functional $f: X \to \mathbb{C}$ is continuous if and only if the preimage $f^{-1}(U)$ of every [open set](/page/Open%20Set) $U \subseteq \mathbb{C}$ is open in $X$. For locally convex spaces, this is equivalent to a bound $|f(x)| \le C \cdot p(x)$ for some continuous seminorm $p$ on $X$. The key point is: **different topologies on $X$ give different notions of [continuity](/page/Continuity), and hence different dual spaces.** A functional that is continuous with respect to one topology may be discontinuous with respect to another. This is precisely what happens in distribution theory. The same underlying vector space $C_c^\infty(\mathbb{R}^n)$ is dense in $\mathcal{S}(\mathbb{R}^n)$, so the elements being tested are "the same," but the topology on $\mathcal{S}$ is weaker than the topology on $\mathcal{D} = C_c^\infty$ (Schwartz [functions](/page/Function) must decay rapidly, while [test functions](/page/Test%20Function) need only have compact support). Dualising reverses the inclusion: more test functions means fewer continuous functionals, so $\mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$. The topology on $X$ controls the size of the dual. ### The Dual Itself Needs a Topology Once we have the dual space $X'$, we need a topology on it to do analysis — to talk about convergence of [sequences](/page/Sequence) of functionals, continuity of maps between dual spaces, and compactness. There is no single "correct" topology on $X'$; different choices serve different purposes. The norm topology (when available) gives the strongest convergence but the weakest compactness. The [weak-* topology](/page/Weak*%20Topology) gives weaker convergence but powerful compactness theorems like [Banach-Alaoglu](/theorems/212). The strong dual topology sits between the two and is needed for the general duality theory of locally convex spaces. The choice of topology on the dual is a tool, not a fixed structure — and understanding which topology to use in which context is one of the core skills in functional analysis. [/motivation] ## The Algebraic Dual We begin with the purely algebraic notion, which requires no topology. [definition:Algebraic Dual] Let $X$ be a vector space over $\mathbb{C}$. The **algebraic dual** of $X$ is the vector space \begin{align*} X^\# := \{f: X \to \mathbb{C} \mid f \text{ is linear}\}, \end{align*} equipped with the vector space operations $(f + g)(x) := f(x) + g(x)$ and $(\alpha f)(x) := \alpha f(x)$ for $f, g \in X^\#$, $\alpha \in \mathbb{C}$, and $x \in X$. [/definition] For finite-dimensional $X$ with $\dim X = n$, the algebraic dual $X^\#$ also has dimension $n$: if $\{e_1, \ldots, e_n\}$ is a basis for $X$, the coordinate functionals $e_i^*(x) := x_i$ (where $x = \sum_{i=1}^n x_i e_i$) form a basis for $X^\#$, called the **dual basis**. In this setting every linear functional is continuous in any norm, so the algebraic and topological duals coincide. The distinction between $X^\#$ and $X'$ is purely an infinite-dimensional phenomenon. [example:The Algebraic Dual Is Strictly Larger In Infinite Dimensions] Let $X = c_{00}$ be the space of sequences with finitely many nonzero terms, equipped with the $\ell^1$ norm $\|x\|_{\ell^1} = \sum_{k=1}^\infty |x_k|$. The topological dual $(c_{00}, \|\cdot\|_{\ell^1})^*$ is $\ell^\infty$: every bounded linear functional has the form $f(x) = \sum_{k=1}^\infty a_k x_k$ for a unique bounded sequence $(a_k) \in \ell^\infty$, with $\|f\| = \sup_k |a_k|$. Now consider the algebraic functional $g: c_{00} \to \mathbb{C}$ defined by $g(e_k) = k$ on the standard basis vectors $e_k$ and extended by linearity. Since $\|e_k\|_{\ell^1} = 1$ but $g(e_k) = k \to \infty$, the functional $g$ is unbounded: $\|g\| = \sup_{\|x\| \le 1} |g(x)| = \infty$. Therefore $g \in X^\# \setminus X^*$ — it is linear but not continuous. [/example] ## The Topological Dual ### Definition for Normed Spaces The simplest and most common setting is that of a normed space, where continuity of a linear functional is equivalent to boundedness. The topology of a normed space $(X, \|\cdot\|)$ gives a clean criterion for which linear functionals belong to the dual: a linear functional $f: X \to \mathbb{C}$ is continuous if and only if the quantity $\sup_{\|x\| \le 1} |f(x)|$ is finite. This quantity then serves as a norm on the dual space, making it a Banach space even if $X$ was not complete. [definition:Topological Dual Of A Normed Space] Let $(X, \|\cdot\|_X)$ be a [normed vector space](/page/Normed%20Vector%20Space) over $\mathbb{C}$. The **topological dual** (or simply **dual space**) of $X$ is \begin{align*} X^* := \{f: X \to \mathbb{C} \mid f \text{ is linear and continuous}\}, \end{align*} equipped with the **operator norm** \begin{align*} \|f\|_{X^*} := \sup_{\substack{x \in X \\ \|x\|_X \le 1}} |f(x)|. \end{align*} The space $(X^*, \|\cdot\|_{X^*})$ is a Banach space (complete normed space), regardless of whether $X$ itself is complete. [/definition] The completeness of $X^*$ follows from the completeness of $\mathbb{C}$: if $(f_n)$ is Cauchy in $X^*$, then for each $x \in X$ the sequence $f_n(x)$ is Cauchy in $\mathbb{C}$ (since $|f_n(x) - f_m(x)| \le \|f_n - f_m\|_{X^*} \|x\|_X$), hence convergent. The pointwise limit $f(x) := \lim_n f_n(x)$ is linear, and a standard $\varepsilon/3$ argument shows $\|f_n - f\|_{X^*} \to 0$. [example:Dual Spaces Of Classical Banach Spaces] **$\ell^p$ spaces.** For $1 \le p < \infty$, the dual of $\ell^p$ is $\ell^q$ where $1/p + 1/q = 1$. The pairing is \begin{align*} f_a(x) = \sum_{k=1}^\infty a_k x_k, \quad a = (a_k) \in \ell^q, \; x = (x_k) \in \ell^p, \end{align*} and the isometric isomorphism $(\ell^p)^* \cong \ell^q$ sends $f \in (\ell^p)^*$ to the sequence $a_k = f(e_k)$. For $p = 2$, this gives $(\ell^2)^* \cong \ell^2$ — a special case of the [Riesz Representation Theorem](/theorems/221). **$L^p$ spaces.** For $1 \le p < \infty$ and a $\sigma$-finite measure space $(E, \mathcal{E}, \mu)$, the dual of $L^p(E, \mu)$ is $L^q(E, \mu)$ with $1/p + 1/q = 1$. The pairing is \begin{align*} T_g(f) = \int_E f(x) g(x) \, d\mu(x), \quad g \in L^q, \; f \in L^p. \end{align*} The case $p = 1$ gives $(L^1)^* \cong L^\infty$. The dual of $L^\infty$ is *not* $L^1$ — it is a strictly larger space of finitely additive measures, which illustrates that the duality $L^p \leftrightarrow L^q$ breaks down at the endpoint. **$C_0(X)$.