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> In article
> <1162284539.979664.169410@xxxxxxxxxxxxxxxxxxxxxxxxxxxx
> >,
> "smn" <smnewberger@xxxxxxxxxxx> wrote:
>
> > The World Wide Wade wrote:
> > > In article
> > >
> <25039834.1162267012389.JavaMail.jakarta@xxxxxxxxxxxxx
> forum.org>,
> > > craig <ctcowan@xxxxxxxxxxx> wrote:
> > >
> > > > Does there exist a smooth, non-zero, compactly
> supported function u on R
> > > > such
> > > > that
> > > >
> > > > v(x) (v defined below) can be extended to a
> smooth function on R. ??
> > > >
> > > >
> > > > Let U denote the open set where u does not
> equal zero and define
> > > >
> > > > v(x):= (u''(x))/ u(x) on U.
> > >
> > > Hint: Suppose u(x) = 0 for x <= 0 and u(x) is
> nonzero for small x
> > > > 0. For such x, use the mean value theorem twice
> to see
> > > |u''(x)/u(x)| >= |u''(x)/[u''(c_x)*x^2]|. That
> indicates v is
> > > blowing up like 1/x^2 near 0.
> > u''(x)/u''(c_x) may tend to 0 so I don't see your
> hint as conclusive.smn
>
> A hint is not supposed to be conclusive, although it
> should be
> helpful. And I think this one is.
This is related to the following question from
my quantuum mechanics hw :
Let V in L^1(R) and smooth. Define Hu:= -u''(x) +V(x) u(x). Show H has no
postive iegenvalues on L^2(R).
ie show there is no non-zero u in L^2 and c>0 such that
-u''(x) + V(x)u(x) = c u(x) . So assuming this then dividing by u(x) we see
that we can't have the above extension property.
Or I guess more correctly we can't extend
c + (u''(x))/u(x) to some function smooth V(x) which is in L^1.
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