Limit at $\infty$ of absolutely continuous functions in $L^2(0,\infty)$

Question

If $f$ is an absolutely continuous function square integrable over $(0,\infty)$, is it necessarily true that $\lim_{x\to\infty}f(x)=0$? I think that I've seen this mentioned in passing in Dunford and Schwartz, but I don't know how to prove it.

Answer 1

It's sufficient that $f$ be uniformly continuous.

Suppose for contradiction that there are arbitrarily large $x$ such that $|f(x)| \geqslant \varepsilon$. Pick $\delta > 0$ so that $|f(x) - f(y)| \leqslant \frac{\varepsilon}{2}$ whenever $|x-y| \leqslant \delta$. Choose an infinite increasing sequence $x_n$ such that $|f(x_i)| \geqslant \varepsilon$ and $x_i, x_j$ are always at least $2 \delta$ apart if $i < j$.

For $|x - x_i| \leqslant \delta$ we have $|f(x)| \geqslant \frac{\varepsilon}{2}$, so

$$\int \limits_{x_i-\delta}^{x_i+\delta} |f(x)|^2 \, \mathrm{d} x \geqslant 2 \delta \cdot \left( \frac{\varepsilon}{2} \right)^2 = \frac{\delta \cdot \varepsilon^2}{2}.$$

That means that the positive region under the graph of $|f|^2$ is infinite, hence $f$ is not square integrable.