On a separable metric space, a sequence uniformly bounded in bounded Lipschitz norm has a pointwise convergent subsequence. Enumerate a countable dense subset and use the Bolzano-Weierstrass theorem and diagonal subsequence argument to obtain convergence there. Uniform control of the Lipschitz constant then makes the subsequence Cauchy at every point. The pointwise closure of the bounded Lipschitz unit ball preserves the original norm bound after rescaling.
Past exam of the mathematics course of the University of Cambridge 2009 ib Paper 4 13E Solution Created 2026-09-24 Updated 2026-10-14
Write and . The quantity in the question is the bounded Lipschitz norm . Both terms are nonnegative and absolutely homogeneous, and each satisfies the triangle inequality. Therefore their sum does too; finite values are preserved by addition and scalar multiplication, making the domain a vector space. If the sum is zero, forces . Thus it is a norm.
For the sequence bounded by one, both and . Given real , choose a nearby rational . ThenFirst make small, then use convergence at to make the last term small. The values at every real form a Cauchy sequence, so converges pointwise to a real function .
To preserve the sum bound rather than merely bound its two terms separately, use every fixed triple :Pass to the pointwise limit and take the independent suprema over and . This gives , the pointwise closure of the bounded Lipschitz unit ball.
To obtain diagonal compactness for bounded Lipschitz functions, enumerate the rationals for an arbitrary sequence in this unit ball. Values at the first rational lie in , so the Bolzano-Weierstrass theorem supplies a convergent subsequence there. Repeatedly extract nested subsequences for successive rationals. The diagonal subsequence argument gives one subsequence converging at every rational. Applying the preceding argument proves there is a pointwise convergent subsequence with limit of bounded Lipschitz norm at most one.