Hrmm, I think I can weaken it. The estimate holds right? If and I think we can chuse xi so large that the sum is as small as we like.
I have a conjecture that's motivated by the following example.
Let be a positive, function that is supported in the interval , normalized so that , and satisfies .
(When I write " " I always mean the Lebegsue integral over the entire real line.)
The functions defined by
are sported in the intervals and their integrals satisfy
.
Finally consider the function defined by
Since the summands are supported on pairwise disjoint intervals, the integral satisfies
,
which converges. By construction for all , so that .
Thus is an example of a non-negative, integrable function (i.e., an -function) which does not vanish at infinity. Its derivative, however, is not so well behaved: by the chain rule and substitution,
,
and the pairwise disjointness of the supports implies that
Conjecture. If both and are integrable on the real line, then . (Let's assume that is at least .)
I've stated this for functions, but I'd be equally happy to see a proof for the class. There one might be able to apply Cauchy-Schwartz and integration by parts in some slick way. Any ideas?
Jerry
I like the idea so far. The estimate you suggested looks good. Adding the hypothesis to conclude that is, well, maybe a little strong, but the problem is moving forward.
Now I'm getting confused: does the condition imply that, for each , there is some satisfying ?
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.
This holds because is dense in .
I was thinking something along the lines of Sobolev spaces: Take and then . By density this holds (a.e.) for functions in . Now if we could extend this density result (I know functions are dense in any open interval, but the problem is taking them with compact support) to any interval of the form then the result would follow by the quote.
@maddas: the Sobolov embedding holds for unbounded intervals...
(Assuming, of course, that I'm translating correctly from French to English, as the only book I have lying around is Brezis' Analyse Fonctionnelle.)There exists a constant , depending only on , such that
@Jose: are you saying that
follows from the density of in ?implies that, for all , there is such that
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!
Yes, sure.
By monotone or dominated convergence theorem, as grows to .
--
And if is continuous and , the usual yields the existence of . For to be integrable, the only possible limit is 0, hence we conclude as .
If is not continuous, the equality is still true (provided ), hence the proof still holds.
I don't think this is valid: Obviously every integrable ( ) piecewise function, with integrable ( ) derivative is weakly differentiable ( ). Now it's easy to prove that if (where is the dimension) then is a function in every bounded neighbourhood of . By analogy we get that is a function in the complement of a bounded neighbourhood of iff . Now take an open ball around and the complement of a ball with a bigger radius than the first. Take such that , and . Now define on the first ball and on the complement of the second. Extend linearly so that your function is piecewise .
Edit: Just noticed that for this argument implies that the function I constructed can't blow up. Your Soboloev embedding works in bounded intervals becuase would then just be either the space of absolutely continous functions if or the space of Hölder continous functions with exponent if , 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.