** For a locally compact Hausdorff space $X$, the dual of $C_0(X)$ (continuous functions vanishing at infinity) is the space of complex Radon measures $\mathcal{M}(X)$, by the Riesz-Markov-Kakutani theorem. This is the measure-theoretic instance of duality: continuous functions are tested against measures. [/example] ### Definition for Locally Convex Spaces For spaces that are not normed — such as $\mathcal{D}(\Omega)$, $\mathcal{S}(\mathbb{R}^n)$, or general [locally convex spaces](/page/Topological%20Vector%20Space) — continuity is characterised by seminorm bounds rather than a single norm. The definition of the topological dual generalises seamlessly. [definition:Topological Dual Of A Locally Convex Space] Let $(X, \{p_\alpha\}_{\alpha \in A})$ be a locally convex topological vector space with topology generated by a family of seminorms $\{p_\alpha\}_{\alpha \in A}$. The **topological dual** of $X$ is \begin{align*} X' := \{f: X \to \mathbb{C} \mid f \text{ is linear and continuous}\}. \end{align*} A linear functional $f: X \to \mathbb{C}$ belongs to $X'$ if and only if there exist finitely many indices $\alpha_1, \ldots, \alpha_m \in A$ and a constant $C > 0$ such that \begin{align*} |f(x)| \le C \cdot \max_{j=1,\ldots,m} p_{\alpha_j}(x) \quad \text{for all } x \in X. \end{align*} [/definition] The seminorm characterisation makes the dual "constructive": to verify that a given linear functional is continuous, one exhibits a finite set of seminorms and a constant. For normed spaces, there is only one seminorm (the norm), and the condition reduces to $|f(x)| \le C\|x\|$, recovering the operator norm definition. ### Why the Topology on $X$ Matters The same underlying set of "test elements" can carry different topologies, producing different duals. This is not an abstract curiosity — it is the mechanism that generates the hierarchy of distribution spaces. [example:The Distribution Theory Chain] Consider the following chain of [topological vector spaces](/page/Topological%20Vector%20Space) and their [topological](/page/Topology) duals, all built from smooth functions on $\mathbb{R}^n$: | Space $X$ | Topology on $X$ | Dual $X'$ | Name | |---|---|---|---| | $\mathcal{D}(\mathbb{R}^n) = C_c^\infty(\mathbb{R}^n)$ | Convergence of all [derivatives](/page/Derivative) on every compact set | $\mathcal{D}'(\mathbb{R}^n)$ | [Distributions](/page/Distribution) | | $\mathcal{S}(\mathbb{R}^n)$ | Convergence in all [Schwartz seminorms](/page/Schwartz%20Space) $\|\phi\|_{\alpha,\beta}$ | $\mathcal{S}'(\mathbb{R}^n)$ | [Tempered distributions](/page/Tempered%20Distributions) | | $\mathcal{E}(\mathbb{R}^n) = C^\infty(\mathbb{R}^n)$ | Convergence of all derivatives on every compact set (but no support restriction) | $\mathcal{E}'(\mathbb{R}^n)$ | Compactly supported distributions | The inclusion of test function spaces goes: \begin{align*} \mathcal{D}(\mathbb{R}^n) \hookrightarrow \mathcal{S}(\mathbb{R}^n) \hookrightarrow \mathcal{E}(\mathbb{R}^n), \end{align*} and each inclusion is continuous (the topology on the larger space is weaker). Dualising reverses the arrows: \begin{align*} \mathcal{E}'(\mathbb{R}^n) \hookrightarrow \mathcal{S}'(\mathbb{R}^n) \hookrightarrow \mathcal{D}'(\mathbb{R}^n). \end{align*} The reversal happens because a functional that is continuous on a larger space (with a weaker topology) must *a fortiori* be continuous on any subspace carrying a stronger topology. But a functional continuous on $\mathcal{D}$ — which only needs to control test functions with compact support — may fail to be continuous on $\mathcal{S}$, where it must also handle test functions with merely polynomial decay. Concretely: the distribution $T_{e^{e^x}}$ defined by $T_{e^{e^x}}(\phi) = \int e^{e^x} \phi(x) \, d\mathcal{L}^1(x)$ belongs to $\mathcal{D}'(\mathbb{R}) \setminus \mathcal{S}'(\mathbb{R})$ — it is continuous on $\mathcal{D}$ (compact support kills the growth) but not on $\mathcal{S}$ (polynomial decay cannot compensate doubly-exponential growth). See the [Tempered Distributions](/page/Tempered%20Distributions) page for the full verification. [/example] ## Topologies on the Dual Space Once the dual space $X'$ is defined as a set, the question arises: what topology should it carry? There is no single answer — different topologies serve different purposes. We describe the three principal choices. ### The Norm Topology (Banach Space Case) When $X$ is a normed space, the operator norm makes $X^*$ into a Banach space. Convergence $f_n \to f$ in the norm topology means $\|f_n - f\|_{X^*} \to 0$, which is **[uniform convergence](/page/Uniform%20Convergence) on the unit ball**: $\sup_{\|x\| \le 1} |f_n(x) - f(x)| \to 0$. This is the strongest natural topology on $X^*$ and gives the best convergence properties — but the unit ball $B_{X^*}$ is compact in this topology *only* if $X^*$ is finite-dimensional (by the Riesz lemma), making it useless for compactness arguments in infinite dimensions. ### The Weak-* Topology The [weak-* topology](/page/Weak*%20Topology) on $X'$ is the coarsest topology making every evaluation map $\mathrm{ev}_x: X' \to \mathbb{C}$, $f \mapsto f(x)$, continuous. Convergence $f_\lambda \to f$ in the weak-* topology means $f_\lambda(x) \to f(x)$ for every $x \in X$ — this is **pointwise convergence** on $X$. The weak-* topology is strictly weaker than the norm topology (in infinite dimensions), but it has a decisive advantage: the [Banach-Alaoglu theorem](/theorems/212) guarantees that the closed unit ball $B_{X^*}$ is compact in the weak-* topology. When $X$ is separable, this compactness is sequential — every bounded sequence in $X^*$ has a weak-* convergent subsequence ([Sequential Banach-Alaoglu](/theorems/496)) — which is the workhorse of existence arguments in PDE theory. ### The Strong Dual Topology For a locally convex space $X$, the **strong dual topology** (or **strong topology**) on $X'$ is defined by the seminorms \begin{align*} p_B(f) := \sup_{x \in B} |f(x)|, \end{align*} where $B$ ranges over all bounded subsets of $X$. Convergence $f_\lambda \to f$ means **uniform convergence on bounded [sets](/page/Set)**. For normed spaces, the bounded subsets are exactly the norm-bounded sets, and the strong dual topology coincides with the operator norm topology. For non-normable spaces like $\mathcal{D}(\Omega)$ or $\mathcal{S}(\mathbb{R}^n)$, the strong dual topology is generally finer than the weak-* topology but may not come from a norm. ### Choosing the Right Topology The three topologies form a hierarchy: weak-* $\subseteq$ strong $\subseteq$ norm (when a norm exists). The choice depends on what one needs: - **Weak-* topology** when compactness is needed (Banach-Alaoglu, existence of weak-* limits of bounded sequences). - **Strong/norm