Bishop–Phelps theorem

Remarks

Proof idea: given $f \in X^*$ and $\varepsilon > 0$, one considers a cone determined by $f$ and $\varepsilon$. A maximality argument, in the spirit of Zorn's lemma or of Ekeland's principle, gives $c \in C$ such that the cone translated to $c$ meets $C$ only at $c$. The Hahn–Banach separation theorem then gives a functional $g$ that attains its supremum on $C$ at $c$ and satisfies $\norm{f - g} \le \varepsilon$.

Example: let $X = c_0$ be the space of real sequences tending to $0$, with the supremum norm. The functional $f(x) = \sum_{n \ge 1} 2^{-n} x_n$ does not attain its norm $1$ on the unit ball, since that would need $x_n = 1$ for all $n$. The truncations $g_N(x) = \sum_{n \le N} 2^{-n} x_n$ attain their norm at $x = (1, \ldots, 1, 0, 0, \ldots)$ and converge to $f$ in norm, as the theorem predicts.

The density cannot be improved to equality in general: by James' theorem, every functional attains its norm on the unit ball only when $X$ is reflexive. The analogue for bounded operators between Banach spaces, in place of functionals, fails in general.

Going deeper

For a student

Norm-attaining functionals correspond to supporting hyperplanes of the unit ball, so the theorem says that support points of closed bounded convex sets are plentiful. It is used in the geometry of Banach spaces, for instance in the study of smoothness and of subdifferentials of convex functions. The complex case follows from the real one, by viewing a complex Banach space as a real one.