How to prove that $|x|^{2}\Delta f+Cf=0$ implies $f=0$ on $\mathbb{R}^{n}$.

Question

How to prove that $|x|^{2}\Delta f+Cf=0$ on $\mathbb{R}^{n}$ implies $f=0$ on $\mathbb{R}^{n}$?Here C is a constant.

I'm reading

Brendle's Convergence of the Yamabe flow in dimension 6 and higher.In this paper he uses the result.And he says the proof can be found in Schoen and Yau's Lectures on Differential Geometry p.209.But I didn't find it.Maybe I used the wrong version of reference book.

Any idea to prove this result will be appreciated.

---

• Brendle, Simon, Convergence of the Yamabe flow in dimension 6 and higher, Invent. Math. 170, No. 3, 541-576 (2007). ZBL1130.53044.

• Schoen, Richard; Yau, Shing-Tung, Lectures on differential geometry, Conference Proceedings and Lecture Notes in Geometry and Topology. 1. Cambridge, MA: International Press. v, 414 p. (1994). ZBL0830.53001.

Answer 1

I quote from page 209:

The following lemma shows that the operator $r^2 \Delta_0 + (n-2)(k+2)(k+1)$ is invertible on $\mathcal P_{k+2}$. Therefore there exists a unique $f\in \mathcal P_{k+2}$ satisfying $(3.2)$. The proof is completed.

Lemma 3.3. The eigenvalues of $r^2\Delta_0$ on $\mathcal P_m$ are $$ \{ \lambda_j = 2j(n-2+2m-2j): j=1,\dots, [m/2]\}.$$ The eigenfunction corresponding to $\lambda_j$ has the form $r^{2j}\psi$, where $\psi\in \mathcal P_{m-2j}$ is a harmonic polynomial.

So $[r^2 \Delta_0 + (n-2)(k+2)(k+1)]f=g$ is uniquely solvable (I presume the uniqueness class is the same as your paper...). In your case $g=0$, $f=0$ is a solution, so it is the only solution.

PS note the existence of eigenvalues means precisely that the result is false for some values of $C$.