Assume that $1\leq p < \infty $, $f \in L^p([0,\infty))$, and $f$ is uniformly continuous. Prove that $\lim_{x \to \infty} f(x) = 0$ .
Uniformly continuous $f$ in $L^p([0,\infty))$
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measure-theory
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0Related question: http://math.stackexchange.com/questions/92105/f-uniformly-continuous-and-int-a-infty-fx-dx-converges-imply-lim-x – 2012-03-25
1 Answers
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Hints: Suppose for a contradiction that $f(x) \not \to 0$. Together with the definition of uniform continuity, conclude that there exist constants $\varepsilon > 0$ and $\delta = \delta(\varepsilon) > 0$ such that for any $M > 0$ there exists $x > M$ for which it holds $ \int_{(x,x + \delta )} {|f(y)|^p \,dy} \ge \bigg(\frac{\varepsilon }{2}\bigg)^p \delta . $ However, this would imply $\int_{[0,\infty )} {|f(x)|^p \,dx} = \infty$, a contradiction.