TITLE: Show that if $f$ is a uniformly continuous function on $\mathbb{R}$ and $f\in L^1(\mathbb{R})$, then $f$ is bounded and $\lim_{|x|\to\infty}f(x)=0$.
QUESTION [0 upvotes]: Show that if $f$ is a uniformly continuous function on $\mathbb{R}$ and $f\in L^1(\mathbb{R})$, then $f$ is bounded and $\lim_{|x|\to\infty}f(x)=0$.
I'm not entirely sure what I should be doing.  Should I construct on interval such that $f$ is bounded as described and then just check the limit?  Any help would be greatly appreciated.

REPLY [1 votes]: It is sufficient to consider positive $x$ and also to assume that $f$ is positive (since only $|f| $ is relevant for the $L^1$ norm). 
Assume $f$ does not converge to $0$. Then there is $\varepsilon > 0$ and a sequence $x_n\rightarrow \infty$ such that $f(x_n) \ge \varepsilon$. Without log of generality $x_{n+1}> x_n+2$. Since $f$ is uniformly continuous, there is $\delta > 0$ such that $|x-y|<\delta \Rightarrow |f(x)-f(y)| < \frac{\varepsilon}{2}$. Wlog $\delta < 1$. In particular, $f\ge \varepsilon/2$ in a $\delta$-neighbourhood of each $x_n$.
By construction, the intervals $(x_n- \delta, x_n+\delta)$ are pairwise disjoint. So
$$\infty >|f|_{L^1} = \int f \ge \sum_1^\infty\int_{(x_n- \delta, x_n+\delta)} f \ge \sum_1^\infty \varepsilon\delta $$
which is a contradiction.
So $\lim_{x\rightarrow\infty} f(x) = 0$ (similar for negative $x$). Since, as a continuous function, $f$ is bounded on each compact interval, $f$ is bounded on all of $\mathbb{R}$
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