2

I have the following problem:

Let $f:\mathbb{R}^n \to \mathbb{R}$ be a continuous function and Lebesgue integrable, i.e. $\int_{\mathbb{R}^n} |f|\ d\mu < \infty$ where $\mu$ is the Lebesgue measure for ${\mathbb{R}^n}$. Then $\lim_{\|x\|\ \to\ \infty} f(x) = 0$.

I have not been able to conclude whether the statement is true or false because in the hipothesis the function $f$ is not uniformly continuous like in the Barbalat's Lemma. So I'm thinking the statement is false, in that case I need a continuous function $f$ but not uniformly also Lebesgue integrable such that doesn't vanishes at $\infty$. Clearly, I have no ideas yet because to me, intuitively, there is no such $f$. Any hint?

PinRod3
  • 359
  • 1
  • 9

1 Answers1

3

Counterexample for $n=1$: define $f : \mathbb R \to \mathbb R$ as follows: for each positive integer $k$, place a triangular "pulse" centered at $x=k$, with height $2$ and base width $1/k^2$. Let $f = 0$ everywhere else. The $k$'th pulse has area $1/k^2$, and the pulses do not overlap, so $f = |f|$ integrates to $\sum_{k=1}^{\infty}1/k^2 = \pi^2/6 < \infty$. Clearly $f$ is continuous but has no limit as $|x| \to \infty$.

By changing the triangle heights to $2k$ and base widths to $1/k^3$, we can maintain the area $1/k^2$ while allowing the heights to grow without bound, so we can even make $f$ unbounded as $|x| \to \infty$.

With appropriate modifications to the pulse shape, we can also make $f$ as smooth as we like, e.g. infinitely differentiable.

Similar counterexamples can be constructed for any $n$.