Normed Spaces and Banach Spaces

A normed space is a vector space $X$ over $\mathbb{R}$ or $\mathbb{C}$ together with a norm $\|\cdot\| : X \to [0, \infty)$ satisfying:

1. $\|x\| = 0 \iff x = 0$ (positive definiteness).
2. $\|\alpha x\| = |\alpha| \cdot \|x\|$ for all scalars $\alpha$ (homogeneity).
3. $\|x + y\| \leq \|x\| + \|y\|$ (triangle inequality).

A norm induces a metric $d(x, y) = \|x - y\|$, making $X$ a metric space.

Example 1. $(C[a, b], \|\cdot\|_\infty)$ with $\|f\|_\infty = \sup_{x \in [a,b]} |f(x)|$.

Example 2. $(C[a, b], \|\cdot\|_1)$ with $\|f\|_1 = \int_a^b |f(x)|\, dx$. This norm is weaker: convergence in $\|\cdot\|_1$ does not imply pointwise convergence.

Example 3. $\ell^p = \{(x_n) : \sum |x_n|^p < \infty\}$ with $\|x\|_p = (\sum |x_n|^p)^{1/p}$ for $1 \leq p < \infty$.

Example 4. $\ell^\infty = \{(x_n) : \sup_n |x_n| < \infty\}$ with $\|x\|_\infty = \sup_n |x_n|$.

A Banach space is a complete normed space (every Cauchy sequence converges).

Theorem 1.1. $\ell^p$ is a Banach space for $1 \leq p \leq \infty$.

Proof. First, $\ell^p$ is a normed space: positive definiteness and homogeneity are immediate, and the triangle inequality is Minkowski’s inequality, proved in §1.8 below.

Let $1 \leq p < \infty$ and let $(x^{(k)})$ be a Cauchy sequence in $\ell^p$, where $x^{(k)} = (x^{(k)}_1, x^{(k)}_2, \dots)$. For each coordinate $i$,

so $(x^{(k)}_i)_k$ is Cauchy in $\mathbb{F}$ and converges; define $x_i = \lim_k x^{(k)}_i$ and $x = (x_i)$.

Given $\varepsilon > 0$, choose $K$ with $\|x^{(k)} - x^{(m)}\|_p \leq \varepsilon$ for all $k, m \geq K$. For any finite $N$ and $k, m \geq K$,

The sum is finite, so we may let $m \to \infty$ inside it: $\sum_{i=1}^{N} |x^{(k)}_i - x_i|^p \leq \varepsilon^p$. This holds for every $N$; take the supremum over $N$ to get $\|x^{(k)} - x\|_p \leq \varepsilon$ for all $k \geq K$. The case $p = \infty$ is identical with sums replaced by suprema. In particular $x^{(k)} - x \in \ell^p$, so $x = x^{(k)} - (x^{(k)} - x) \in \ell^p$, and $x^{(k)} \to x$ in norm. $\blacksquare$

Note the mechanism, which recurs throughout functional analysis: Cauchy in norm forces coordinatewise convergence; the candidate limit is then shown to lie in the space and to be the norm limit, by letting a finite partial sum absorb the limit.

Theorem 1.2. $L^p(\mu)$ is a Banach space for $1 \leq p \leq \infty$.

Proof. The norm axioms again reduce to Minkowski (the integral version, proved identically to §1.8 with sums replaced by integrals; see the $L^p$ page in the measure theory track). For completeness, let $(f_n)$ be Cauchy in $L^p$, $1 \leq p < \infty$. Extract a subsequence with

Let $h = \sum_{k=1}^{\infty} |g_{k+1} - g_k|$, a function in $[0, \infty]$. By monotone convergence and Minkowski,

so $h$ is finite almost everywhere. On the full-measure set $\{h < \infty\}$, the telescoping series $\sum_k (g_{k+1}(x) - g_k(x))$ converges absolutely, hence converges; define

and set $g = 0$ on the null set where the series diverges. Then $|g - g_1| \leq h$ gives $g \in L^p$, and for each $k$,

