Hahn–Banach and Duality

A bounded linear functional is a measurement: a continuous linear way of turning a vector into a number. Evaluating a function at a point, integrating it against a weight, taking a Fourier coefficient: these are all linear functionals. The dual space $X^*$ is the space of all of them, and much of functional analysis consists of studying a space through its dual. Two questions come first. Are there enough functionals to tell vectors apart? And what does the dual of a familiar space look like?

The first question is answered by the Hahn–Banach theorem, in two forms. The analytic form extends functionals from a subspace without increasing their norm. The geometric form, easier to picture and just as important, says that two disjoint convex sets can be separated by a hyperplane. The second question has satisfying answers: the dual of $L^p$ is $L^q$ for conjugate exponents (with one exception), and the dual of $C(K)$ is the space of measures. That last fact is why limits of probability densities, the objects of 3A.4 Measures, Probability and Weights, can be measures, and it reappears at the frontier of Ricci flow research.

By the end of this chapter you will be able to:

• state and use the analytic Hahn–Banach theorem, and find norming functionals;
• prove that a point outside a closed convex set in $\mathbb{R}^n$ can be separated from it by a hyperplane, and state the general version;
• identify the duals of $\ell^p$, $L^p$ and $C(K)$, and explain which spaces are reflexive;
• use separating hyperplanes to price assets by no arbitrage, and to classify data.

3.1Prices from separation

3.2The analytic Hahn–Banach theorem

A function $p : X \to \mathbb{R}$ on a real vector space is sublinear if $p(x + y) \leq p(x) + p(y)$ and $p(\lambda x) = \lambda p(x)$ for $\lambda \geq 0$. A norm is sublinear.

Proof architecture. One step: to extend $f$ from $M$ to $M + \mathbb{R}x_0$, one must choose a value $F(x_0) = c$ with $f(m) + tc \leq p(m + tx_0)$ for all $m \in M$ and real $t$. Dividing by $|t|$, this says $c$ must lie between $\sup_m(f(m) - p(m - x_0))$ and $\inf_m(p(m + x_0) - f(m))$, and sublinearity is exactly what makes the supremum at most the infimum (Exercise 3.8). All steps: extend one dimension at a time. In a space with a countable dense set (a separable space, such as $L^p$ for $p < \infty$) ordinary induction and continuity suffice; in general, Zorn's lemma (equivalent to the axiom of choice, 2A.8 Infinite Sets) provides a maximal extension, which must be defined everywhere.

The case used most is $p(x) = C\|x\|$.

In concrete spaces the norming functional of (2) can usually be written down. In $L^p$ with $1 < p < \infty$, it is $g \mapsto \int g\,\frac{|f|^{p-1}\overline{\operatorname{sign}f}}{\|f\|_p^{p-1}}$, the case of equality in Hölder's inequality (Exercise 3.9). The theorem is needed when there is no formula, and to know that enough functionals exist in general.

3.3Separating convex sets

The hyperplane $\{v\cdot x = c\}$ touches $C$ at $p$ and separates it from $x_0$. If $x_0$ is a boundary point of $C$ instead, a limiting argument gives a supporting hyperplane: one that passes through $x_0$ with $C$ on one side (Figure 3.1).

In a general normed space the nearest point need not exist, but the conclusion survives, with a continuous linear functional in place of $v\cdot{}$.

Proof architecture (Brezis, Theorems 1.6 and 1.7). For (1), reduce to separating a point from an open convex set $C$ containing $0$, and apply the analytic form with the Minkowski gauge $p(x) = \inf\{t > 0 : x/t \in C\}$, which is sublinear and comparable to the norm. For (2), thicken $B$ slightly into an open convex set that still misses $A$, and apply (1).

Linear programming is a third use. Minimising a linear cost over a convex polyhedron of feasible plans has a dual problem whose variables are prices, the "shadow prices" of the constraints: the marginal value of relaxing each constraint by one unit. The duality theorem of linear programming, that the optimal costs of the two problems agree, is a separation theorem for polyhedra (Farkas's lemma, Exercise 3.10).

