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Assume that $1\leq p < \infty $, $f \in L^p([0,\infty))$, and $f$ is uniformly continuous. Prove that $\lim_{x \to \infty} f(x) = 0$ .

Zarrax
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Yeonjoo Yoo
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    Here's a sketch: suppose there was $\epsilon>0$ and arbitrarily large $x_1, x_2, \dots \to \infty$ such that $f(x) > \epsilon$. Then uniform continuity shows that $f(x) > \epsilon$ in small (but bounded-below) neighborhoods of each, so... – Akhil Mathew Aug 04 '11 at 04:30
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    Yeonjoo Yoo: Would you please elaborate on where you got stuck, or perhaps let us know if you work out the details from Akhil's sketch? (Incidentally, there is a minor typo in the first sentence of Akhil's comment where it should say $f(x_k)>\epsilon$ for all $k$. More importantly in the second sentence you'll have better luck showing $f(x)>\epsilon/2$ in small (but with length bounded below) intervals at each $x_k$. More room is helpful because no lower bound was assumed on $f(x_k)-\epsilon$.) – Jonas Meyer Aug 04 '11 at 04:49
  • (Another minor typo: Akhil and I meant $|f(x)|$ where we wrote $f(x)$.) – Jonas Meyer Aug 04 '11 at 05:10
  • Related question: http://math.stackexchange.com/questions/92105/f-uniformly-continuous-and-int-a-infty-fx-dx-converges-imply-lim-x – Nate Eldredge Mar 25 '12 at 14:00

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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.

Shai Covo
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