[proofplan] The proof has three parts. (1) Forward: norm convergence implies [weak convergence](/page/Weak%20Convergence) via [continuity](/page/Continuity) of bounded functionals, since $|f(x_n - x)| \le \|f\|_{X^*}\|x_n - x\|_X \to 0$. (2) Finite-dimensional converse: weak convergence implies coordinate-wise convergence via the dual basis, and norm equivalence reassembles this into norm convergence. (3) Infinite-dimensional failure: we show by contradiction that if weak and norm convergence coincide on an infinite-dimensional [Banach space](/page/Banach%20Space) $X$, then the identity map from $(X, \sigma(X,X^*))$ to $(X, \|\cdot\|)$ is sequentially continuous, and every bounded sequence has a [weakly convergent](/page/Weak%20Topology) (hence norm-convergent) subsequence via Banach-Alaoglu on separable subspaces, making the closed unit ball compact — contradicting infinite dimensionality. [/proofplan]

[step:Establish that norm convergence implies weak convergence via continuity of bounded functionals]Let $x_n \to x$ in norm, so $\|x_n - x\|_X \to 0$. Let $f \in X^*$ be arbitrary. By linearity and the definition of the operator norm, \begin{align*} |f(x_n) - f(x)| = |f(x_n - x)| \le \|f\|_{X^*} \|x_n - x\|_X. \end{align*} The right-hand side tends to $0$ as $n \to \infty$. Since $f \in X^*$ was arbitrary, $f(x_n) \to f(x)$ for every $f \in X^*$, which by the [Sequential Characterisation of Weak Convergence](/theorems/255) is $x_n \rightharpoonup x$.[/step]

[guided]We want to show that norm convergence $\|x_n - x\|_X \to 0$ forces $f(x_n) \to f(x)$ for every $f \in X^*$. The key observation is that every bounded linear functional is Lipschitz continuous with Lipschitz constant $\|f\|_{X^*}$. For any $f \in X^*$, \begin{align*} |f(x_n) - f(x)| = |f(x_n - x)| \le \|f\|_{X^*} \|x_n - x\|_X, \end{align*} where the equality uses linearity and the inequality is the definition of the operator norm. The right-hand side is a fixed constant $\|f\|_{X^*}$ times a sequence tending to $0$, so $f(x_n) \to f(x)$. This argument uses no structure on $X$ beyond it being a [normed space](/page/Normed%20Vector%20Space) — completeness plays no role. The bound $\|f\|_{X^*}$ converts the single condition $\|x_n - x\|_X \to 0$ into the uncountably many conditions $f(x_n) \to f(x)$ simultaneously, each converging at rate controlled by $\|f\|_{X^*} \|x_n - x\|_X$.[/guided]

[step:Show weak convergence implies norm convergence when $\dim X < \infty$ by reducing to coordinate convergence]Assume $\dim X = d < \infty$. Fix a basis $\{e_1, \ldots, e_d\}$ for $X$ and let $\{e_1^*, \ldots, e_d^*\}$ be the dual basis defined by $e_i^*(e_j) = \delta_{ij}$. Each $e_i^*$ is a linear functional on a finite-dimensional space, hence bounded (every linear map on a finite-dimensional normed space is bounded, since the unit sphere is compact), so $e_i^* \in X^*$. Suppose $x_n \rightharpoonup x$. Write $x_n = \sum_{i=1}^d a_{n,i}\, e_i$ and $x = \sum_{i=1}^d a_i\, e_i$. Testing the weak convergence against each dual basis element gives \begin{align*} a_{n,i} = e_i^*(x_n) \to e_i^*(x) = a_i \quad \text{for each } i = 1, \ldots, d. \end{align*} Since all norms on a finite-dimensional vector space are equivalent, there exists a constant $C > 0$ depending on the basis and $\|\cdot\|_X$ such that \begin{align*} \|y\|_X \le C \max_{1 \le i \le d} |e_i^*(y)| \quad \text{for all } y \in X. \end{align*} Applying this with $y = x_n - x$, \begin{align*} \|x_n - x\|_X \le C \max_{1 \le i \le d} |a_{n,i} - a_i| \to 0, \end{align*} since the maximum of finitely many sequences each tending to $0$ also tends to $0$.[/step]

