Recently, I have read a paper by Chen and Li published in the journal CPAA [here or here] about the super polyharmonic property of solutions for some partial differential systems.

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 ofin , 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 .

Step 1.1.

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

In particular,

and therefore

From the second equation, , we have

Integrating yields

and

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.

Step 1.2.

For the sake of simplicity, let us denote

and

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

which implies

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.

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