Does a function that fulfills the set of criteria stated in the title of the question exist?
Actually, my question could be restated like this: Does Lipschitz continuity always imply differentiability almost everywhere?
2026-02-22 19:45:38.1771789538
Nowhere-differentiable Lipschitz-continuous function
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Yes. It seem to contradict Rademacher's theorem, but actually it doesn't as you haven't put in all requirements for Rademacher's theorem in your criteria.
The Rademacher's theorem requires the domain (in which the function is Lipschitz) must be open (or have an open subset), if it fails that it doesn't have to be differentiable almost everywhere. Also if the set has measure zero then differentiability almost everywhere wouldn't imply differentiability anywhere anyway.
A trivial counter example would be the function $f:\{0\}\to\{0\}$ defined as $f(0)=0$. It's obviously Lipschitz, but nowhere differentiable. It however fails to fulfil the requirements of Rademacher's theorem.