So the subsequence converges in $L^p$ to $g$. Finally, a Cauchy sequence with a convergent subsequence converges: $\|f_n - g\|_p \leq \|f_n - f_{n_k}\|_p + \|f_{n_k} - g\|_p$, choosing $k$ with $n_k$ large. The case $p = \infty$ is simpler: Cauchy in $\|\cdot\|_\infty$ is uniform Cauchy, and a uniform limit of essentially bounded measurable functions is essentially bounded. $\blacksquare$

Theorem 1.3. $(C[a, b], \|\cdot\|_\infty)$ is a Banach space, but $(C[a, b], \|\cdot\|_1)$ is not.

Proof. Supremum norm. Let $(f_n)$ be Cauchy in $\|\cdot\|_\infty$. For each fixed $x$, the real sequence $(f_n(x))$ satisfies $|f_n(x) - f_m(x)| \leq \|f_n - f_m\|_\infty$, so it is Cauchy; define $f(x) = \lim_n f_n(x)$. Given $\varepsilon > 0$, choose $N$ with $|f_n(x) - f_m(x)| < \varepsilon$ for all $x$ and all $n, m \geq N$; holding $n$ and $x$ fixed and letting $m \to \infty$ gives $|f_n(x) - f(x)| \leq \varepsilon$ for all $x$; the convergence is uniform. A uniform limit of continuous functions is continuous: given $x_0$ and $\varepsilon$, pick $n$ with $\|f_n - f\|_\infty \leq \varepsilon$ and use the continuity of $f_n$ at $x_0$ in the $\varepsilon/3$ bound

So $f \in C[a, b]$ and $f_n \to f$ in $\|\cdot\|_\infty$.

$L^1$ norm. Not complete; the explicit Cauchy sequence converging to a discontinuous limit is constructed in Practice Problem 1 below. $\blacksquare$

Theorem 1.4. All norms on a finite-dimensional vector space are equivalent.

Proof. Let $X$ have dimension $n$ with basis $e_1, \dots, e_n$, and let $\|\cdot\|_\bullet$ be the auxiliary norm $\bigl\|\sum_i a_i e_i\bigr\|_\bullet = \sum_i |a_i|$. It suffices to show every norm $\|\cdot\|$ on $X$ is equivalent to $\|\cdot\|_\bullet$: equivalence of norms is transitive, and both inequalities below compose.

Upper bound. For $x = \sum_i a_i e_i$,

Lower bound. By the upper bound, $\bigl| \|x\| - \|y\| \bigr| \leq \|x - y\| \leq C\|x - y\|_\bullet$, so $a \mapsto \bigl\|\sum_i a_i e_i\bigr\|$ is continuous as a function on $\mathbb{F}^n$ with the $\|\cdot\|_\bullet$ metric. The set $S = \{a : \|a\|_\bullet = 1\}$ is compact (closed and bounded in finite dimensions; Heine—Borel), and the function is strictly positive on $S$: if $\bigl\|\sum a_i e_i\bigr\| = 0$ then $\sum a_i e_i = 0$ (positive definiteness), so all $a_i = 0$ (basis), contradicting $\|a\|_\bullet = 1$. A continuous function on a compact set attains its infimum, so $m = \min_S \|\cdot\| > 0$. For $x \neq 0$ the vector $x / \|x\|_\bullet$ lies on $S$, so

$\blacksquare$

Corollary 1.5. Every finite-dimensional normed space is a Banach space.

Proof. Let $X$ be finite-dimensional. A $\|\cdot\|$-Cauchy sequence is $\|\cdot\|_\bullet$-Cauchy by Theorem 1.4, i.e. each coordinate is Cauchy in $\mathbb{F}$, hence convergent ($\mathbb{R}$ and $\mathbb{C}$ are complete). Convergence in $\|\cdot\|_\bullet$ implies convergence in $\|\cdot\|$, again by Theorem 1.4. $\blacksquare$

One consequence used constantly: every finite-dimensional subspace of a normed space is closed, being complete in its own right (a complete subspace of a metric space is closed; its complement is open, since a point at distance zero from it is its limit point).

