A question about Lp spaces

Question

Suppose $f\in L^p[0,\infty)$ for some $p>0$. Also suppose that $f$ is absolutely continuous and $f'\in L^1$. Show that $\lim_{x\to\infty}f(x)=0$. Since $f$ is absolutely continuous thus we have $|f(x)-f(0)|\le\int^x_0 |f'(t)| dt\le||f'||_{L^1}$. Thus we can only show that $f$ is bounded. How does it vanish at infinity?

Answer 1

Hint: Show that for all $\epsilon > 0$, there is an $M$ such that $|f(x)| < \epsilon$ for $x > M$ by applying a similar form of your inequality and using that $f \in L^p$ for some $p > 0$ (I am assuming finite.).

Answer 2

First we show that $\lim_{x \to +\infty} |f(x)|$ exists! To this End, (WLOG) assume $\{x_n\} ,~ \{y_n\}$ are strictly increasing sequences with $x_1 =y_1$ and $x_n, ~ y_n \to +\infty$ and $x_n \leq y_n \leq x_{n+1} $ such that

$$f(x_n) \to \alpha = \liminf_{x \to +\infty} |f(x)| , \quad f(y_n) \to \beta = \limsup_{x \to +\infty} |f(x)| $$
(Note that $\beta < + \infty$ since you already proved that $f$ is bounded).

Now since $f$ is absolutely continuous we have

$$f(y_n) -f(x_n)=\int^{y_n}_{x_n} f'(t) dt$$

Right side converges to zero since $f' \in L^1$ but left side converges to $\beta - \alpha$, therefore $$\alpha = \beta =\lim_{x \to +\infty} |f(x)|$$

Now since $f \in L^p$, we have $ \alpha =0$. Altogether shows that $$\lim_{x \to +\infty} f(x)=0$$