# Lectures on Lipschitz analysis by Heinonen J.

By Heinonen J.

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Exercise. 24). 7. 8. Let σ be a polyhedral k-chain in Rn and let ω ∈ Fk (Rn ). 25) | ω, σ | ≤ |σ| · ||ω|| . In particular, the dual norm of σ as an element of Fk (Rn ) does not exceed its flat norm. Proof. Let τ be a polyhedral (k + 1)-chain in Rn . 24), that ω, σ = ω, σ − ∂τ + ω, ∂τ ≤ |σ − ∂τ | · ||ω||∞ + |τ | · ||dω||∞ ≤ (|σ − ∂τ | + |τ |) · ||ω|| . 25) and the proposition follows. We need some further results in order to prove that the dual norm of a polyhedral chain bounds its flat norm. 23), because polyhedral chains are dense in Fk (Rn ) by definition, and because Fk (Rn ) is a Banach space.

One can also show that Lipschitz maps pull back flat forms to flat forms. Let us next see why such an assertion is not trivial. Thus, let f : Rn → Rm be Lipschitz and let ω = a1 dx1 + · · · + am dxm be a flat 1-form in Rm . 12) f ∗ ω(x) := a1 (f (x))df1 + · · · + am (f (x))dfm , and is obviously a flat 1-form in Rn . 12) is easily meaningless a priori. Namely, f can map a set of positive measure to a point, where the values of the coefficients ai are not well defined. To counter this example, one may argue that at such points the differential dfi must vanish, and we can set f ∗ ω(x) = 0.

If there are two line segments involved, then their mutual location becomes relevant. For example, the flat norm of the 2-chain σ := [0, e1 ] + [e1 + εe2 , εe2 ] , ε > 0, in R2 is at most 3ε. 7. Exercise. (a) Let σ = [e, 0]+[0, e ] for two unit vectors e, e . Show that |σ| = |σ| = 2 if and only if e = −e . (b) Suppose that σ1 , . . , σN are similarly oriented disjoint k-simplexes in a k-dimensional affine subspace of Rn . Prove that |σ| = |σ| if N σ= λi σi , λi ≥ 0 . 7 (a) seems difficult to determine.