Theorem 1.6 (Riesz’s Lemma). Let $X$ be a normed space and $Y$ a proper closed subspace. For every $0 < \theta < 1$, there exists $x \in X$ with $\|x\| = 1$ and $d(x, Y) \geq \theta$.

Proof. Pick any $x_0 \in X \setminus Y$. Since $Y$ is closed, $X \setminus Y$ is open, so some ball around $x_0$ misses $Y$ and $\alpha := d(x_0, Y) > 0$. Choose $y_0 \in Y$ with $\|x_0 - y_0\| \leq \alpha / \theta$; possible since $\theta < 1$ makes $\alpha/\theta > \alpha$. Set

For any $y \in Y$,

and $y_0 + \|x_0 - y_0\| y \in Y$ (a subspace absorbs scaling and addition), so the last quantity is at least $\alpha / \|x_0 - y_0\| \geq \theta$. $\blacksquare$

Corollary 1.7. The closed unit ball of a normed space is compact if and only if the space is finite-dimensional.

Proof. ($\Leftarrow$) If $X$ is finite-dimensional, fix a linear isomorphism with $\mathbb{F}^n$. By Theorem 1.4 the closed unit ball is closed in, and contained in a multiple of, the Euclidean unit ball under this identification; closed and bounded in $\mathbb{F}^n$; hence compact by Heine—Borel.

($\Rightarrow$) Contrapositive. Suppose $X$ is infinite-dimensional and build a sequence in the unit ball with no Cauchy subsequence. Pick $x_1$ with $\|x_1\| = 1$. Inductively, having chosen $x_1, \dots, x_n$, let $Y_n = \operatorname{span}\{x_1, \dots, x_n\}$; a finite-dimensional subspace, hence closed (Corollary 1.5), and proper (as $X$ is infinite-dimensional). Riesz’s lemma with $\theta = 1/2$ produces $x_{n+1}$ with $\|x_{n+1}\| = 1$ and $d(x_{n+1}, Y_n) \geq 1/2$.

For $i < j$ the point $x_i$ lies in $Y_{j-1}$, so $\|x_j - x_i\| \geq d(x_j, Y_{j-1}) \geq 1/2$. No subsequence is Cauchy, so the closed unit ball is not (sequentially) compact; and in a metric space compactness and sequential compactness coincide. $\blacksquare$

This corollary is the first taste of what makes infinite dimensions genuinely different: in finite dimensions, closed and bounded is compact; in infinite dimensions it never is.

Let $X$ be a normed space and $Y \subseteq X$ a closed subspace. The quotient space $X / Y$ consists of equivalence classes $[x] = x + Y$ with the quotient norm:

Theorem 1.8. If $X$ is a Banach space and $Y$ is a closed subspace, then $X/Y$ is a Banach space.

Proof. The quotient norm is a norm. Homogeneity and positivity of the infimum are clear, and the triangle inequality follows since for any $y, y' \in Y$,

so taking infima over $y$ and $y'$ separately gives $\|[x+x']\| \leq \|[x]\| + \|[x']\|$. Finally, $\|[x]\| = 0$ means $x$ is a limit of points of $Y$, i.e. $x \in \overline{Y} = Y$ (closedness is exactly what is used here), so $[x] = 0$.

