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Suppose you have the following PDE: find $u \in H^1(\Omega)$ such that

$$-\Delta u = f, \\ \frac{\partial u}{\partial n} = 0. $$

Further assume a solvability condition

$$\int_\Omega f ~\mathrm{d}\mathbf{x} = 0.$$

It is known, that this problem is solvable up to a constant. So speaking in terms of operators this means that the operator $A\colon H^1(\Omega) \to \widetilde{H}^{-1}(\Omega)$ defined by $$(A u, v) := \int_\Omega \nabla u \nabla v ~ \mathrm{d}\mathbf{x}$$ is not invertible. However one can always build up the moore-penrose pseudoinverse $A^\dagger$. So the minimum-norm-solution is given by $$u = A^\dagger f$$.

My question is, whether $A^\dagger$ is also bounded and maybe coercive (or semi-elliptic) ?


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Doesn't the pseudo-inverse of the Neumann-Laplacian just gives the "zero-mean solution"? – Dirk Sep 17 '13 at 11:21
Yes I think so. I'm just curious about the properties of the pseudo-inverse. Since for an bounded linear elliptic(coercive) operator there holds that the inverse operator is elliptic(coercive) as well. So my question is if one can make statements about the pseudoinverse as well. – Elias Ka Sep 17 '13 at 11:23

The operator $A$ is a continuous operator from $H^1(\Omega)$ to the mean value free subspace of its dual. This subspace is closed, such that the closed range theorem applies: there is a continuous operator $B$ from this subspace to $H^1(\Omega)$, such that $AB f = f$ for all $f$ in this subspace.

Note that $B$ is not uniquely defined. In order to select $A^\dagger$ from the possible choices for $B$, you can for instance require that for any $f$ holds \begin{gather} \int_\Omega Af\,dx = 0. \end{gather}

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Definition of Pseudo-Inverse: Let $$A\colon X\to Y $$ be a bounded linear operator between two hilbert-spaces. Further define $$\widetilde{A}\colon \mathrm{ker}(A)^\perp \to \mathrm{ran}(A)$$. Then we define $$A^\dagger\colon \mathrm{ran}(A)\otimes \mathrm{ran}(A)^\perp \subset Y \to X \\ y \mapsto \widetilde{A}^{-1} P_{\mathrm{ran}(A)} y$$ Here $P_{\mathrm{ran}(A)}$ is the projector onto (the closure of) the range of $A$. – Elias Ka Sep 17 '13 at 12:36
Then $A^\dagger$ is continuous iff the range of $A$ is closed – Guido Kanschat Sep 17 '13 at 12:43

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