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Mark McClure
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I believe the Takakgi function satisfies this property. According to The Takagi Function: A Survey, the Takagi function $T$ satisfies $$T(x+h)-T(x) = O(h\log(1/|h|) \: \text{ as } \: h\to0$$ and this is the best possible estimate.

Mark McClure
  • 2.1k
  • 14
  • 18