[proofplan]
We prove the two directions separately. The forward direction (reflexive implies norm-attaining) is a direct application of weak compactness: [reflexivity](/page/Reflexive%20Space) makes the closed unit ball weakly compact (by the [Characterisation of Reflexivity](/theorems/977)), and a [continuous](/page/Continuity) linear functional attains its supremum on a weakly compact set. The converse (norm-attaining implies reflexive) is a deep result due to James (1964); we state it as a claim and note that the proof requires basic sequence techniques beyond the scope of this argument.
[/proofplan]
[step:Show that reflexivity implies every functional attains its norm]
Assume $X$ is reflexive. Let $f \in X^*$ with $f \neq 0$ (the case $f = 0$ is immediate: take any $x_0$ with $\|x_0\| \le 1$). We must find $x_0 \in X$ with $\|x_0\|_X \le 1$ and $f(x_0) = \|f\|_{X^*}$.
By the [Characterisation of Reflexivity](/theorems/977) (implication $(1) \Rightarrow (2)$), the closed unit ball $\overline{B}_X := \{x \in X : \|x\|_X \le 1\}$ is compact in the [weak topology](/page/Weak%20Topology) $\sigma(X, X^*)$. The functional $f: X \to \mathbb{R}$ is weakly continuous, since $f \in X^*$ and the weak topology is defined as the coarsest topology making all elements of $X^*$ continuous.
Since $\overline{B}_X$ is $\sigma(X, X^*)$-compact and $f$ is $\sigma(X, X^*)$-continuous, the image $f(\overline{B}_X) \subset \mathbb{R}$ is compact (the continuous image of a compact set is compact). In particular, $f(\overline{B}_X)$ is a [compact subset](/page/Compact%20Space) of $\mathbb{R}$, so it attains its supremum: there exists $x_0 \in \overline{B}_X$ with
\begin{align*}
f(x_0) = \sup\{f(x) : x \in \overline{B}_X\}.
\end{align*}
We verify that this supremum equals $\|f\|_{X^*}$. By definition, $\|f\|_{X^*} = \sup\{|f(x)| : x \in \overline{B}_X\}$. Since $\overline{B}_X$ is symmetric ($x \in \overline{B}_X$ implies $-x \in \overline{B}_X$) and $|f(x)| = \max(f(x), f(-x))$, we have
\begin{align*}
\sup_{x \in \overline{B}_X} f(x) = \sup_{x \in \overline{B}_X} |f(x)| = \|f\|_{X^*}.
\end{align*}
Therefore $f(x_0) = \|f\|_{X^*}$ with $\|x_0\|_X \le 1$.
[guided]
The argument is an instance of the general principle that a continuous function on a compact set attains its extrema. The key step is identifying the correct topology.
**Why weak topology?** The norm topology on $\overline{B}_X$ does NOT make it compact in infinite dimensions (the closed unit ball of an infinite-dimensional [normed space](/page/Normed%20Vector%20Space) is never norm-compact). However, reflexivity provides weak compactness — this is the content of the [Characterisation of Reflexivity](/theorems/977), which requires $X$ to be a Banach space (completeness is essential) and states that reflexivity is equivalent to weak compactness of $\overline{B}_X$.
**Why is $f$ weakly continuous?** By definition, the weak topology $\sigma(X, X^*)$ is the initial topology generated by the family $\{g : g \in X^*\}$. Every $g \in X^*$ is therefore $\sigma(X, X^*)$-continuous. In particular, $f \in X^*$ is $\sigma(X, X^*)$-continuous.
**The supremum equals the norm.** We must verify that $\sup_{x \in \overline{B}_X} f(x) = \|f\|_{X^*}$, not merely $\sup_{x \in \overline{B}_X} |f(x)| = \|f\|_{X^*}$. The distinction matters because a priori $f$ might achieve $|f(x)| = \|f\|_{X^*}$ only with a negative value. But the ball is symmetric: if $f(x) < 0$ and $|f(x)| = \|f\|_{X^*}$, then $f(-x) = -f(x) = \|f\|_{X^*} > 0$ with $-x \in \overline{B}_X$. So the supremum of $f$ (not $|f|$) on $\overline{B}_X$ equals $\|f\|_{X^*}$.
[/guided]
[/step]
[step:State the converse (James, 1964)]
[claim:James' Theorem — Converse]
Let $X$ be a Banach space. If every $f \in X^*$ attains its norm on $\overline{B}_X$ (i.e., for every $f \in X^*$ there exists $x_0 \in \overline{B}_X$ with $f(x_0) = \|f\|_{X^*}$), then $X$ is reflexive.
[/claim]
The proof of the converse is due to R. C. James (1964) and is substantially deeper than the forward direction. The argument proceeds by contrapositive: assuming $X$ is not reflexive, one constructs a functional $f \in X^*$ that does *not* attain its norm. The construction relies on **basic sequence techniques** — specifically, the existence of a basic sequence in any infinite-dimensional Banach space, and a carefully orchestrated inductive procedure to build a non-norm-attaining functional from the basic sequence. These techniques go beyond the functional-analytic tools (Hahn-Banach, weak compactness, canonical embeddings) used in the rest of this article.
The original proof appears in:
R. C. James, *Characterizations of reflexivity*, Studia Math. **23** (1964), 205--216.
A streamlined exposition can be found in J. Diestel, *Sequences and Series in [Banach Spaces](/page/Banach%20Space)*, Springer GTM 92, Chapter V.
[/step]
