Zero limit

Let $f:[0, +\infty) \rightarrow \mathbb{R}$ be an integrable and uniformly continuous function. Prove that $\lim \limits_{x \rightarrow +\infty} f(x)=0$. Can the result hold if the function is just continuous instead of uniformly continuous? Give a brief explanation.

Solution

Suppose , on the contrary, that the limit is not zero. Pick an $a$ and an $x_n$ sequence such that for each $n$ both $x_n \geq x_{n_1} +1$ and $f(x_n)>a \quad (1)$ hold. Since $f$ is uniformly continuous there exists a $\delta>0$ such that:

$$\left | x-y \right |<\delta \Rightarrow \left | f(x)-f(y) \right |< \frac{a}{2} \tag{2}$$

However, in the interval $I_n = \left [ x_n - \frac{\delta}{2}, x_n + \frac{\delta}{2} \right ]$ we have that $\displaystyle \left | \int \limits_{I_n} f(x)\, {\rm d}x \right |> \frac{a\delta}{2}$ which comes as a sequence of $(1), \; (2)$. This is a contradiction. Hence $\lim \limits_{x \rightarrow +\infty} f(x)=0$.

The drop of uniformly continuity cannot be achieved. For example, take $f(x)=x \sin x^4$ which is continuous , the integral converges but the limit is not zero because for the sequence $x_n = \left [ \left ( 2n+1 \right )\frac{\pi}{4} \right ]^{1/4}$ we note that $f(x_n) \rightarrow +\infty$.