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Thread: [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 $\displaystyle \phi$ 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
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  2. #2
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    Hrmm, I think I can weaken it. The estimate $\displaystyle |f(x)|\le|f(\xi)|+\int_\xi^x|f'|\le|f(\xi)|+\int_\ xi^\infty|f'|$ holds right? If $\displaystyle \liminf_{x\to\infty}|f|=0$ and $\displaystyle f'\in L^1$ 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 $\displaystyle \liminf_{x\to\infty} |f(x)|=0$ to conclude that $\displaystyle \lim_{x\to\infty} |f(x)| = 0$ is, well, maybe a little strong, but the problem is moving forward.

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

    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 $\displaystyle \int f < \infty$ meant $\displaystyle \int_R^\infty f \to 0$ as $\displaystyle R\to\infty$ :/
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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 $\displaystyle \int |g| < \infty$ imply that, for each $\displaystyle \varepsilon > 0$, there is some $\displaystyle \xi$ satisfying $\displaystyle \int_\xi^\infty |g| < \varepsilon$ ?
    This holds because $\displaystyle C_0^{\infty } (\mathbb{R} )$ is dense in $\displaystyle L^p$.

    I was thinking something along the lines of Sobolev spaces: Take $\displaystyle f\in C_0^{\infty } (\mathbb{R} )$ and $\displaystyle x_0 \in \mathbb{R} \setminus supp(f)$ then $\displaystyle |f(x)|\leq \int_{x_0}^{x} |f'(y)|dy \leq \| f'\|_{L^1}$. By density this holds (a.e.) for functions in $\displaystyle W^{1,1} (\mathbb{R} )$. Now if we could extend this density result (I know $\displaystyle C^{\infty }$ functions are dense in any open interval, but the problem is taking them with compact support) to any interval of the form $\displaystyle (a,\infty )$ then the result would follow by the quote.
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  7. #7
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    @Jose27. Does the Sobolev embedding $\displaystyle ||f||_{L^\infty(a,\infty)} \le C ||f||_{W^{1,1}(a,\infty)}$ hold?
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  8. #8
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    @maddas: the Sobolov embedding holds for unbounded intervals...
    There exists a constant $\displaystyle C$, depending only on $\displaystyle |I|\leq \infty$, such that

    $\displaystyle \|u\|_{L^\infty(I)} \leq C\|u\|_{W^{1,p}(I)},
    \quad u\in W^{1,p}(I).$
    (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

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

    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 $\displaystyle \int |g| < \infty$ imply that, for each $\displaystyle \varepsilon > 0$, there is some $\displaystyle \xi$ satisfying $\displaystyle \int_\xi^\infty |g| < \varepsilon$ ?
    Yes, sure.

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

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

    If $\displaystyle f'$ is not continuous, the equality $\displaystyle \int_0^\xi f'(u)du=f(\xi)-f(0)$ is still true (provided $\displaystyle \int_0^\xi |f'(u)|du<\infty$), hence the proof still holds.
    Last edited by Laurent; Jun 9th 2010 at 05:56 AM.
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  10. #10
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    Sweet!

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

    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 ($\displaystyle L^p$) piecewise $\displaystyle C^1$ function, with integrable ($\displaystyle L^p$) derivative is weakly differentiable ($\displaystyle W^{1,p}$). Now it's easy to prove that if $\displaystyle \alpha > 1-\frac{n}{p}$ (where $\displaystyle n$ is the dimension) then $\displaystyle f_{\alpha}(x)= |x-x_0|^{\alpha }$ is a $\displaystyle W^{1,p}$ function in every bounded neighbourhood of $\displaystyle x_0$. By analogy we get that $\displaystyle f$ is a $\displaystyle W^{1,p}$ function in the complement of a bounded neighbourhood of $\displaystyle x_0$ iff $\displaystyle \alpha <1- \frac{n}{p}$. Now take an open ball around $\displaystyle x_0$ and the complement of a ball with a bigger radius than the first. Take $\displaystyle r,s$ such that $\displaystyle r>1-\frac{n}{p}$, and $\displaystyle s<1-\frac{n}{p}$. Now define $\displaystyle f_r$ on the first ball and $\displaystyle f_s$ on the complement of the second. Extend linearly so that your function is piecewise $\displaystyle C^1$.

    Edit: Just noticed that for $\displaystyle n=1$ this argument implies that the function I constructed can't blow up. Your Soboloev embedding works in bounded intervals becuase $\displaystyle W^{1,p}$ would then just be either the space of absolutely continous functions if $\displaystyle p=1$ or the space of Hölder continous functions with exponent $\displaystyle 1-\frac{1}{p}$ if $\displaystyle p>1$, 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; Jun 9th 2010 at 12:44 PM.
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