Completeness. Let $([x_n])$ be Cauchy in $X/Y$. Extract a subsequence with $\|[x_{n_{k+1}}] - [x_{n_k}]\| < 2^{-k}$. Since $\|[x_{n_{k+1}}] - [x_{n_k}]\| = d(x_{n_{k+1}} - x_{n_k},\, Y)$, choose $y_{k+1} \in Y$ with

Now define $z_1 = x_{n_1}$ and inductively $z_{k+1} = z_k + (x_{n_{k+1}} - x_{n_k} - y_{k+1})$. The increments are absolutely summable, so $(z_k)$ converges in the Banach space $X$; write the limit as $z$. Passing to quotients, the recursion telescopes:

the last step by induction from $[z_1] = [x_{n_1}]$. Hence

so the subsequence converges, and a Cauchy sequence with a convergent subsequence converges, as in Theorem 1.2. $\blacksquare$

Proposition 1.9. The quotient map $\pi : X \to X/Y$, $\pi(x) = [x]$, is a bounded linear operator with $\|\pi\| = 1$ (for $Y$ a proper closed subspace).

Proof. Linearity is inherited from the vector space quotient. Since $0 \in Y$, $\|[x]\| = d(x, Y) \leq \|x\|$, so $\|\pi\| \leq 1$. Conversely, Riesz’s lemma provides, for every $\theta < 1$, a unit vector $x$ with $\|x\| = 1$ and $\|[x]\| = d(x, Y) \geq \theta$; so $\|\pi\| \geq \theta$ for all $\theta < 1$, whence $\|\pi\| \geq 1$. Combining: $\|\pi\| = 1$.

Note the sup is generally not attained: Riesz’s lemma gives $d(x, Y) \geq \theta$ for each $\theta < 1$ but not necessarily $= 1$. In finite dimensions the unit sphere is compact and the sup is attained; in infinite dimensions it can genuinely fail to be. $\blacksquare$

The dual space $X^*$ of a normed space $X$ is the space of all bounded linear functionals $f : X \to \mathbb{F}$, equipped with the operator norm:

Theorem 1.10. The dual space $X^*$ is always a Banach space, regardless of whether $X$ is complete.

Proof. Linearity of the candidate limit. Let $(f_n)$ be Cauchy in the operator norm. For each fixed $x$, $|f_n(x) - f_m(x)| \leq \|f_n - f_m\| \|x\|$, so $(f_n(x))$ is Cauchy in $\mathbb{F}$; define $f(x) = \lim_n f_n(x)$. Taking limits in the linearity identity $f_n(\alpha x + \beta y) = \alpha f_n(x) + \beta f_n(y)$ gives linearity of $f$.

Boundedness. Cauchy sequences are bounded in norm: $\|f_n\| \leq C$ for all $n$. So $|f_n(x)| \leq C\|x\|$ for every $n$ and $x$; let $n \to \infty$: $|f(x)| \leq C\|x\|$, so $\|f\| \leq C$.

Convergence in norm. Given $\varepsilon > 0$, choose $N$ with $\|f_n - f_m\| \leq \varepsilon$ for $n, m \geq N$. Fix $n \geq N$ and $x$ with $\|x\| \leq 1$. Then for every $m \geq N$,

Let $m \to \infty$: the first term tends to $0$, giving $|f(x) - f_n(x)| \leq \varepsilon$ for all $\|x\| \leq 1$. Taking the supremum over such $x$: $\|f - f_n\| \leq \varepsilon$. $\blacksquare$

Note where completeness of $X$ would have been used; nowhere. The dual is complete because the scalar field is, and the operator norm controls convergence uniformly on the unit ball.

Examples of dual spaces:

• $(\ell^p)^* \cong \ell^q$ where $1/p + 1/q = 1$ for $1 \leq p < \infty$.
• $(c_0)^* \cong \ell^1$, where $c_0$ is the space of sequences converging to $0$.
• $(L^p(\mu))^* \cong L^q(\mu)$ for $1 \leq p < \infty$ and $1/p + 1/q = 1$.

Theorem 1.11. Every normed space $X$ has a completion: a Banach space $\tilde{X}$ and an isometric embedding $i : X \to \tilde{X}$ with dense image. The completion is unique up to isometric isomorphism.

