How can I prove that the operator $$(I+\lambda G)$$ is invertible, where $\lambda >0$ and $G$ is the Green function of an elliptic operator $A$ in a bounded domain $\Omega$?

$\Omega$ can be very smooth. For the operator, we can use the Laplacian or the fractional Laplacian with exterior Dirichlet condition.

  • $\begingroup$ Can you be more explicit regarding the operator A and the domain $\Omega$ ? Regularity of the coefficients and of the boundary of $\Omega$ ? $\endgroup$
    – user69642
    Commented Feb 22, 2021 at 22:29
  • $\begingroup$ Formally, if $G = L^{-1}$, then $(I+\lambda G)^{-1} = (LG + \lambda G)^{-1} = G^{-1} (L + \lambda I)^{-1} = L (L + \lambda I)^{-1}$, where $(L + \lambda)^{-1}$ is the resolvent operator, it maps the entire space into the domain of $L$. $\endgroup$ Commented Feb 22, 2021 at 22:32
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    $\begingroup$ G is a positive and compact operator. Isn't it sufficient? $\endgroup$ Commented Feb 22, 2021 at 22:34
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    $\begingroup$ If G is the inverse Laplacian, it is not a positive operator. On the other hand, a sign change is your friend ... $\endgroup$ Commented Feb 23, 2021 at 2:57
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    $\begingroup$ @RomainGicquaud Let's take the fractional Laplacian, so the relevant spaces are $H^{2s}$ and $L^2$ $\endgroup$
    – Hiro
    Commented Feb 25, 2021 at 11:13

2 Answers 2


Here is a sketch of an approach that can can be generalized to much more extensive settings. A good reference is Michael E. Taylor's book Partial Differential Equations, Part 1, chapter 5, but it is quite standard and can be found probably in most books about linear elliptic PDEs.

Let $D:C^{\infty}(\Omega)\rightarrow\mathfrak{X}(\Omega)$ be a first order differential operator that assigns to each smooth function a vector field. Denote by $D^{*}:\mathfrak{X}(\Omega)\rightarrow C^{\infty}(\Omega)$ its $L^{2}$-dual, fully charactrized by operations on compactly supported functions. This gives rise to a boundary operator $\sigma:C^{\infty}(\Omega)\rightarrow \mathfrak{X}(\Omega)|_{\partial\Omega}$ such that,

$$ (Du,v)=(u,D^{*}v)+(\sigma u,v)_{\partial\Omega} $$ e.g. when $D=\nabla$ then $D^{*}=\nabla^{*}$ is the divergence operator and $\sigma u=u|_{\partial\Omega} N$. The result is a second-order differential operator $D^{*}D:C^{\infty}(\Omega)\rightarrow C^{\infty}(\Omega)$. In the case where $D=\nabla$, we have Poincare's inequality which reads for $u$ with $\sigma u=0$: $$ (D^{*}Du,u)=(Du,Du)\geq C\|u\|_{L^{2}}^{2} $$ This is the sense in which $D^{*}D$ is a positive definite operator on $L^2(\Omega)$. The problem here is that $D^{*}D$ is not actually defined on the whole of $L^{2}(\Omega)$, only on the dense subset $H^{2}(\Omega)$. There are several ways to contain this, also taking into account the demand on the boundary $\sigma u=0$. The most general one, I think, is the method of unbounded operators, which you can find either Rudin's Functional Analysis chapter 13 or in the Appendix of Taylor's book. In any case, once you establish the manner in which $D^{*}D:L^{2}(\Omega)\rightarrow L^{2}(\Omega)$ is positive definite, which goes through the fact that it is "self adjoint", then you can produce a left inverse for it (up to finite dimensional effects, maybe), with nonnegative eigenvalues. This inverse, which is the so called Green Function, must be compact by Relich's theorem since it takes functions from $L^{2}(\Omega)$ to the domain of $D^{*}D$, which is $H^{2}(\Omega)$. This in turn implies that $I+\lambda G$ is always invertible.


An extension of my comment above: this is a very general result.

Suppose that $-L$ generates a strongly continuous semigroup $T_t$ of contractions on some Banach space $X$. Let $$R_\lambda = \int_0^\infty e^{-\lambda t} T_t dt$$ be the resolvent, and suppose that $$G = \lim_{\lambda \to 0^+} R_\lambda$$ (with a strong limit) is well-defined on some subspace $Y$. Then: $$ R_\lambda - R_\mu + (\lambda - \mu) R_\lambda R_\mu = 0 $$ on $X$, and hence $$ R_\lambda - G + \lambda R_\lambda G = 0 $$ on $Y$. It follows that $$(I - \lambda R_\lambda)(I + \lambda G) = I - \lambda (R_\lambda - G + \lambda R_\lambda G) = I$$ on $Y$, and so $I + \lambda G$ is "invertible": there is a bounded operator $I - \lambda R_\lambda$ on $X$ which acts as the right inverse of $I + \lambda G$ on $Y$.

Note that if $L$ is the (fractional) Laplacian in a bounded domain (or, more generally, any positive self-adjoint operator satisfying the Poincaré inequality) acting on $X = L^2$, then $Y = X$, and so $I - \lambda R_\lambda$ is a true inverse of $I + \lambda G$.


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