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Let $A = -\Delta$ be the Laplace-Dirichlet operator, $D(A) = H_0^1(\Omega) \cap H^2(\Omega)$ where $\Omega \subset \mathbb{R}^d$ is a bounded domain. It is known that there exists a Green's function for $A$, i.e. $G \colon L^2(\Omega) \to D(A)$ such that $Av = f \Longleftrightarrow v = Gf$.

The problem:

I would like to prove that $G$ is bounded, i.e. $\sup_{f \in L^2,\ ||f||_2 = 1}||Gf||_2 = M < \infty$.

What I tried to do:

Let $\lambda \in \sigma(A) \Longrightarrow \lambda < 0$. (this is another a well-known result). So $0 \in \rho(A)$ and therefore $G = R(0 \ ; A)$. Then $||G|| = ||R(0 \ ; A)|| < \infty$, being the resolvent bounded by definition.

The question:

Does it work?

Thank you for your precious help and time!

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I have corrected the answer, the original answer was not right. –  Yimin Mar 23 '13 at 20:03

2 Answers 2

up vote 1 down vote accepted

The problem has something to do with the local integrability of the Green function of elliptic equation in dimension $n$.

And we know by using Schauder estimate, if the Green function has some singularity in the domain $\Omega$, then the growth of Green function when approaching the singularity is like $|x|^{2-n}$, when $n\ge 3$, and $\ln |x|$, when $n=2$.

Thus if we want $Gf\in L^2(\Omega)\subset L^2_{loc}(\mathbf{R}^n)$ for any $f\in L^2(\Omega)$, we need the Green function to be in $L^2_{loc}(\mathbf{R}^n)$, thus Green function has to be square integrable around the singularity.

which means around singularities,

[UPDATE, THANKS TO Jose27, I forgot the integral is for $Gf$, I calculated another thing. Now it is right.]

$$\int_{\Omega} \left(\int_0^{\epsilon} (r^{2-n}) r^{n-1}dr\right)^2 dx = \int_{\Omega}\left( \int_0^{\epsilon} rdr\right)^2 dx$$

So $Gf$ is square integrable around each singularity.

when $n\ge 4$, this is not integrable. So $n\le 3$. loc

when $n=3$, we can see the Green function belongs to $L^2_{loc}(\mathbf{R}^3)$, thus your statement $\|G\|\lt \infty$ can be true.

Now we can see the $L^2$ integrability works for all dimensions. It is obviously that $n=2$, the Green function belongs to $L^p_{loc}(\mathbf{R}^2)$

Remark: If Green function $G$ is in $L^1_{loc}$, then $G$ defines a bounded map from $L^2_{loc}$ to $L^2_{loc}$.

More general case is Schur test.[Thanks Jose27 again.]

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One thing I have doubts on: Since $|G_1(x,y)|\leq |\Phi(x-y)|$ ($\Phi$ is the fundamental solution of $\Delta$ and $G_1$ the Green function of $\Omega$) and $$\int_{\Omega} |\Phi(x-y)| dy \leq C$$ uniformly in $x$, a variant of Schur's test gives that $\int_{\Omega} G_1(x,y)f(y)dy$ defines a bounded operator in $L^2(\Omega)$ and this operator agrees with $G$ when $f$ is, say, smooth. Where does the restriction on the dimension come into play? Notice also that the $G$ in the question is not a "function" but an operator. –  Jose27 Mar 23 '13 at 19:26
Another way of looking at it is that $G$ is the inverse of the Laplacian which is a compact, self-adjoint operator, regardless of the dimension. –  Jose27 Mar 23 '13 at 19:34
@Jose27 You are right, it works for each dimension, I made a mistake in the post, I forgot the double integral, thanks for pointing out my error! –  Yimin Mar 23 '13 at 19:58

I dont know if your method works, but you can do this: $Gf=v$ is the same as $$\int_\Omega \nabla (Gf)\nabla \phi=\int_\Omega f\phi,\ \forall\ \phi\in H_0^1$$

If you take $\phi = Gf$ in the last equality, you get the inequality $$\|Gf\|_{H_0^1}\leq\|f\|_2$$

From the last inequality, you can conclude that $$\|Gf\|_2\leq\|f\|_2$$

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