On the behavior at infinity of an integrable function

Question

For $f\in L^1(\mathbb{R}),$ in general we can not say $\lim\limits_{|x|\rightarrow \infty}f(x)=0.$ (In fact there are counter examples of continuous function). Suppose in addition we assume that $f$ is uniformly continuous on $\mathbb{R}$ then does the integrability (in the sense of Lebesgue) of $f$ imply $\lim\limits_{|x|\rightarrow \infty}f(x)=0?$

Answer 1

If $f$ is uniformly continous and $\lim_{x\to \infty} f(x) \neq 0$ or does not exists then there exists a increasing sequence $(x_n )$ such that $x_n > x_{n-1} +1$ and $|f(x_n )| >\varepsilon $ for some $\varepsilon >0$ since $f$ is uniformly continous then there exists $1>\delta >0$ such that $|x-y|<\delta$ implies $||f(x)| -|f(y)||\leq \frac{\varepsilon}{2} $ thus if $x\in (x_k -\delta , x_k +\delta )$ then $|f(x)| \geq \frac{\varepsilon}{2} ,$ hence $$\int_{\mathbb{R} } |f(x) | dx \geq \sum_{k=1}^{\infty} \int_{x_k -\delta}^{x_k+\delta} |f(x) |dx =\infty$$