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Quote: Originally Posted by silversand The first one is not uniformly continous( I don't think) Here is number two Since g(x) is differentable and the derivative is bounded on all of . so for some let set So now if We need to show that Now by the Mean Value theorem on

Quote: Originally Posted by TheEmptySet The first one is not uniformly continous( I don't think) Here is number two Since g(x) is differentable and the derivative is bounded on all of . so for some let set So now if We need to show that Now by the Mean Value theorem on the first one is not uniformly cont on b/c the derivative is . so as contradicting the fact that the derivative must be bounded.