Show that $ f (x) \to 0$ as $|x| \to \infty$.

Question

I am learning Measure Theory .However I got stuck on follow

Let $f $ be a uniformly continuous real valued function on the real line $\Bbb R.$

Assume that $f $ is integrable with respect to the Lebesgue measure on $\Bbb R$. Show that $ f (x) \to 0$ as $|x| \to \infty$.

My try:

As $f$ is integrable then $\int_{\Bbb R} |f|<\infty $. In order to prove the above I have to find a suitable $G>0$ such that $|f(x)|<\epsilon$ whenever $x<-G$ and $x>G$.

But I can't proceed anymore.Neither I could use the fact that $f$ is uniformly continuous.

Answer 1

Hints:

1. Show that for any compact set $K$ that $$\lim_{|x| \to \infty} \int_{x+K} |f(y)| \, \lambda(dy)=0.$$
2. For fixed $\epsilon>0$ choose $\delta>0$ such that $$|f(y)-f(x)| \leq \epsilon \qquad \text{for all} \, \, x,y \in \mathbb{R}, |x-y| \leq \delta.$$ If we set $K := \overline{B_{\delta}(0)}$, then this is equivalent to $$|f(y)-f(x)| \leq \epsilon \qquad \text{for all} \, \, x \in \mathbb{R}, y \in x+K.$$ Now combine the identity $$|f(x)| = \frac{1}{\lambda(K+x)} \int_{K+x} |(f(x)-f(y))+f(y)| \, \lambda(dy)$$ with the triangle inequality to conclude that $$\lim_{|x| \to \infty} |f(x)| =0.$$

Answer 2

By contradiction. Without loss of generality, suppose there exists $r>0$ and a sequence $(x_n)_{n\in N}$ tending to $\infty$ as $n\to \infty,$ such that $f(x_n)>r$ for all $n\in N.$

For each $n,$ there exists $y>x_n$ such that $f(y)=f(x_n)/2,$ otherwise the continuity of $f$ implies that for some n we have $f(y)>r/2$ for all $y>x_n,$ and the integral diverges.

Let $y_n=\inf \{\;y>x_n: f(y)=f(x_n)/2\;\}.$ By the continuity of $f,$ we have $y\in [x_n,y_n]\implies f(y)\geq f(x_n)/2>r/2\; \implies \;\int_{x_n}^{y_n}f(y) dy >r(y_n-x_n)/2.$

We may choose a subsequence $(k_n)_{n\in N}$ of $N$ such that $y_{k_n}<x_{k_{n+1}}.$ For brevity let $x_{k_n}=x'_n$ and $y_{k_n}=y'_n.$ For $f$ to be integrable we must have $\sum_{n\in N}r(y'_n-x'_n)/2<\infty, $ hence $y'_n-x'_n\to 0$ as $n\to \infty.$

The uniform continuity of $f$ implies that for some $d>0$ we have $|y-x|<d\implies |f(y)-f(x)|<r/4.$ But for sufficiently large $n$ we have $|y'_n-x'_n|<d,$ giving $r/4> |f(y'_n)-f(x'_n)| = f(x'_n)/2>r/2,$ a contradiction.

Answer 3

Here is a proof by contradiction that is easier to follow than the one posted.

For the sake of contradiction suppose that there is some $\epsilon_0$ such that $\forall A>0, \exists x\geq A, |f(x)|>\epsilon_0$.

Since $f$ is uniformly continuous, there exists $\delta >0$ such that $\forall x,y, |x-y|\leq \delta \implies |f(x)-f(y)|\leq \epsilon_0/2$.

The first statement yields the construction of an increasing sequence $x_n$ such that the intervals $[x_n-\delta, x_n+\delta]$ are disjoint and $|f(x_n)|>\epsilon_0 $.

Now, let $n\in \mathbb N$ be arbitrary and consider some $y\in[x_n-\delta, x_n+\delta] $.

The inequality $|f(x_n)|-|f(y)|\leq |f(x_n)-f(y)|\leq \epsilon_0/2 $ yields $$|f(y)|\geq |f(x_n)|-\epsilon_0/2\geq \epsilon_0/2$$

Therefore, over each of the intervals $[x_n-\delta, x_n+\delta] $, we have $|f|\geq \epsilon_0/2$ .

Note that $\displaystyle \int_{\mathbb R} |f|\geq \sum_n \int_{x_n-\delta}^{x_n+\delta} |f| \geq \sum_n \epsilon_0 \delta = \infty$

Contradiction.

Proceed similarly to prove that $f\to \infty $ as $x\to -\infty$