Auerbach's lemma
Auerbach’s lemma says that in any finite-dimensional normed space you can choose a basis that is as nicely behaved as possible with respect to the dual space.
- In an n-dimensional normed space (V, ||·||) there exists a basis e1, …, en such that each ei has norm 1, and the corresponding dual basis e1*, …, en* in V* also has norm 1 and satisfies ei*(ej) = δij. Such a basis is called an Auerbach basis.
- If V is an inner product space, you can simply take any orthonormal basis; its dual is given by the inner product, so the same property holds.
- Geometric view: there is a linear image of R^n that makes the unit cross-polytope (the unit ball for the ℓ1 norm) sit inside the image and the image sit inside the unit cube (the unit ball for the ℓ∞ norm).
- Sketch of the proof by induction on dimension n: pick a unit vector en in V. The unit ball is convex and symmetric, so there exists a supporting hyperplane P at en. Define a dual vector en* in V* with en*(en) = 1 and whose kernel ker(en*) is a hyperplane, i.e., an (n−1)-dimensional normed space. Apply the induction hypothesis to ker(en*) to get an Auerbach basis for that subspace, then add en to extend to a basis of V and arrange the duals accordingly.
- Corollary (projection): If V is an n-dimensional subspace of a normed space X, there exists a projection P: X → V with operator norm ||P|| ≤ n. Take an Auerbach basis e1, …, en of V with dual e1*, …, en*, extend each ei* to fi in X* with ||fi|| = 1, and define P(x) = ∑ fi(x) ei. Then P is a projection onto V and its norm is at most n.
This page was last edited on 3 February 2026, at 15:15 (CET).