# f uniformly Lipschitz, integral from 0 to inf of |f(x)| < inf imply |f(x)| converges to 0 as x -> inf. [duplicate]

Suppose f: [0,inf) -> R is uniformly Lipschitz and the integral from 0 to inf of |f(x)| < inf. Then|f(x)| converges to 0 as x -> inf. Prove this by proving the contrapositive. ie. negate the limit portion and show via the properties given by f that the integral will be infinite.

I'm lost here. For the limit, I think I'd need to choose an epsilon so that the limit does not converge to 0. Then, I somehow use that f is Lipschitz (not totally sure what that means aside from |f(x) - f(y)| < M |x-y|) to show that the integral is infinite (don't totally know how to show that, either). Any help would be greatly appreciated.

## marked as duplicate by Martin R, Community♦Nov 23 '15 at 6:21

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• Is mine the same, since it has a different form of continuity in the question? – user269711 Nov 23 '15 at 5:43
• Lipschitz continuity implies uniform continuity, so the other question is slightly more general. – Martin R Nov 23 '15 at 5:49
• Ah, thank you very much! – user269711 Nov 23 '15 at 6:21