Proof. Construction. Let $\mathcal{C}$ be the set of Cauchy sequences in $X$, with $(x_n) \sim (y_n)$ iff $\|x_n - y_n\| \to 0$ (an equivalence relation; check transitivity via the $\varepsilon/2$ argument). Set $\tilde{X} = \mathcal{C}/{\sim}$ and define

The limit exists because $|\,\|x_n\| - \|x_m\|\,| \leq \|x_n - x_m\|$ (reverse triangle inequality) makes $(\|x_n\|)$ Cauchy in $\mathbb{F}$. It is well defined: replacing $(x_n)$ by an equivalent sequence changes each $\|x_n\|$ by at most $\|x_n - y_n\| \to 0$. The norm axioms pass to the limit from each coordinate.

The embedding. $i(x) = [(x, x, x, \dots)]$ is linear, and $\|i(x)\| = \lim_n \|x\| = \|x\|$: isometric. Its image is dense: given $\xi = [(x_n)] \in \tilde{X}$ and $\varepsilon > 0$, pick $N$ with $\|x_n - x_m\| < \varepsilon$ for $n, m \geq N$; then

Completeness. Let $(\xi_k)$ be Cauchy in $\tilde{X}$. Choose representatives $x^{(k)} \in X$ with $\|\xi_k - i(x^{(k)})\| < 1/k$ (density). Then for $k, l$ large,

so $(x^{(k)})$ is Cauchy in $X$; let $\xi = [(x^{(k)})] \in \tilde{X}$. A three-term estimate with the same ingredients gives $\|\xi_k - \xi\| \to 0$. So $\tilde{X}$ is complete.

Uniqueness. Given two completions $i : X \to \tilde{X}$ and $j : X \to \bar{X}$, define $\Phi$ on the dense subspace $i(X)$ by $\Phi(i(x)) = j(x)$. On $i(X)$, $\Phi$ is isometric:

An isometry on a dense subspace of a metric space extends uniquely to an isometry on the whole space (the extension sends $z \mapsto \lim \Phi(z_n)$ along any sequence in the subspace converging to $z$; well defined by isometry and Cauchy). The extension $\Phi : \tilde{X} \to \bar{X}$ is surjective: its image is complete; being isometric to a complete space; hence closed in $\bar{X}$, and it contains the dense set $j(X)$. $\blacksquare$

Example. The completion of $(C[a, b], \|\cdot\|_1)$ is $L^1[a, b]$.

Infinite-dimensional normed spaces have properties that contrast sharply with finite-dimensional ones:

• The closed unit ball is not compact (Riesz’s lemma).
• There exist discontinuous linear operators (requires the axiom of choice).
• Not every linear subspace is closed.
• The weak topology differs from the norm topology.

Lemma 1.12.0 (Young’s inequality). For $1 < p < \infty$ and $1/p + 1/q = 1$, all $a, b \geq 0$:

Proof. If $a = 0$ or $b = 0$ both sides vanish. Otherwise fix $b$ and minimise $\varphi(a) = a^p/p + b^q/q - ab$ over $a \geq 0$. Then $\varphi'(a) = a^{p-1} - b$, which vanishes at $a = b^{1/(p-1)} = b^{q/p}$, and $\varphi''(a) = (p-1)a^{p-2} > 0$, so this is the global minimum (with $\varphi(0) = b^q/q > 0$ confirming the boundary is worse). The value there:

Since $q/p + 1 = q(1/p + 1/q) = q$, the minimum is $b^q - b^q = 0$. So $\varphi(a) \geq 0$ everywhere. $\blacksquare$

Theorem 1.12 (Hölder’s Inequality). For $1 \leq p, q \leq \infty$ with $1/p + 1/q = 1$:

Proof. Case $1 < p < \infty$. If $\|x\|_p = 0$ or $\|y\|_q = 0$ the inequality is trivial. Otherwise normalise: $\hat{x} = x / \|x\|_p$ and $\hat{y} = y / \|y\|_q$ satisfy $\|\hat{x}\|_p = \|\hat{y}\|_q = 1$, and both sides scale by the same factor $\|x\|_p \|y\|_q$, so it suffices to prove $\sum |\hat{x}_n \hat{y}_n| \leq 1$.