topology** when quantitative estimates are needed (operator norm bounds, Banach space structure of the dual). - **Intermediate topologies** arise in specific contexts: the topology of uniform convergence on compact sets is used on spaces of holomorphic functions, and Mackey topologies play a role in the general duality theory of locally convex spaces. [example:Weak-* Versus Norm Convergence In $\ell^1$] Let $X = c_0$ (sequences converging to zero) with the $\sup$ norm. Its dual is $X^* = \ell^1$. Consider the sequence $f_n = e_n \in \ell^1$ (the standard basis vectors). In the weak-* topology on $\ell^1 = (c_0)^*$: for any $x = (x_k) \in c_0$, \begin{align*} f_n(x) = x_n \to 0 \quad \text{as } n \to \infty \quad (\text{since } x \in c_0). \end{align*} So $f_n \overset{*}{\rightharpoonup} 0$ in the weak-* topology. But $\|f_n\|_{\ell^1} = 1$ for every $n$, so $\|f_n - 0\|_{\ell^1} = 1 \not\to 0$ — the sequence does *not* converge in the norm topology. The weak-* topology is strictly weaker. This gap between weak-* and norm convergence is not a deficiency — it is the source of compactness. The sequence $(f_n)$ is bounded in $\ell^1$, so by [Sequential Banach-Alaoglu](/theorems/496) (using the [separability](/page/Separable) of $c_0$), it has a weak-* convergent subsequence. In the norm topology, no subsequence converges (since $\|f_n - f_m\|_{\ell^1} = 2$ for $n \neq m$), so the unit ball is not norm-compact. [/example] ## The Riesz Representation Theorem: When $X \cong X^*$ For a general Banach space, the dual $X^*$ is a different space from $X$ — it may be strictly larger (as with $c_0^* = \ell^1 \neq c_0$) or carry a different structure. [Hilbert spaces](/page/Hilbert%20Space) are the exception: the inner product provides a canonical identification of $H$ with $H^*$, so the dual "is" the space itself. [quotetheorem:221] The isomorphism $\Phi: H^* \to H$ sends each functional to the unique vector that represents it. This is the reason Hilbert spaces are the natural setting for variational problems: the [Lax-Milgram theorem](/page/Second-Order%20Elliptic%20Equations) and the theory of weak solutions rely on identifying a functional (the right-hand side of a PDE) with an element of the space (the solution). For non-Hilbert Banach spaces, no such canonical identification exists, and the duality theory is more involved. The Riesz theorem also explains why the $L^p$ duality is "perfect" for $1 < p < \infty$: these spaces are reflexive ($X^{**} \cong X$), and for $p = 2$ the duality is isometric via the inner product. The failure of reflexivity for $L^1$ and $L^\infty$ (the dual of $L^\infty$ is strictly larger than $L^1$) is one of the principal technical difficulties in the theory of conservation laws and measure-valued solutions. ## Reflexivity and the Double Dual Given a normed space $X$, we can form the dual $X^*$, and then the dual of the dual $X^{**} = (X^*)^*$. There is always a canonical embedding $J: X \to X^{**}$ defined by evaluation: for $x \in X$, the functional $J(x) \in X^{**}$ acts on $f \in X^*$ by \begin{align*} J(x)(f) := f(x). \end{align*} This map is always an isometric embedding — $\|J(x)\|_{X^{**}} = \|x\|_X$ — but it need not be surjective. [definition:Reflexive Space] A Banach space $X$ is **reflexive** if