I have a conjecture that's motivated by the following example.

Let $\displaystyle \phi$ be a positive, $\displaystyle C^\infty$ function that is supported in the interval $\displaystyle (-1/2, 1/2)$, normalized so that $\displaystyle \int \phi~dx = 1$, and satisfies $\displaystyle \phi(0) = \alpha > 0$.

(When I write "$\displaystyle \int~dx$" I always mean the Lebegsue integral over the entire real line.)

The functions defined by

$\displaystyle \phi_n(x) = \phi(n^2x), \quad n=1,2,\ldots$

are sported in the intervals $\displaystyle (-1/(2n^2), 1/(2n^2))$ and their integrals satisfy

$\displaystyle \int \phi_n~dx = \frac{1}{n^2}\int \phi(y)~dy = \frac{1}{n^2}$.

Finally consider the function defined by

$\displaystyle \psi(x) = \sum_{n=1}^n \phi_n(x-n).$

Since the summands are supported on pairwise disjoint intervals, the integral satisfies

$\displaystyle \int \psi~dx = \sum_n \int \phi_n(x-n)~dx

= \sum_{n=1}^\infty n^{-2}$,

which converges. By construction $\displaystyle \psi(n) = \alpha$ for all $\displaystyle n=1,2\ldots$, so that $\displaystyle \lim_{x\to\infty}\psi(x) \neq 0$.

Thus $\displaystyle \psi$ is an example of a non-negative, integrable function (i.e., an $\displaystyle L^1$-function) which does not vanish at infinity. Its derivative, however, is not so well behaved: by the chain rule and substitution,

$\displaystyle \int |\phi_n'|~dx = \int n^2|\phi'(x)|~dx = \int |\phi'(y)|~dy > 0$,

and the pairwise disjointness of the supports implies that

$\displaystyle \int |\psi'|~dx = \sum_{n=1}^\infty \int~|\phi_n'|~dx =\infty.$

Conjecture.If both $\displaystyle |f|$ and $\displaystyle |f'|$ are integrable on the real line, then $\displaystyle \lim_{x\to\infty} f(x)=0$. (Let's assume that $\displaystyle f$ is at least $\displaystyle C^1$.)

I've stated this for $\displaystyle L^1$ functions, but I'd be equally happy to see a proof for the $\displaystyle L^2$ class. There one might be able to apply Cauchy-Schwartz and integration by parts in some slick way. Any ideas?

Jerry