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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. 1967 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 kbs;t keep its continui ty while making it zero outside tesla 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 Stonereluctant because their notations are so lousy and crossrefer 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 ell 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 bytheby. 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 < fEC(D) S 0 on 6D) *Pg>^ 1!1 191j ''l! E >1, ! n=(Oll), a1 . f(x) = xsinx xsinl tf 1% &ui, A ,, ;L :z ;f ; S (ook :2 3:) 1r 0re ka, tT4. rOs r 't Wv Cf N 0 \ sea $ S) Tk Lx, 6l?6e Jos ^"'f rC'sprov/ \ ( August ), 1991 Ruth: 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 Hcont.), 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 Hcontinuity 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 Hcont. (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,, K.L. 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 (JuneJuly) in Greece this summer. The point that I tried to make in my last letter is very simple. You had asked the following: 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 u=Fh 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 Dsee 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. Sincerely, _ (<^ ^ o  992M& 2f "1Pfrm A'tr 2f 2IP V *^ a((yfv)6 o '* *: I(A O L^/W Vf rir~o A Jt1C U u. t ~I ) ? 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