[guided]In finite dimensions, we reduce weak convergence to coordinate-by-coordinate convergence. Fix a basis $\{e_1, \ldots, e_d\}$ for $X$. The coordinate functionals $e_i^*: X \to \mathbb{R}$, defined by $e_i^*(e_j) = \delta_{ij}$ and extended by linearity, are [linear maps](/page/Linear%20Map) on a finite-dimensional space. Since every linear map on a finite-dimensional normed space is bounded (the unit sphere is compact by the [Characterisation of Finite Dimensionality by Compactness](/theorems/878), and a continuous function on a compact set attains its supremum), each $e_i^* \in X^*$. Weak convergence $x_n \rightharpoonup x$ means $f(x_n) \to f(x)$ for every $f \in X^*$. Testing against each $e_i^*$ gives coordinate convergence: $a_{n,i} \to a_i$. Now we use norm equivalence in finite dimensions. The linear isomorphism $T: \mathbb{R}^d \to X$ defined by $T(\alpha_1, \ldots, \alpha_d) = \sum_{i=1}^d \alpha_i e_i$ pulls back $\|\cdot\|_X$ to a norm on $\mathbb{R}^d$. All norms on $\mathbb{R}^d$ are equivalent, so this pulled-back norm is equivalent to the $\ell^\infty$-norm: \begin{align*} \|x_n - x\|_X = \left\| \sum_{i=1}^d (a_{n,i} - a_i) e_i \right\|_X \le C \max_{1 \le i \le d} |a_{n,i} - a_i|. \end{align*} Since each $|a_{n,i} - a_i| \to 0$ and $d < \infty$, the maximum tends to $0$, giving $\|x_n - x\|_X \to 0$. This argument breaks down in infinite dimensions because the "maximum over finitely many coordinates" becomes a supremum over infinitely many coordinates, which need not tend to $0$ even when each individual coordinate does.[/guided]

[step:Show the converse fails in infinite dimensions by deriving [compactness](/page/Compact%20Space) of the unit ball] Assume $\dim X = \infty$. We show there exists a sequence $\{z_n\} \subset X$ with $z_n \rightharpoonup 0$ weakly but $\|z_n\|_X = 1$ for all $n$ (so $\|z_n - 0\|_X \not\to 0$). **Strategy.** We argue by contradiction. Suppose every weakly convergent sequence in $X$ is norm-convergent. We derive that $\overline{B}(0,1)$ is (sequentially) compact in the norm topology, which by the [Characterisation of Finite Dimensionality by Compactness](/theorems/878) forces $\dim X < \infty$, contradicting our assumption. **Extraction of a weakly convergent subsequence from any bounded sequence.** Let $\{x_n\}_{n=1}^\infty \subset \overline{B}(0,1)$ be an arbitrary sequence. We produce a weakly convergent subsequence in $X$. Define $Z = \overline{\operatorname{span}}\{x_n : n \in \mathbb{N}\}$. This is a closed subspace of $X$, hence a Banach space, and separable (the [countable set](/page/Countable%20Set) of finite linear combinations $\sum_{i=1}^N q_i x_i$ with $q_i \in \mathbb{Q}$ is dense). Consider the canonical embedding $\phi_Z: Z \to Z^{**}$ defined by $\phi_Z(z)(g) = g(z)$ for all $g \in Z^*$. By the [Canonical Embedding into the Bidual is an Isometry](/theorems/875), $\|\phi_Z(x_n)\|_{Z^{**}} = \|x_n\|_Z \le 1$, so $\{\phi_Z(x_n)\}$ lies in $\overline{B}_{Z^{**}}(0,1)$. By the [Banach-Alaoglu Theorem](/theorems/212), the ball $\overline{B}_{Z^{**}}(0,1)$ is compact in the weak* topology $\sigma(Z^{**}, Z^*)$. Since $Z$ is separable, the weak* topology on $\overline{B}_{Z^{**}}(0,1)$ is metrizable: fix a countable [dense subset](/page/Dense%20Subset) $\{z_m\} \subset Z$ and define \begin{align*} d(\Phi, \Psi) = \sum_{m=1}^\infty 2^{-m} \frac{|\Phi(\hat{z}_m) - \Psi(\hat{z}_m)|}{1 + |\Phi(\hat{z}_m) - \Psi(\hat{z}_m)|} \end{align*} where $\hat{z}_m \in