# Ngô Quốc Anh

## March 10, 2011

### log K is harmonic implies that K is contant

Filed under: PDEs — Ngô Quốc Anh @ 18:29

In this short note we present a result in a paper due to Edward M. Fan [here]. To be precise, we prove

Given a sphere $(\mathbb S^2,g_0)$ with standard metric, if $K(x)>0$ and $\Delta \ln \big(K(x)\big)=0$, then $K$ must be a constant.

Proof. To prove the result, we shall use the following well-known formula $\displaystyle \Delta \ln K=\frac{\Delta K}{K}-\frac{|\nabla K|^2}{K^2}.$

Therefore, the fact that $\ln K$ is harmonic implies that $\displaystyle\frac{\Delta K}{K}=\frac{|\nabla K|^2}{K^2}.$

Since $K>0$, multiplying both sides by $K^2$, we get $\displaystyle K\Delta K=|\nabla K|^2.$

Integrating both sides on $\mathbb S^2$ using standard volume form, we get $\displaystyle \int_{\mathbb S^2}K\Delta K dv_{g_0}=\int_{\mathbb S^2}|\nabla K|^2dv_{g_0}.$

Now integration by parts shows that $\displaystyle -\int_{\mathbb S^2}|\nabla K|^2dv_{g_0}=\int_{\mathbb S^2}|\nabla K|^2dv_{g_0}.$

This in turn shows that $|\nabla K|=0.$

We conclude that $K(x)$ must be a constant function.

## 2 Comments »

1. Where exactly does the proof go wrong when considering manifolds which are not the sphere with the round metric?

Comment by OrbiculaR — March 11, 2011 @ 1:05

• Hi, thanks for your interest in my blog. Apparently, the conclusion still holds as long as the manifolds are closed in the sense that they are compact without boundary.

Comment by Ngô Quốc Anh — March 11, 2011 @ 2:25

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