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
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
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.