3.4Dual spaces

The dual of a normed space $X$ is $X^* = L(X, \mathbb{R})$ (or $\mathbb{C}$), with the operator norm. It is always a Banach space (4A.1 Banach Spaces and Bounded Operators), even if $X$ is not complete. For the spaces of Book 3A the duals are concrete.

Proof architecture. Hölder gives $\|\phi_g\| \leq \|g\|_q$, and the norming function of Exercise 3.9 gives equality. For the converse, given $\phi$, define a set function $\nu(E) = \phi(1_E)$ on sets of finite measure. It is countably additive (continuity of $\phi$ plus dominated convergence) and absolutely continuous with respect to $\mu$, so the Radon–Nikodym theorem (3A.4 Measures, Probability and Weights) gives a density $g$ with $\phi(1_E) = \int_Eg\,d\mu$; one then checks $g \in L^q$ and extends from indicators to all of $L^p$ by density. Brezis gives a different proof for $1 < p < \infty$ through uniform convexity.

The case $p = \infty$ is the exception. Every $g \in L^1$ gives a functional on $L^\infty$, but not every functional on $L^\infty$ comes from an $L^1$ function: Hahn–Banach produces, for example, a "limit at $0$" functional on $L^\infty(0, 1)$ that extends $f \mapsto \lim_{x\to0}f(x)$ from the functions that have such a limit, and no integrable $g$ can represent it (Exercise 3.12). For sequences, $(\ell^p)^* = \ell^q$ for $1 \leq p < \infty$, and the dual of $c_0$ (sequences tending to $0$, with the sup norm) is $\ell^1$.

(Stated, not proved here; Kreyszig §4.4 treats $C[a, b]$, which was Riesz's original 1909 theorem, with Stieltjes integrals.) The Dirac mass $\delta_{x_0}$, evaluation at a point, is the functional $f \mapsto f(x_0)$, of norm $1$: it lives naturally in $C(K)^*$ even though it has no density (3A.4 Measures, Probability and Weights).

3.4.1Reflexivity

Each $x \in X$ defines a functional on $X^*$, namely $\phi \mapsto \phi(x)$, and by Corollary 3.2 this embeds $X$ isometrically into its bidual $X^{**}$. If the embedding is onto, $X$ is reflexive. Hilbert spaces are reflexive (4A.4 Hilbert Spaces and Lax–Milgram); so is $L^p$ for $1 < p < \infty$, since $(L^p)^{**} = (L^q)^* = L^p$. But $L^1$, $L^\infty$ and $C(K)$ (for infinite $K$) are not.

Why care? Reflexivity is exactly the condition under which bounded sequences have weakly convergent subsequences (4A.6 Weak Convergence and the Direct Method), the replacement for the compactness that infinite dimensions take away. That is why existence theory in PDE prefers $L^p$ and Sobolev spaces with $1 < p < \infty$, and handles $p = 1$ and $p = \infty$ separately and with care.

3.5History

Eduard Helly proved an extension theorem for functionals on $C[a, b]$ in 1912. Hans Hahn (1927) and Stefan Banach (1929) independently proved the general theorem now named after them. Hermann Minkowski developed separation and supporting hyperplanes for convex bodies around 1900, and Gyula Farkas published his lemma on linear inequalities in 1902. Frigyes Riesz identified the dual of $C[a, b]$ in 1909 and the duals of $L^p$ in 1910; the representation for general compact spaces is due to Andrey Markov (1938) and Shizuo Kakutani (1941). Linear programming duality grew out of work by Leonid Kantorovich (1939), John von Neumann and George Dantzig (1947). Harrison, Kreps and Pliska's papers on arbitrage and martingales appeared in 1979 and 1981.

3.6Exercises

Solution

Replicate with $\Delta$ shares and $B$ in cash: $120\Delta + B = 20$, $90\Delta + B = 0$, so $\Delta = \frac23$, $B = -60$, at cost $100\cdot\frac23 - 60 = \frac{20}3$. If the option sells for $8$, sell it and buy the replicating portfolio: $8 - \frac{20}{3} = \frac43$ profit now, and tomorrow the portfolio exactly pays what the option owes.

Solution

$\phi(1_{(0,\varepsilon)}) = 1$ for every $\varepsilon$, but $\int_0^\varepsilon g \to 0$ by dominated convergence.