Let f:R→R be a function satisfying ∣f(x+y)−f(x)−f(y)∣<1 for all x,y∈R. Prove that f(2008x)−2008f(x)<1 for all x∈R.
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∣f(2008x)−2008f(x)∣=k=1∑2007(f((k+1)x)−f(x)−f(kx))≤k=1∑2007∣f((k+1)x)−f(x)−f(kx)∣<2007. Replacing x with 2008x and simplifying, one gets 2008f(x)−f(2008x)<20082007<1.