Proof. [01HR]
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Proof.
It is clear that for all . Conversely it is a standard consequence of Rademacher’s theorem that . For each we also have . We now claim that there exists such that
for all and all , which will conclude the proof. Indeed the supremum in the right-hand side is a lower semicontinuous function of . As a consequence it achieves its infimum on the compact set , and this infimum cannot be zero since spans for each . The claim follows by homogeneity. ∎