Let us consider the following equation

for and . In this entry, by using boothstrap argument, we show that

Theorem. If positive function solves the equation, then .

In the process of proving the result, we need the following result

Proposition. Let be a non-negative function and set.

For , there exist positive constants and depending only on and such that for any with , and satisfying

we have

.

As a consequence, we have

Corollary. For , , and , let , and satisfy.

Then for some , .

*Proof*. For some we have, by the PDE,

.

For any , we write

.

Take

and

.

Since

by the Holder inequality

.

Obviously . For any , we have by Corollary for some . Since any point can be taken as the origin, we have proved that for all .

By the Holder inequality . Note that we can differentiate under the integral sign for so

.

Therefore, the remaining task is to prove

.

Clearly, . Since is arbitrary, . Back to the decomposition by using the regularity theory for integral equations we see that is at least Holder continuous in . Since is arbitrary, is at least Holder continuous in .

It now follows from this entry that is -Holder continuous in . Note that since is locally Holder continuous in , is bounded from above. Since we have no strictly positive lower bound, we find that cannot be -Holder continuous for arbitrary . Thanks to a corollary in this entry we find that is -Holder continuous in . Now a regularity result tells us that

which implies by the decomposition

.

In other words, the regularity of further improves. We now differentiate to further improve the regularity of . We have

.

Eventually we have via a bootstrap argument.

Hôm nay có thời gian lên mạng. Rất vui vì nhìn thấy QA vẫn cập nhật tin tức hằng ngày lên đây. Khi nào cưới thông báo nhé. Chúc vui khỏe.

Comment by Lê Minh Đức — September 1, 2010 @ 2:09

Cảm ơn sếp, nhất định là sẽ thông báo cho mọi người rồi 😀

Comment by Ngô Quốc Anh — September 1, 2010 @ 2:11