Consider this PDE on a symmetric tensor $h$ on $S^2$:

$$\Delta \text{tr}(h) - div(div(h)) + \text{tr}(h) = f$$ where $f \in L^2(S^2)$ and $\Delta$, $div$ and $\text{tr}$ are with respect to the round metric on $S^2$.

I wish to show that there exists at least one solution to this.

If we assume for simplicity that $h = \frac{1}{2} \text{tr}(h) g_{S^2}$, then the PDE becomes $$\Delta \text{tr}(h) + 2\text{tr}(h) = 2f$$ which doesn't have a solution for every $f$ since $-2$ is an eigenvalue of $\Delta$ (I am assuming that $\Delta + \lambda$ is not surjective if $-\lambda$ is an eigenvalue; is this correct?). I am not sure how to approach this.

One approach is decomposing $h$ into its trace part and a conformal Lie derivative of a vector field $X$: $h = \frac{1}{2} \text{tr}(h) g_{S^2} + \mathcal{L}_{conf}X$. Then the PDE becomes:

$$\frac{1}{2}\Delta \text{tr}(h) - div(\Delta_{conf}X) + \text{tr}(h) = f$$ where $\Delta_{conf}$ is the conformal laplacian on vector fields.

I am not able to continue.

Any help is appreciated.