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Let $f$ an integrable function (i.e. $L^1(\mathbb R)$). Suppose $f$ is uniformly continuous. Prove that $$\lim_{|x|\to \infty }f(x)=0.$$

If suppose first that $\ell:=\lim_{|x|\to \infty }f(x)$ exist. By contradiction, I suppose $\ell\neq 0$, and suppose WLOG that $\ell>0$. Using continuity, there is $M>0$ such that $f(x)>\frac{\ell}{2}$ for all $x\geq M$. Therefore, $$\int_{\mathbb R} f(x)dx=\int_{|x|>M}f(x)+\int_{[-M,M]}f(x)\geq \frac{\ell}{2}\int_{|x|>M}dx+C=+\infty ,$$ what contradict $f$ integrable.

Q1) Is it correct ?

Q2) How can I do if I don't suppose that $\lim_{|x|\to \infty }f(x)$ exist ?

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Suppose there exists $x_{n}\rightarrow \infty $ and $\epsilon >0$ such that $|f(x_{n})|>\varepsilon $ for all $n$ and $x_{n}\rightarrow \infty $. By going to a subequence we may suppose $x_{n+1}>x_{n}+1$ for each $n$. Suppose $|f(x)-f(y)| < \epsilon /2$ whenever $|x-y|<\delta$. The intervals $(x_{n},x_{n}+\delta),n=k,k+1,...$ are disjoint Now $ \int_{x_{n}}^{x_{n}+\delta}f(y)dy\geq \int_{x_{n}}^{x_{n}+\delta}[f(x_{n})-\epsilon /2]dy>\frac{\varepsilon }{2}\delta$. Summing these intervals we see that $\int_{0}^{\infty }f(x)dx=\infty $, a contradiction.

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    $\begingroup$ +1 but it's slightly more complicated, because $f$ is not necessarily always positive. But you can choose infinitely many intervals where it keeps the same sign, by pigeonhole principle. Alternatively, even without considering the integral on all intervals, the function $x\to\int_0^xf(u)du$ can't be convergent. Also, you don't suppose the inequality $|f(x)-f(y)|<\epsilon/2$, it's granted by uniform continuity (there is a $\delta$ that works). You don't need either $x_{n+1}>x_n+1$, but $x_{n+1}>x_n+\delta$ (it's enough to find a subsequence, and it's possible because $x_n\to+\infty$). $\endgroup$ – Jean-Claude Arbaut May 28 '18 at 12:54
  • $\begingroup$ @Jean-ClaudeArbaut I just missed two absolute value signs. I wanted to use the fact that $|f(y)| \geq |f(x_n)|- |f(x_n)-f(x)|$. There is no need to know when $f$ is positive and when it is negative. $\endgroup$ – Kavi Rama Murthy May 29 '18 at 4:55
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Unfortunately, your hypothesis of $\lim_{|x|\to \infty }f(x)$ exist is very strong, since for continuous function, they can be integrable over $\mathbb R$ but $\limsup_{x\to \infty }f(x)$ can even be infinite.


Hint

Let $\varepsilon>0$. There is $\delta\in (0,1)$ s.t. $$|x-y|<\delta\implies |f(x)-f(y)|<\frac{\varepsilon}{2}.$$ Since $f$ is integrable, using monotone convergence, $$\lim_{n\to \infty }\int_{-n}^n |f|=\int_{\mathbb R}|f|.$$ Therefore, there is $N$ s.t. $$\int_{\mathbb R\backslash [-N,N]}|f|<\delta\varepsilon.$$ Suppose that there is $|x|>N+1$ s.t. $|f(x)|<\varepsilon$, an try to get a contradiction.

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The idea of the argument is this: an integrable function that does not tend to zero may have thinner and thinner pikes, and the integral on each pike needs to tend to zero. If the function is uniformly continuous, these pikes can't get thinner and thinner, and the function can't be integrable, as the integral on pikes does not tend to zero.

I'll consider $\Bbb R^+$ here, but the same argument works with $\Bbb R^-$, hence $\Bbb R$.

Suppose $f$ is uniformly continuous on $\Bbb R^+$ and $f(x)$ does not tend to zero as $x\to+\infty$. Let's prove it's not integrable.

Since $f$ does not converge to zero, there is an $\varepsilon>0$ such that you can define an increasing sequence $x_n\to+\infty$ such that $|f(x_n)|>\varepsilon$.

Since $f$ is uniformly continuous, for this $\varepsilon$, there is a $\delta>0$ such that $|x-y|<\delta\implies|f(x)-f(y)|<\varepsilon/2$. That means that on every interval $]x_n-\delta,x_n+\delta[$, $|f(x)|>\varepsilon/2$. Hence the area under the curve on this interval is larger that $\delta\varepsilon>0$, and that does not depend on $n$.

Therefore $\int_{x_n-\delta}^{x_n+\delta}f(u)du$ does not tend to zero as $n\to+\infty$, and the function $h(x)=\int_0^{x}f(u)du$ can't converge as $x\to+\infty$, hence the function $f$ is not integrable.

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