Problem 3

Exercise 2.3 (Integration and uniform continuity)  

1. Show that if $f\in L^1(\mathbb{R})$ and $f$ is uniformly continuous, then $\lim_{x\to\infty} f(x)=0$.
2. Was the assumption of uniform continuity necessary to conclude that $f$ decays?

Proof. Part (a): to show that $\lim_{x\to\infty} f(x)=0$, suppose otherwise. First select $\delta>0$ such that $\vert x-y\vert<\delta$ implies $\vert f(x)-f(y)\vert<\epsilon$. Let $x_1=0$ and for each $n=2,3,\dots$ select $x_n>x_{n-1}+\delta$ such that $\vert f(x_n)\vert>2\epsilon$. If $x\in B_\delta(x_n)$, then $\vert f(x)-f(x_n)\vert<\epsilon$, so that $\vert f(x)\vert>\epsilon$.

Therefore, the integral on a single ball is positive:

$$\int_{B_\delta(x_n)}\vert f(x)\vert \geq \epsilon\delta$$ (2.16)

There are infinitely many of these balls contained in the real line, so this shows

$$\int\vert f\vert \geq \sum_{n=1}^\infty \int_{B_\delta(x_n)}\vert f(x)\vert=\infty$$ (2.17)

Part (b): The assumption of uniform continuity is necessary to conclude that $f$ decays, as the following function demonstrates:

$$f(x) = \begin{cases} 1 \quad x\in \mathbb{Q}\\ 0 \quad x\in \mathbb{R}\setminus\mathbb{Q} \end{cases}$$ (2.18)

Certainly, $\int f = 0$, because $\mu(\mathbb{Q})=0$ but $\lim_{x\to\infty} \neq 0$ and in fact this limit does not exist. $\qedsymbol$