the canonical embedding $J: X \to X^{**}$ is surjective: every bounded linear functional on $X^*$ arises from evaluation at some $x \in X$. [/definition] Reflexivity has profound consequences. In a reflexive space, the weak topology (generated by $X^*$) and the weak-* topology on $X^{**}$ coincide on $J(X) = X^{**}$, and the Banach-Alaoglu theorem applied to $X^{**}$ gives weak compactness of the unit ball of $X$ itself. This is the **Eberlein-Šmulian theorem**: in a reflexive Banach space, every bounded sequence has a weakly convergent subsequence. [example:Reflexive And Non-Reflexive Spaces] **Reflexive:** All $L^p$ spaces with $1 < p < \infty$ are reflexive. The double dual chain is $L^p \xrightarrow{J} (L^q)^* \cong L^p$, where $1/p + 1/q = 1$, and the canonical embedding $J$ is surjective. Hilbert spaces ($p = 2$) are a special case. **Non-reflexive:** $L^1$ is not reflexive: $(L^1)^* = L^\infty$, but $(L^\infty)^*$ strictly contains $L^1$ (it includes singular functionals like Banach limits). Similarly, $c_0$ is not reflexive: $c_0^* = \ell^1$ and $(\ell^1)^* = \ell^\infty \supsetneq c_0$. The failure of reflexivity for $\ell^1$ can be seen concretely. The functional $\Lambda \in (\ell^1)^* = \ell^\infty$ defined by $\Lambda(x) = \sum_{k=1}^\infty x_k$ is bounded on $\ell^1$ ($|\Lambda(x)| \le \|x\|_{\ell^1}$), so $\Lambda \in (\ell^1)^*$. But $\Lambda$ corresponds to the constant sequence $(1, 1, 1, \ldots) \in \ell^\infty$. The double dual functional $\Phi \in (\ell^\infty)^*$ that "evaluates at $(1,1,1,\ldots)$" — meaning $\Phi(a) = \sum_{k=1}^\infty a_k$ for $a \in \ell^1 \subset \ell^\infty$ — cannot be extended to all of $\ell^\infty$ as evaluation at a single element $x \in c_0$ (since $(1,1,\ldots) \notin c_0$). So the canonical embedding $J: c_0 \to (\ell^1)^* = \ell^\infty$ misses the constant sequence: $J(c_0) \subsetneq \ell^\infty$. [/example] ## The Duality Reversal Principle The distribution theory example illustrates a general principle that is worth isolating. [theorem:Duality Reversal For Continuous Inclusions] Let $X$ and $Y$ be locally convex topological vector spaces with a continuous linear injection $\iota: X \hookrightarrow Y$ (meaning $\iota$ is injective, linear, and continuous). Then the transpose map \begin{align*} \iota^*: Y' &\to X' \\ f &\mapsto f \circ \iota \end{align*} is a well-defined linear map. If $\iota(X)$ is dense in $Y$, then $\iota^*$ is injective: $Y'$ embeds into $X'$. [/theorem] The proof of well-definedness is immediate: if $f \in Y'$ is continuous on $Y$ and $\iota: X \to Y$ is continuous, then $f \circ \iota: X \to \mathbb{C}$ is a composition of continuous maps, hence continuous, so $f \circ \iota \in X'$. For injectivity when $\iota(X)$ is dense: if $f \circ \iota = 0$, then $f$ vanishes on the dense subspace $\iota(X)$, so by continuity $f = 0$ on all of $Y$. This abstract theorem explains the reversal of inclusions in distribution theory. The continuous inclusions $\mathcal{D}(\mathbb{R}^n) \hookrightarrow \mathcal{S}(\mathbb{R}^n) \hookrightarrow \mathcal{E}(\mathbb{R}^n)$ are dense (smooth compactly supported functions are dense in both $\mathcal{S}$ and $\mathcal{E}$), so the duality reversal gives injections $\mathcal{E}' \hookrightarrow \mathcal{S}' \hookrightarrow \mathcal{D}'$. The density condition is essential — without it, $\iota^*$ can have a kernel, and the dual inclusion may fail to be injective. [example:Duality Reversal For Sobolev Embeddings] The Sobolev embedding $H^1(\mathbb{R}^n) \hookrightarrow L^2(\mathbb{R}^n)$ is continuous and dense (smooth compactly supported functions are dense in both spaces). Identifying $L^2$ with its dual via the [Riesz theorem](/theorems/221), the duality reversal gives the **Gelfand triple** (or rigged Hilbert space): \begin{align*} H^1(\mathbb{R}^n) \hookrightarrow L^2(\mathbb{R}^n) \cong (L^2(\mathbb{R}^n))^* \hookrightarrow (H^1(\mathbb{R}^n))^* = H^{-1}(\mathbb{R}^n). \end{align*} The space $H^{-1}(\mathbb{R}^n)$ — the dual of $H^1$ — is a space of distributions: it contains objects like $\delta_0$ (in dimension $n \ge 3$, where $\delta_0 \in H^{-s}$ for $s > n/2$) that are rougher than $L^2$ functions. The chain $H^1 \hookrightarrow L^2 \hookrightarrow H^{-1}$ is the framework for weak solutions of elliptic PDEs: the right-hand side $f$ lives in $H^{-1}$, the solution $u$ lives in $H^1$, and the bilinear form $B[u, v] = (f, v)_{H^{-1} \times H^1}$ is the [weak formulation](/page/Second-Order%20Elliptic%20Equations). [/example] ## Problems [problem] Let $X = C([0,1])$ with the supremum norm. Define $f_n: X \to \mathbb{R}$ by $f_n(g) = g(1/n)$. Show that $f_n \in X^*$ with $\|f_n\|_{X^*} = 1$, that $f_n \overset{*}{\rightharpoonup} f_0$ in the weak-* topology where $f_0(g) = g(0)$, and that $\|f_n - f_0\|_{X^*} = 2$ for all $n$ (so the convergence is not in the norm topology). [/problem] [solution] **Step 1: $f_n \in X^*$ with $\|f_n\| = 1$.** The map $f_n(g) = g(1/n)$ is linear and satisfies $|f_n(g)| = |g(1/n)| \le \|g\|_\infty$, so $\|f_n\| \le 1$. Testing on the constant function $g \equiv 1$: $f_n(g) = 1$, so $\|f_n\| = 1$. **Step 2: Weak-* convergence.** For any $g \in C([0,1])$, continuity at $0$ gives $f_n(g) = g(1/n) \to g(0) = f_0(g)$ as $n \to \infty$. This is exactly $f_n \overset{*}{\rightharpoonup} f_0$. **Step 3: No norm convergence.** We must show $\|f_n - f_0\| = 2$. By the triangle inequality, $\|f_n - f_0\| \le \|f_n\| + \|f_0\| = 2$. For the reverse bound, construct $g_n \in C([0,1])$ with $\|g_n\|_\infty = 1$, $g_n(1/n) = 1$, and $g_n(0) = -1$: take $g_n$ to be linear on $[0, 1/n]$ with $g_n(0) = -1$, $g_n(1/n) = 1$, and $g_n \equiv 1$ on $[1/n, 1]$. Then \begin{align*} (f_n - f_0)(g_n) = g_n(1/n) - g_n(0) = 1 - (-1) = 2, \end{align*} so $\|f_n - f_0\| \ge 2$. Combined: $\|f_n - f_0\| = 2$ for all $n$. [/solution] [problem] Let $H$ be an infinite-dimensional separable Hilbert space with orthonormal basis $\{e_k\}_{k=1}^\infty$. Using the Riesz isomorphism $H^* \cong H$, show that $e_k \rightharpoonup 0$ weakly in $H$ (i.e., $(e_k, x)_H \to 0$ for every $x \in H$) but $\|e_k\| = 1$ does not converge to $0$. Conclude that the unit ball in $H$ is weakly sequentially compact (by reflexivity and [Sequential Banach-Alaoglu](/theorems/496)) but not norm-compact. [/problem] [solution] **Step 1: [Weak convergence](/page/Weak%20Convergence) to zero.