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Consider the PDE $$\Delta f + \lambda f = g$$

on $S^2$, where $\Delta$ is with respect to the round metric, $g \in L^2(S^2)$ and $\lambda>0$. I wish to study the existence and uniqueness of this PDE for different values of $\lambda$.

It is known that if $\lambda = 0$, then we have existence if and only if $\int_{S^2} g = 0$, and we have uniqueness up to constants. Also, we have existence and uniqueness in the case $\lambda<0$.

It is also known that $\Delta$ has nonpositive eigenvalues of the form $-l(l+1)$ for nonnegative integers $l$ with the eigenfunctions being the spherical harmonics.

If $\lambda>0$ and $\lambda \neq l(l+1)$ for any positive integer $l$, do we get existence and uniqueness? Say if I choose $\lambda = 1$, do we get existence and uniqueness?

Any reference is appreciated.

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    $\begingroup$ Since $\Delta$ has compact resolvents, the only spectral values of $\Delta$ are eigenvalues. $\endgroup$
    – Jochen Glueck
    Commented Dec 10, 2021 at 19:15
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    $\begingroup$ As @JochenGlueck says... and the same applies to any compact Riemannian manifold, for the same reason. Of course, in general, it's less explicit. $\endgroup$
    – paul garrett
    Commented Dec 10, 2021 at 19:29
  • $\begingroup$ So what does that mean in terms of the existence and uniqueness of that PDE? $\endgroup$
    – Laithy
    Commented Dec 10, 2021 at 19:30
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    $\begingroup$ @Laithy: The operator $\Delta + \lambda: \operatorname{dom}(\Delta) \to L^2$ is bijective if and only if $-\lambda$ is not in the spectrum of $\Delta$. This is immediate from the definition of the spectrum. (I think this question would have been a better fit for Mathematics StackExchange.) $\endgroup$
    – Jochen Glueck
    Commented Dec 10, 2021 at 20:55
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    $\begingroup$ @Laithy: Yes, $H^2(S)$ is the right domain. The estimate you mention is also true. if $-\lambda$ is not in the spectrum, then $(\Delta + \lambda)^{-1}$ is a bounded operator from $L^2$ to $H^2$. $\endgroup$
    – Jochen Glueck
    Commented Dec 11, 2021 at 0:03

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