For every $N$ and every $n \leq N$, Young’s inequality with $a = |\hat{x}_n|$, $b = |\hat{y}_n|$ gives $|\hat{x}_n \hat{y}_n| \leq |\hat{x}_n|^p / p + |\hat{y}_n|^q / q$. Summing over $n \leq N$:

The partial sums are increasing and bounded, so the full series converges with sum $\leq 1$.

Case $p = 1$, $q = \infty$: $|x_n y_n| \leq |x_n| \|y\|_\infty$ termwise, then sum. Case $p = \infty$ symmetrically. $\blacksquare$

Theorem 1.13 (Minkowski’s Inequality). For $1 \leq p \leq \infty$:

Proof. $p = 1$ is the ordinary triangle inequality in $\mathbb{F}$, summed; $p = \infty$ is the supremum version. Let $1 < p < \infty$ and $q$ its conjugate. The case $x + y = 0$ is trivial; otherwise expand:

Apply Hölder to each sum. Since $(p-1)q = p$, the Hölder factor for the second argument is

so

If $\|x + y\|_p > 0$, divide both sides by $\|x + y\|_p^{p/q}$; the exponent satisfies $p - p/q = p(1 - 1/q) = 1$, leaving $\|x + y\|_p \leq \|x\|_p + \|y\|_p$. $\blacksquare$

These inequalities prove that $\ell^p$ and $L^p$ are normed spaces. The integral versions follow by the identical arguments with sums replaced by integrals; see the $L^p$ page in the measure theory track, where the completeness of $L^p$ is also treated.

Problem 1. Show that $C[a, b]$ with $\|f\|_1 = \int_a^b |f(x)|\, dx$ is not complete.

Solution. Consider $f_n(x) = \begin{cases} 0 & a \leq x \leq (a+b)/2 - 1/n \\ \text{linear} & \text{in the transition} \\ 1 & (a+b)/2 + 1/n \leq x \leq b \end{cases}$. This is a Cauchy sequence in $\|\cdot\|_1$ but converges to the discontinuous step function. $\blacksquare$

Problem 2. Prove that $\ell^p \subset \ell^q$ for $1 \leq p < q \leq \infty$.

Problem 3. Show that $\|x\|_\infty = \lim_{p \to \infty} \|x\|_p$ for $x \in \ell^p \cap \ell^\infty$.

Problem 4. Prove that the dual of $c_0$ is $\ell^1$.

A normed space $X$ carries the weak topology $\sigma(X, X^*)$, the coarsest topology making all $f \in X^*$ continuous. A sequence converges weakly ($x_n \rightharpoonup x$) if $f(x_n) \to f(x)$ for every $f \in X^*$.

The dual space $X^*$ carries the weak-* topology $\sigma(X^*, X)$, the coarsest topology making all evaluation maps $x \mapsto f(x)$ continuous.

Theorem 1.14 (Banach-Alaoglu). The closed unit ball of $X^*$ is compact in the weak-* topology.

Proof. Let $B = \{f \in X^* : \|f\| \leq 1\}$. For each $x \in X$ let $D_x = \{z \in \mathbb{F} : |z| \leq \|x\|\}$, a closed disc; compact by Heine—Borel in $\mathbb{F} \cong \mathbb{R}$ or $\mathbb{R}^2$. Consider the product $P = \prod_{x \in X} D_x$ and the map

The weak-* topology on $B$ is exactly the subspace topology inherited from $P$: both are initial topologies for the same family of coordinate maps $f \mapsto f(x)$. So $B$ is compact iff $\Phi(B)$ is compact as a subspace of $P$.