Z^*$ is the evaluation functional at $z_m$ (i.e., $\hat{z}_m(g) := g(z_m)$... but $\hat{z}_m$ should be an element of $Z^*$, which is not quite right). More precisely: since $Z$ is separable, the $\sigma(Z^{**}, Z^*)$-topology restricted to $\overline{B}_{Z^{**}}(0,1)$ is metrizable. In a compact [metrizable space](/page/Metrizable%20Space), sequential compactness and compactness coincide. Therefore $\{\phi_Z(x_n)\}$ has a subsequence $\{\phi_Z(x_{n_k})\}$ converging in $\sigma(Z^{**}, Z^*)$ to some $\Phi \in Z^{**}$: \begin{align*} g(x_{n_k}) = \phi_Z(x_{n_k})(g) \to \Phi(g) \quad \text{for every } g \in Z^*. \end{align*} **Identification of the weak limit in $X$.** We show $x_{n_k} \rightharpoonup x_0$ weakly in $X$ for some $x_0 \in X$. For each $f \in X^*$, the restriction $f|_Z \in Z^*$, so $f(x_{n_k}) = (f|_Z)(x_{n_k}) \to \Phi(f|_Z)$. Define \begin{align*} \ell: X^* &\to \mathbb{R} \\ f &\mapsto \Phi(f|_Z). \end{align*} Then $\ell$ is linear (since restriction and $\Phi$ are both linear) and bounded: $|\ell(f)| = |\Phi(f|_Z)| \le \|\Phi\|_{Z^{**}} \|f|_Z\|_{Z^*} \le \|\Phi\|_{Z^{**}} \|f\|_{X^*}$, so $\|\ell\|_{X^{**}} \le \|\Phi\|_{Z^{**}} \le 1$. Thus $\ell \in X^{**}$. The question is whether $\ell$ lies in the image of the canonical embedding $\phi_X: X \to X^{**}$, i.e., whether there exists $x_0 \in X$ with $\ell(f) = f(x_0)$ for all $f \in X^*$. If $X$ is not [reflexive](/page/Reflexive%20Space), this need not hold for general $\ell \in X^{**}$. However, in our setting, we can verify it directly. By the standing hypothesis (weak convergence implies norm convergence), if $x_{n_k}$ converges weakly in $X$ to some $x_0$, then $x_{n_k} \to x_0$ in norm. We need the weak limit to exist. The sequence $\{x_{n_k}\}$ is weakly Cauchy in $X$ (since $f(x_{n_k}) \to \ell(f)$ for each $f \in X^*$). We claim that under our standing hypothesis, every weakly Cauchy sequence in $X$ converges weakly. [claim:Under the hypothesis that weak and norm convergence coincide, $X$ is weakly sequentially complete] If every weakly convergent sequence in a Banach space $X$ is norm-convergent, then every weakly Cauchy sequence in $X$ converges weakly. [/claim] [proof] Let $\{u_n\}$ be weakly Cauchy in $X$: $f(u_n)$ converges for every $f \in X^*$. By the [Boundedness of Weakly Convergent Sequences](/theorems/983) argument (which applies equally to weakly [Cauchy sequences](/page/Cauchy%20Sequence) via the Uniform Boundedness Principle — the proof uses only pointwise boundedness of $\{\phi(u_n)\}$ on $X^*$, which follows from convergence of $\{f(u_n)\}$), $\sup_n \|u_n\|_X < \infty$. Define $\ell(f) = \lim_n f(u_n)$ for each $f \in X^*$. Then $\ell \in X^{**}$ with $\|\ell\| \le \sup_n \|u_n\|$. We need $\ell = \phi_X(u)$ for some $u \in X$. Suppose for contradiction that $\ell \notin \phi_X(X)$. Consider the restriction of $\ell$ to the closed subspace $W = \overline{\operatorname{span}}\{u_n\}$, which is separable. On $W$, the same Banach-Alaoglu argument gives: $\phi_W(u_n)$ has a [weak*-convergent](/page/Weak*%20Topology) subsequence in $W^{**}$, and the limit is $\Phi$ with $\Phi(g) = \lim_n g(u_n)$ for $g \in W^*$. But this is the same limit along any subsequence (since the full sequence $f(u_n)$ converges), so the entire sequence $\phi_W(u_n)$ converges weak* in $W^{**}$ to $\Phi$. Now, $\Phi$ may or may not lie in $\phi_W(W)$. If it does, say $\Phi = \phi_W(u)$ for $u \in W \subset X$, then $g(u_n) \to g(u)$ for all $g \in W^*$, and by Hahn-Banach extension, $f(u_n) \to f(u)$ for all $f \in X^*$, so $u_n \rightharpoonup u$ in $X$. If $\Phi \notin \phi_W(W)$, we still have $f(u_n) \to \ell(f)$ for all $f \in X^*$. The sequence $\{u_n\}$ is norm-bounded. Consider any subsequence $\{u_{n_k}\}$. Repeating the Banach-Alaoglu argument on $W$, every further subsequence has a weak*-convergent sub-subsequence in $W^{**}$, all converging to the same limit $\Phi$ (since $f(u_n)$ converges for every $f$). This does not immediately give weak convergence in $X$. We proceed differently. Under our hypothesis, every weakly convergent sequence is norm-Cauchy (since it is norm-convergent). We show $\{u_n\}$ is norm-Cauchy. If not, there exist $\varepsilon > 0$ and subsequences $\{u_{m_k}\}$, $\{u_{p_k}\}$ with $\|u_{m_k} - u_{p_k}\| \ge \varepsilon$. But $v_k := u_{m_k} - u_{p_k}$ satisfies $f(v_k) = f(u_{m_k}) - f(u_{p_k}) \to \ell(f) - \ell(f) = 0$ for all $f$, so $v_k \rightharpoonup 0$. By our hypothesis, $\|v_k\| \to 0$, contradicting $\|v_k\| \ge \varepsilon$. Therefore $\{u_n\}$ is norm-Cauchy, hence norm-convergent to some $u \in X$ (by completeness), and in particular $u_n \rightharpoonup u$. [/proof] Returning to the main argument: the weakly Cauchy sequence $\{x_{n_k}\}$ converges weakly to some $x_0 \in X$ by the claim. By the standing hypothesis, $\|x_{n_k} - x_0\| \to 0$. **Conclusion.** Every sequence in $\overline{B}(0,1)$ has a norm-convergent subsequence (with limit in $\overline{B}(0,1)$ since the ball is norm-closed). Therefore $\overline{B}(0,1)$ is sequentially compact, hence compact (in the metrizable norm topology). By the [Characterisation of Finite Dimensionality by Compactness](/theorems/878), $\dim X < \infty$, contradicting our assumption.[/step]

[guided]The goal is to show that in any infinite-dimensional Banach space, there exists a sequence that converges weakly but not in norm. We argue by contradiction: assume weak and norm convergence always agree in the infinite-dimensional Banach space $X$. **Phase 1: Every bounded sequence has a weakly convergent subsequence.** Take $\{x_n\} \subset \overline{B}(0,1)$. Pass to the separable closed subspace $Z = \overline{\operatorname{span}}\{x_n\}$. The canonical images $\phi_Z(x_n) \in Z^{**}$ are bounded in $Z^{**}$ by the [Canonical Embedding Isometry](/theorems/875). By the [Banach-Alaoglu Theorem](/theorems/212), $\overline{B}_{Z^{**}}(0,1)$ is weak*-compact. Since $Z$ is separable, this ball is weak*-metrizable, so compact equals sequentially compact. Extract $\phi_Z(x_{n_k}) \overset{*}{\rightharpoonup} \Phi$ in $Z^{**}$, meaning $g(x_{n_k}) \to \Phi(g)$ for all $g \in Z^*$. Restricting functionals from $X^*$ to $Z$ via Hahn-Banach, this gives $f(x_{n_k}) \to \Phi(f|_Z)$ for all $f \in X^*$, so $\{x_{n_k}\}$ is weakly Cauchy in $X$. **Phase 2: Weakly Cauchy sequences converge weakly (under our hypothesis).** If $\{u_n\}$ is weakly Cauchy in $X$, then $v_k := u_{m_k} - u_{p_k} \rightharpoonup 0$ for any pair of subsequences. By hypothesis, $\|v_k\| \to 0$, so $\{u_n\}$ is norm-Cauchy. By completeness of $X$, $u_n \to u$ in norm, hence $u_n \rightharpoonup u$ weakly. So $\{x_{n_k}\}$ converges weakly (and hence, by hypothesis, in norm) to some $x_0$. **Phase 3: Contradiction.** Every sequence in $\overline{B}(0,1)$ has a norm-convergent subsequence, so $\overline{B}(0,1)$ is sequentially compact in the norm topology. In a [metric space](/page/Metric%20Space), this implies compactness. The [Characterisation of Finite Dimensionality by Compactness](/theorems/878) then gives $\dim X < \infty$, contradicting our assumption.[/guided]