In this notes, we consider their result in a very particular case – the single equations. We shall prove the following.
Theorem 2.1. Let be a positive integer and . For each positive solution of
in , there holds
Proof. For simplicity, we write
We must show that for all . For simplicity, we divide the proof into two steps.
Step 1. Proving .
Assume the contradiction, we then have two possible cases
Case 1. There is some such that .
Case 2. and there is a point such that .
As can be easily seen, in Case 2, is a minimum point for , in particular, . However, this is a contradiction since
Hence, it suffices to consider Case 1. Without loss of generality, we may assume .
Let us introduce the following notation
called the average of with respect to a ball with radius centered at . Sometimes, if , we simply write . Using this notation, we have the following well-known identity
Using the Jensen inequality,
one immediately has
Using the first inequality, we obtain
By integrating over , we get
From the second equation, , we have
Hence, there is some such that . Take a point with as the new center to make average, then from
we now have
Since the system
still holds for the new center , we also have
As before, this helps us to write
But now we can conclude that
Using we can consider . However, in order to make sure that our inequalities hold for all , we need to find a new center. Hence, there is a sequence of points such that
As a consequence of the preceding inequality, must be even.
For the sake of simplicity, let us denote
First, we observe that we can make to be as large as we want in . Indeed, if we let
then still solves
Since , we know that for all . It follows that . Hence, we can choose sufficiently large such that is as large as we want. Therefore, we can use instead of . Having this observation, for any given to be determined later, we can assume
where is chosen in such a way that . Clearly,
Integrating twice gives
Again by integrating twice, we arrive at
By repeating this procedure, we eventually obtain
with . We now denote
we then have . By mathematical induction, we easily prove that
for any . We now choose and then such that
By choosing sufficiently large and by mathematical induction, we can show that
Having all these preparation,
as . This is a contradiction.
Step 2. Proving for .
Suppose for some , at some point and for all . Going through Step 1.1 above, the signs of are alternating. Since , we must have . In particular, . Using the equation , we obtain . Integrating
shows that somewhere. This is also a contradiction.