$P$ is compact by Tychonoff’s theorem (Theorem 8.4 in the topology track). And $\Phi(B)$ is closed in $P$: a point $(z_x) \in P$ lies in $\Phi(B)$ iff

and each of these conditions is closed; the map $(z_\bullet) \mapsto z_{x+y} - z_x - z_y$ is a finite combination of coordinate projections, hence continuous, and the condition is the preimage of $\{0\}$. (Positivity of the norm needs no separate condition: $|z_x| \leq \|x\|$ is built into $D_x$, and it reproduces $\|f\| \leq 1$ by taking suprema.) A closed subset of a compact space is compact, so $\Phi(B)$; hence $B$; is compact. $\blacksquare$

A normed space is separable if it contains a countable dense subset.

Examples: $\ell^p$ is separable for $1 \leq p < \infty$. $\ell^\infty$ is not separable. $C[a, b]$ is separable (polynomials with rational coefficients are dense).

Theorem 1.15. If $X^*$ is separable, then $X$ is separable. The converse does not hold: $\ell^1$ is separable but $(\ell^1)^* \cong \ell^\infty$ is not.

Proof. Let $\{f_n\}$ be a countable dense subset of the unit sphere $\{f \in X^* : \|f\| = 1\}$; separability of $X^*$ gives a countable dense set in all of $X^*$, and intersecting with spheres (or normalising) yields one in the sphere. For each $n$ pick $x_n$ with $\|x_n\| = 1$ and $|f_n(x_n)| \geq 1/2$ (the unit norm is a supremum, so values arbitrarily close to $1$ are attained).

Claim: the rational span $D$ of $\{x_n\}$ is dense in $X$. $D$ is countable. Suppose not dense; then some $y \in X$ and $\delta > 0$ have $d(y, \operatorname{span}\{x_n\}) \geq \delta$. The distance functional extends: by Hahn—Banach (Theorem 1.3 of the fundamental-theorems page) there is $f \in X^*$ with $\|f\| = 1$, $f|_{\operatorname{span}\{x_n\}} = 0$, and $f(y) = \delta$.

Density of $\{f_n\}$ in the sphere gives a subsequence $f_{n_k} \to f$ in operator norm. But for every $k$,

since $f(x_{n_k}) = 0$. The right side tends to $0$; contradiction. So $D$ is dense and $X$ is separable. $\blacksquare$

A Banach space $X$ is reflexive if the natural embedding $J : X \to X^{**}$ defined by $J(x)(f) = f(x)$ is surjective.

Examples: $\ell^p$ is reflexive for $1 < p < \infty$. $\ell^1$ and $\ell^\infty$ are not reflexive. Every finite-dimensional space is reflexive.

Theorem 1.16. A Banach space is reflexive if and only if its closed unit ball is weakly compact.

Proof. ($\Leftarrow$) Suppose $B_X$ is weakly compact. The canonical embedding $J : X \to X^{**}$ is weak-to-weak-* continuous: for each $f \in X^*$, the composition $J(\cdot)(f) = f(\cdot)$ is weakly continuous by definition of $\sigma(X, X^*)$. Moreover the weak-* topology of $X^{**}$ restricted to $J(X)$ is the weak topology of $X$ transported by $J$; both are initial topologies for the same set of functionals, $\{f : f \in X^*\}$.

By Goldstine’s theorem (a consequence of Hahn—Banach; see the fundamental-theorems page), $J(B_X)$ is weak-* dense in $B_{X^{**}}$. As a weakly compact set, $J(B_X)$ is compact in the subspace topology; which is the weak-* topology just identified; and a compact subset of a Hausdorff space is closed in the ambient space. So $J(B_X)$ is weak-* closed, and closed and dense in $B_{X^{**}}$ forces $J(B_X) = B_{X^{**}}$. Scaling: every $x^{**} \in X^{**}$ with $x^{**} \neq 0$ is a multiple of a point of $B_{X^{**}} = J(B_X)$, so $J$ is surjective.