Ruth Williams, 1991


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Ruth Williams, 1991
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Williams, Ruth
Chung, Kai Lai
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Folder: Ruth Williams, 1991


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Mathematics -- History -- 20th century

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University of Florida
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March 14, 1991

Ruth: Thank you very much for your reply. It id good as far it goes.
But there is a better way. The following theorem must be true but at
the moment I cannot find a simple proof.
Th. If f is bounded and continuous in an open connectedd) set Do
the G f belongs to C2(D) and its Laplacian is -2f.
For R (not trivial) I gave the proof on pp. 196-7 of Lectures.
Lasfyear when I lectured in your old city Melbournes I found to my
delight that the method used there (Schwarz's famous lemma) can be
extended to any dimension and so yields a proof of the Th. But le me
delay reexamining that proof until later. As you must also have rea-
lizeds the only difficulty is to obtain an extension of f from D to a
larger openfset bounded like the D you cited k-bs;t keep its continui-
ty while making it zero outside te-sla compact set. Then my results
on pp. 192 and 194 M&. my last letter) apply and all goes smoothly. But
if f is meM'ely continuous bounded in D it may not converge at all at the
boundary and such an extension seems unlikely. Other methods must be
used such as yours in the particular case where f is 0 subja to (vani-
shing)at iD (as you pointed out in your generalization). I wonder if
your method will extend to the Th. above? Probably not because you do
not have the differentiability needed to use Ito.
Now this morning very very reluctantly I tried to find the Tth. in Port-
Stone---reluctant because their notations are so lousy and cross-refer-
ences so scarce. On p. 119, T.6.6 should contain the Th, above but in
the middle of its proof they say "wlog a f ( o) can be assumed to vanish
outside some compact subset of D". Their argument for this "wlog" was
S-= given but I do not understand it. Can you explain it or ask Por to do
so. After all, all my trouble stems exactly from problem wh mak-
ing f vanish outside ANY compact (they would make it vanish outside a
SUBcompact!) Perhaps I misread the text again because their horrible
misnotation, but you can check itawA e-ll me '
Now so far as my original problem is concerned, .it is precisely the
eigenequation AJ=X~ and the condition eC (D) (inherent interpretation
as you told mes not the exterior one which is stronger) that make the
J satisfy the conditions for f in the Th. [The vanishing part may be
Seakehed to just converging everywhere to finite limits on 6D).
Conclusion: unless you. an explain the proaf o their
DooK we must give another correct proof o the T. ~ 11 in'i myeir

;ielbourne notes by-the-by. By the meantime, can you not consult your
brilliant colleagues such as Ron or Pat to see they know a good proof
of Th. ? I wonder if it is given in Rao's book, but that will probab-
ly give me goosepimples. on the other hand, if we allow the use of
Green's FUNCTIONS and their "known" properties, don't you think you can
find an analysttin La Jolla to give a proof of the Th. (certainly known)
analytically? I won't try here because my past experience gave me pause.
Please keep this in mind at the Seminar where you can ask not only Port
but a whole hunch of "experts". Allow them to assume D to be smooth!
Zx < fE-C(D) S 0 on 6D) *Pg>^ -1!1 191j ''l! E >1, !
n=(Oll), a-1 .
f(x) = xsinx -xsinl

tf 1% &ui, A ,,-

;L :z ;f ; S (ook :2

1r 0re ka, tT4. rOs r 't Wv Cf N 0 \ sea

$ S) Tk-- Lx, 6l?6e Jos ^"'f rC'sprov/ \ (

August ), 1991
Since 60 = XJ and 6 is c2(D), 6GD(6 )=-260 would be trivial
if 60 can be extended to Rd to be C2 (Or H-cont.), by the lemma
in my Lectures regarding U(60). So the only real problem is that
extension across the bouidary. iow if we apply the above to the sub-
domain Dn which wlog (.) may be taken to be regular, then I think
the possibility of extension (even to be 02) is a standard technique
used in PDE. I seem to recall seeing it stated at the very beginning
of Hormander's second book, as a "resolution of unity". The trouble
is that I have never studied that kind of thing myself, and do not know
a neat version of it which may be applied here. Can you state and
prove such a versions unless you can find it in the book?
You said that your proof "does not use 60 = X0" and apparently
does not use Ito's formula either (at least it can be avoided). Then
you should clarify that matter by stating the minimum assumption you
need on (bounded) D, Ond 0 [sic] which will imply:
(1) 6GC(() = -2) _- 'k; if4 1p
This minimum assuption must be different from the classical one
requiring the H-continuity of 6 in Rd. What is it? That 6 is C2(D)
and 56 be merely bounded measurable in D? If true that is a nice lit-
tle supplement to the classical result, because 6 being in 02(D) by
the standard notation'Eiisonly that 60 is continuous in Do not neces-
sarily bounded. So the point seems to be that under the additional
condition of the boundedness of 60 we can then extend 0 to be at least
H-cont. (C2?) in Rd. Please state this explicitly.4 Ae o ale aspu
Therefore the crux is to state Ns appropriate extension lemma.
Consult HArmander or some other good text to see what is available.

Sq -9-

-4o W-,,


April 19, 1991
Durham, NC

Dear Professor Chung,

How are you doing? How is the beautiful California?

Stephanos Venakides is doing fine. He became a full professor recently but his grant
is not doing very well. In fact, he is at MSRI (Berkeley) and is coming back here in two
weeks. I am planning to spend two months (June-July) in Greece this summer.

The point that I tried to make in my last letter is very simple. You had asked the
Given f E Cc(D), where D is a smooth bounded domain in Rd (d > 1), is there an
F E C2 such that
AF = -2f in D?
I just said that the answer seems negative in general because, for any such F one can set


where h is harmonic in D and h = F on 9D. Then h E CO(D) and u satisfies

, Xr---% U R'

Au = -2f in D,

u = 0 on 0D.

~A cf4T

Thus u = Uf is the unique solution of (1). In general, the solution of (1) is not strong
(i.e. C2) if f is not Holder (for example, if f(x) = (In |x|)-1 near x = 0 E D-see Gilbarg
& Trudinger p. 17 and p. 54). In any case u E C,'a(D), for any a < 1.

Since u is not C2 and h, being harmonic, is COO in D, we conclude that F = u + h
cannot be C2. In other words, any F that satisfies your requirement can be written as a
sum of a harmonic function h and a very specific function u = Uf. Thus F is exactly as
smooth as Uf and it is well known that the fact that f E C,(D) does not guarantee that
Uf is in C2.


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