Posted by Student on December 6, 2007, 2:13 pm
What's wrong with the following counterexample? If F=[-1,1] for y>0 sufficiently large then \delta(x+y) = y+x-1, which then implies that \delta(x+y)/y -> 1 as y -> \infty. Similar arguments hold for y<0. Thus, \delta(x+y) \neq o(\abs{y}).
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