** Let $x \in H$. By Parseval's identity, $\sum_{k=1}^\infty |(e_k, x)_H|^2 = \|x\|_H^2 < \infty$. Since the [series](/page/Series) converges, the general term must tend to zero: $(e_k, x)_H \to 0$ as $k \to \infty$. Via the Riesz isomorphism, the functional $f_k := (\cdot, e_k)_H \in H^*$ satisfies $f_k(x) = (x, e_k)_H \to 0$ for every $x$, so $f_k \overset{*}{\rightharpoonup} 0$. **Step 2: No norm convergence.** $\|e_k\|_H = 1$ for every $k$, so $\|e_k - 0\|_H = 1 \not\to 0$. **Step 3: [Weak sequential compactness](/theorems/214).** Since $H$ is reflexive (by the [Riesz Representation Theorem](/theorems/221), $H^{**} \cong H^* \cong H$), the Eberlein-Šmulian theorem guarantees that every bounded sequence in $H$ has a weakly convergent subsequence. The sequence $(e_k)$ is bounded ($\|e_k\| = 1$), and we have shown it converges weakly to $0$. **Step 4: Non-compactness in norm.** For $j \neq k$: $\|e_j - e_k\|_H^2 = \|e_j\|^2 + \|e_k\|^2 = 2$ (by orthogonality), so $\|e_j - e_k\|_H = \sqrt{2}$ for all $j \neq k$. No subsequence can be Cauchy in the norm topology, so the unit ball is not norm-compact. [/solution] [problem] Let $\iota: \mathcal{S}(\mathbb{R}^n) \hookrightarrow L^2(\mathbb{R}^n)$ be the natural inclusion. Using the Riesz isomorphism $(L^2)^* \cong L^2$ and the duality reversal, show that every $g \in L^2(\mathbb{R}^n)$ defines a tempered distribution, and compute the relevant seminorm bound. [/problem] [solution] **Step 1: The inclusion $\iota$ is continuous.** For $\phi \in \mathcal{S}(\mathbb{R}^n)$, choose $N > n/2$. Then $|\phi(x)| \le \|\phi\|_{N,0} (1 + |x|)^{-N}$ where $\|\phi\|_{N,0} = \sup_x (1+|x|)^N |\phi(x)|$, and $(1+|x|)^{-N} \in L^2(\mathbb{R}^n)$ since $2N > n$. Therefore \begin{align*} \|\phi\|_{L^2}^2 = \int_{\mathbb{R}^n} |\phi(x)|^2 \, d\mathcal{L}^n(x) \le \|\phi\|_{N,0}^2 \int_{\mathbb{R}^n} (1+|x|)^{-2N} \, d\mathcal{L}^n(x) = C_N^2 \|\phi\|_{N,0}^2, \end{align*} confirming $\|\iota(\phi)\|_{L^2} \le C_N \|\phi\|_{N,0}$. **Step 2: Duality reversal.** The transpose $\iota^*: (L^2)^* \to \mathcal{S}'$ sends $f \in (L^2)^*$ to $f \circ \iota \in \mathcal{S}'$. Via the Riesz isomorphism, each $g \in L^2$ corresponds to the functional $T_g(h) = \int g \bar{h} \, d\mathcal{L}^n$, and the composition gives \begin{align*} (T_g \circ \iota)(\phi) = \int_{\mathbb{R}^n} g(x) \overline{\phi(x)} \, d\mathcal{L}^n(x) \quad \text{for } \phi \in \mathcal{S}(\mathbb{R}^n). \end{align*} This is a tempered distribution (since $\iota^*$ maps into $\mathcal{S}'$ by the duality reversal theorem). **Step 3: Seminorm bound.** By the Cauchy-Schwarz inequality and Step 1: \begin{align*} |T_g(\phi)| \le \|g\|_{L^2} \|\phi\|_{L^2} \le C_N \|g\|_{L^2} \|\phi\|_{N,0}. \end{align*} This is the [characterisation bound](/theorems/456) with $C = C_N\|g\|_{L^2}$, $N$ as chosen, and $M = 0$ (no derivatives of $\phi$ needed). The embedding $L^2(\mathbb{R}^n) \hookrightarrow \mathcal{S}'(\mathbb{R}^n)$ is continuous: if $g_k \to g$ in $L^2$, then $T_{g_k}(\phi) \to T_g(\phi)$ for every $\phi \in \mathcal{S}$. [/solution] ## References - W. Rudin, *Functional Analysis*, 2nd ed. (1991). - J. B. Conway, *A Course in Functional Analysis*, 2nd ed. (1990). - H. Brezis, *Functional Analysis, [Sobolev Spaces](/page/Sobolev%20Space) and Partial Differential Equations* (2011). - F. Trèves, *Topological Vector Spaces, Distributions and Kernels* (1967). - L. Hörmander, *The Analysis of Linear Partial Differential Operators I* (1983).