# question on absolute continuous function.

I am given $f,f' \in L^1(\mathbb{R})$, and f is absolute continuous, I want to show that:

$$\lim_{|x|\rightarrow \infty} f(x)= 0$$

Not sure how to show this, I know that $f(x)=\int_0^x f'(t) \, dt+f(0)$, and I can assume without loss of generality that $f(0)=0$, any help?

-
The standard way is to approach $f$ by compactly-supported functions in $W^{1,1}$. –  Siminore Jun 15 '12 at 13:12

Because $f^'\in L^1$ you know that $\lim_{x\rightarrow \infty} f(x)$ exists (note $|\int_x^y f^'| \le \int_x^y|f^'|$ and $\int_x^\infty|f^'|\rightarrow 0$ if $x\rightarrow\infty$). Since $f\in L^1$ the limit can only be zero, otherwise $\int|f|$ would not exist.

(of course the reasoning is the same for $x\rightarrow-\infty$)

-

For each $x_1 < x_2$ we have $$f(x_2) - f(x_1) = \int_{x_1}^{x_2}f'(t)dt$$ Since $f'\in L^1(\mathbb R)$ we can conclude that for $\epsilon > 0$, and $x_1$ great enough, $$|f(x_2) - f(x_1)| < \epsilon$$ Therefore $\lim_{x\to +\infty} f(x)$ exists and since $f\in L^1(\mathbb R)$ it must be $0$.

-

For every $x,s \in \mathbf{R}$ we have $$|f(x)| = \left|f(s)+\int_s^x f'(t)dt\right| \le |f(s)|+\int_s^x |f'(t)|dt.$$ Integrating the above inequality over $x\le s \le x+1$ we get \begin{eqnarray} |f(x)| &\le& \int_{x}^{x+1}|f(s)|ds+\int_{x}^{x+1}(\int_s^x |f'(t)|dt)ds\cr &\le& \int_{x}^{x+1}|f(s)|ds+\int_{x}^{x+1}(\int_{x}^{x+1}|f'(t)|dt)ds\cr &=& \int_{x}^{x+1}|f(s)|ds+\int_{x}^{x+1}|f'(t)|dt. \end{eqnarray} Thus we have $$|f(x)| \le a(x):=\int_{x}^{\infty}|f|+\int_{x}^{\infty}|f'| \ \ \forall x \in \mathbf{R}.$$ Since $f, f' \in L^1(\mathbf{R})$ it follows that $a(x) \to 0$ as $x \to \infty$. Similarly, one shows after integrating over $x-1 \le s \le x$ that $$|f(x)| \le b(x):=\int_{-\infty}^x|f|+\int_{-\infty}^x |f'| \ \ \forall x \in \mathbf{R},$$ and for the same reason as before we have $b(x) \to 0$ as $x \to -\infty$.

-