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I know that the function f: [-1,1] defined by f(x) = x^2sin(1/x^2) for x ≠ 0, f(0) = 0 is continuous and its derivative f'(x) = 2xsin(1/x^2)-2/xcos(1/x^2) for x ≠ 0, f'(0) = 0 is unbounded on [-1,1]. But this function isn't absolutely continuous...

Any help would be much appreciated : )