# Math Help - [SOLVED] Integrable functions vanishing at infinity.

1. ## [SOLVED] Integrable functions vanishing at infinity.

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

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

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

The functions defined by

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

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

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

Finally consider the function defined by

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

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

$\int \psi~dx = \sum_n \int \phi_n(x-n)~dx
= \sum_{n=1}^\infty n^{-2}$
,

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

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

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

and the pairwise disjointness of the supports implies that

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

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

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

Jerry

2. Hrmm, I think I can weaken it. The estimate $|f(x)|\le|f(\xi)|+\int_\xi^x|f'|\le|f(\xi)|+\int_\ xi^\infty|f'|$ holds right? If $\liminf_{x\to\infty}|f|=0$ and $f'\in L^1$ I think we can chuse xi so large that the sum is as small as we like.

3. I like the idea so far. The estimate you suggested looks good. Adding the hypothesis $\liminf_{x\to\infty} |f(x)|=0$ to conclude that $\lim_{x\to\infty} |f(x)| = 0$ is, well, maybe a little strong, but the problem is moving forward.

Now I'm getting confused: does the condition $\int |g| < \infty$ imply that, for each $\varepsilon > 0$, there is some $\xi$ satisfying $\int_\xi^\infty |g| < \varepsilon$ ?

I though results like that required the domain to have finite measure--? I'll have a look at Folland later.

As a caveat, I proposed this problem to some colleagues a while back, all Ph.D.'s in math (like myself), but none of us found a solution, at least not during the coffee break. I'm not suggesting that it's a research level problem, but it may be harder than it looks.

4. Oh. I assumed that $\int f < \infty$ meant $\int_R^\infty f \to 0$ as $R\to\infty$ :/

5. What makes analysis/measure theory so difficult is the same thing that makes it so seductive.

I'll think about this some more and get back to you; right now I gotta do some work.

Jerry

6. Originally Posted by drjerry
does the condition $\int |g| < \infty$ imply that, for each $\varepsilon > 0$, there is some $\xi$ satisfying $\int_\xi^\infty |g| < \varepsilon$ ?
This holds because $C_0^{\infty } (\mathbb{R} )$ is dense in $L^p$.

I was thinking something along the lines of Sobolev spaces: Take $f\in C_0^{\infty } (\mathbb{R} )$ and $x_0 \in \mathbb{R} \setminus supp(f)$ then $|f(x)|\leq \int_{x_0}^{x} |f'(y)|dy \leq \| f'\|_{L^1}$. By density this holds (a.e.) for functions in $W^{1,1} (\mathbb{R} )$. Now if we could extend this density result (I know $C^{\infty }$ functions are dense in any open interval, but the problem is taking them with compact support) to any interval of the form $(a,\infty )$ then the result would follow by the quote.

7. @Jose27. Does the Sobolev embedding $||f||_{L^\infty(a,\infty)} \le C ||f||_{W^{1,1}(a,\infty)}$ hold?

8. @maddas: the Sobolov embedding holds for unbounded intervals...
There exists a constant $C$, depending only on $|I|\leq \infty$, such that

$\|u\|_{L^\infty(I)} \leq C\|u\|_{W^{1,p}(I)},
(Assuming, of course, that I'm translating correctly from French to English, as the only book I have lying around is Brezis' Analyse Fonctionnelle.)

@Jose: are you saying that

$\|g\|_{L^1(\mathbf{R})} <\infty$ implies that, for all $\varepsilon >0$, there is $a>0$ such that $\|g\|_{L^1(a,\infty)} < \varepsilon$
follows from the density of $C_0^1(\mathbf{R})$ in $L^1(\mathbf{R})$ ?

If so, I still don't see a proof, and none of the pictures in my head are clear enough to produce a counter-example. Could you give some more details?

Thanks!

9. Originally Posted by drjerry
Now I'm getting confused: does the condition $\int |g| < \infty$ imply that, for each $\varepsilon > 0$, there is some $\xi$ satisfying $\int_\xi^\infty |g| < \varepsilon$ ?
Yes, sure.

By monotone or dominated convergence theorem, $\int_\xi^\infty |g|=\int_0^\infty {\bf 1}_{[\xi,\infty)}|g|\to \int 0 = 0$ as $\xi$ grows to $+\infty$.

--
And if $f'$ is continuous and $\int |f'|<\infty$, the usual $\int_0^\infty f'(u)du=\lim_{\xi\to\infty}\int_0^\xi f'(u)du=\lim_{\xi\to\infty} (f(\xi)-f(0))$ yields the existence of $\lim_{x\to\infty} f(x)$. For $f$ to be integrable, the only possible limit is 0, hence we conclude $f(x)\to 0$ as $x\to\infty$.

If $f'$ is not continuous, the equality $\int_0^\xi f'(u)du=f(\xi)-f(0)$ is still true (provided $\int_0^\xi |f'(u)|du<\infty$), hence the proof still holds.

10. Sweet!

First, thanks for the dominated convergence theorem argument on the vanishing of $\int_\xi^\infty |g|$ as $\xi\to \infty$.

Your proof of the conjecture is lovely! That the existence of the limit at infinity suffices is something that I was overlooking -- too simple.

11. Originally Posted by drjerry
@maddas: the Sobolov embedding holds for unbounded intervals...
I don't think this is valid: Obviously every integrable ( $L^p$) piecewise $C^1$ function, with integrable ( $L^p$) derivative is weakly differentiable ( $W^{1,p}$). Now it's easy to prove that if $\alpha > 1-\frac{n}{p}$ (where $n$ is the dimension) then $f_{\alpha}(x)= |x-x_0|^{\alpha }$ is a $W^{1,p}$ function in every bounded neighbourhood of $x_0$. By analogy we get that $f$ is a $W^{1,p}$ function in the complement of a bounded neighbourhood of $x_0$ iff $\alpha <1- \frac{n}{p}$. Now take an open ball around $x_0$ and the complement of a ball with a bigger radius than the first. Take $r,s$ such that $r>1-\frac{n}{p}$, and $s<1-\frac{n}{p}$. Now define $f_r$ on the first ball and $f_s$ on the complement of the second. Extend linearly so that your function is piecewise $C^1$.

Edit: Just noticed that for $n=1$ this argument implies that the function I constructed can't blow up. Your Soboloev embedding works in bounded intervals becuase $W^{1,p}$ would then just be either the space of absolutely continous functions if $p=1$ or the space of Hölder continous functions with exponent $1-\frac{1}{p}$ if $p>1$, actually I think this embedding works on an arbitrary interval, but it only implies the boundedness of the functions in case the domain is bounded.