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

  1. #1
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    [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 <br />
= \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
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  2. #2
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    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.
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  3. #3
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    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.
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  4. #4
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    Oh. I assumed that \int f < \infty meant \int_R^\infty f \to 0 as R\to\infty :/
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  5. #5
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    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
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  6. #6
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    Quote Originally Posted by drjerry View Post
    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.
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  7. #7
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    @Jose27. Does the Sobolev embedding ||f||_{L^\infty(a,\infty)} \le C ||f||_{W^{1,1}(a,\infty)} hold?
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  8. #8
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    @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)}, <br />
\quad u\in 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!
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  9. #9
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    Quote Originally Posted by drjerry View Post
    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.
    Last edited by Laurent; June 9th 2010 at 06:56 AM.
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  10. #10
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    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.
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  11. #11
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    Quote Originally Posted by drjerry View Post
    @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.
    Last edited by Jose27; June 9th 2010 at 01:44 PM.
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