I'm concerned with the following
Proposition: If a compact manifold $M$ satisfies $$Rc + \textstyle\frac{1}{2}\mathcal{L}_Xg = \lambda g $$
where $\lambda$ is a constant (i.e. $M$ is a compact Ricci soliton), then in fact $$Rc + \nabla^2f = \lambda g $$
so $X = \nabla f + K$ where $K$ is a Killing vector.
(Quick note: I use $\nabla f$ to indicate grad $f$. I know, I know; technically $\nabla f=df$, but $\nabla f$ is too good as notation to waste on something that already has a name, and grad $f$ is unforgivably grotesque.)
One can see this fact quoted in, for example, the second paragraph of this paper: http://www.mathem.pub.ro/bjga/v19n1/B19-1-cd-924.pdf
The reference made there is to Perelman's first paper on the Ricci flow: https://arxiv.org/pdf/math/0211159.pdf
Perelman never comes out and states this theorem explicitly as far as I know, but the outline of his reasoning for it seems to occur in Remark 3.2 on page 9. Apparently it hinges on the monotonicity of his entropy functional, but I'm not much of an expert on this stuff.
I had the following idea for an alternate approach to this proposition:
First, solve the elliptic equation $$div(X)=\Delta f$$ for the function f. (In coordinates this is $\nabla_i X^i=g^{ij}\nabla_i\nabla_j f$, just so my conventions are clear.) Now write $$X=\nabla f+(X-\nabla f)=\nabla f+K$$
It is clear that $div(K)=0$. Given that we know that the above proposition is true, it must be the case that $K$ is a Killing vector. (This is because a Killing vector is divergence-free, and the decomposition of a vector field $X$ into a gradient and a divergence-free vector field is unique, which is easy to see.)
Unfortunately, I haven't been able to prove that this divergence-free vector field $K$ is a Killing field. One attempt I made begins by integrating: $$\int |\textstyle\frac{1}{2}\mathcal{L}_K g|^2 dV = \int <K^{\flat}, div(Rc+\nabla^2 f-\lambda g)>dV$$ $$=\int<K^{\flat}, div\nabla^2 f>dV$$ $$= \int <K^{\flat}, Rc(\nabla f)> dV$$ $$= -\textstyle\frac{1}{2}\int \mathcal{L}_K g(K, \nabla f) dV $$
(I've omitted details, but I used a bunch of integration by parts, $div(K)=0$, the second Bianchi identity, and the main equation. The details are for the reader if they are interested.).
So, for example, to prove $\mathcal{L}_K g=0$ it is sufficient to prove $\mathcal{L}_K g(K, \nabla f)=0$, though I can't see any reason why this might be true either.
My question is: Does anyone have any suggestion as to how we might prove $K$ is Killing? I would prefer that the proof be "elliptic" in the sense of relying on the equations as they are and not introducing the parabolic methods of the Ricci flow as Perelman did. Any thoughts at all would be welcome.
