(Uniform) continuous function $f : [0,\infty) \rightarrow \mathbb R$ is Lebesgue integrable, then $\lim_{x\rightarrow \infty} f(x) =0$?

Question

I've studied real-analysis on my own. However, it is too harsh to solve the problem which makes me confused.

Especially, checking true or false is hard, because it is hard for me to think some kinds of counter examples.

Here's my question.

Check true or false, and give me a reason if it's true, give me a counter example if it's false. :

1. If continuous function $f : [0,\infty) \rightarrow \mathbb R$ is Lebesgue integrable, then $\lim_{x\rightarrow \infty} f(x) =0$.
2. If Uniformly continuous function $f : [0,\infty) \rightarrow \mathbb R$ is Lebesgue integrable, then $\lim_{x\rightarrow \infty} f(x) =0$.

Could someone give me a nice solution or CE?

Answer 1

To disprove the continuous case use this .

Suppose $f$ is uniformly continuous. Then $|f|$ is also uniformly continuous and $\lim_{x\to\infty}f(x)\neq 0$ . Then $\lim\sup_{x\to\infty}|f(x)|\neq 0$

Thus there exists some $\epsilon>0$ and a sequence $x_{n}\to \infty$ such that $|f(x_{n})|>\epsilon$ .Now by uniform continuity, there exists $\delta>0$ such that $|x-y|\leq \delta\implies |f(x)-f(y)|\leq \frac{\epsilon}{2}$ .

So consider for each $n$ , the intervals $[x_{n}-\frac{\delta}{2},x_{n}+\frac{\delta}{2}]$ . Then $|f(x)|\geq |f(x_{n})|-|f(x)-f(x_{n})|\geq \frac{\epsilon}{2}$ . Thus $\inf_{x\in[x_{n}-\frac{\delta}{2},x_{n}+\frac{\delta}{2}]}|f(x)|>\frac{\epsilon}{2}$ for each $n$.

Thus $$\int_{\Bbb{R}}\sum_{n=1}^{\infty}\mathbf{1}_{[x_{n}-\frac{\delta}{2},x_{n}+\frac{\delta}{2}]}\cdot|f(x)|\,\,d\lambda\leq \int_{\Bbb{R}}|f(x)|\,d\lambda<\infty$$

But the left hand side is greater than $\sum_{n=1}^{\infty}\frac{\epsilon}{2}\lambda([x_{n}-\frac{\delta}{2},x_{n}+\frac{\delta}{2}])=\sum_{n=1}^{\infty}\frac{\epsilon}{2}\delta=\infty$ which is a contradiction

Answer 2

This holds in any dimension $f$, that is, if $f\in L_1(\mathbb{R}^d)$ and $f$ is uniformly continuous, then $\lim_{|x|\rightarrow\infty}f(x)=0$.

Suppose there is a sequence $x_n$ such that $|x_n|\nearrow\infty$ and $f(x_n)|>\varepsilon$ for some $\varepsilon>0$ (That is, assume $\lim_{|x|\rightarrow\infty}f(x)\neq0$). Uniform continuity implies that there is $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon/2$. For any $p\in\mathbb{R}^d$, $B(p;\delta)$nbe the open ball centered at $p$ of radius $\delta$.

Since $f\in L_1$, $\lim_n\int_{B(x_n;\delta)}|f|=0$ by dominates convergence since $|x-x_n|\geq||x|-|x_n||\xrightarrow{n\rightarrow\infty}0$.

On $B(x_n;\delta)$, $|f(y)|=|f(y)-f(x_n)+f(x_n)|\leq |f(x_n)|-|f(x_n)-f(y)|>\varepsilon/2$. Hence $$\int_{B(x_0;\delta)}|f|\geq\varepsilon/2 |B(x_n;\delta)|=\delta^d\omega_d\varepsilon/2$$ there $\omega_d$ is the volume of the $B(0;1)$. This is not possible!