# Show that if $f$ is a uniformly continuous function on $\mathbb{R}$ and $f\in L^1(\mathbb{R})$, then $f$ is bounded and $\lim_{|x|\to\infty}f(x)=0$.

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.

• No, you are supposed to show that $f$ is bounded even on unbounded set. The fact that it is bounded on any finite interval follows immediately from continuity (which implies uniform continuity on any closed bounded interval). Dec 6, 2015 at 8:43
• I don't follow. Could you elaborate? Dec 6, 2015 at 8:47

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 $\mathbb{R}$