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This is sort of an idle question, and I'll admit I didn't think very hard about it.

Let $H^1 = H^1(\mathbb{R}^n)$ be the Sobolev space with norm $||f||_{H^1}^2 = ||f||_{L^2(\mathbb{R}^n)}^2 + ||\nabla f||_{L^2(\mathbb{R}^n)}^2$. Suppose $\psi \in C^1(\mathbb{R}^n)$ is a bounded function with bounded gradient. Then the multiplication operator $Af = \psi f$ is a bounded operator on $H^1$ (with norm at most $\sqrt{||\psi||_\infty^2 + ||\nabla \psi ||_\infty^2}$ ).

What is the adjoint of $A$?

That is, given $g \in H^1$ we wish to find $h$ such that $$(\psi f, g) + (\nabla(\psi f),\nabla g) = (f,h) + (\nabla f,\nabla h)$$ for all $f \in H^1$. I tried some simple manipulations on this expression but didn't get anywhere.

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I didn't find the solution, but if we assume that the adjoint has the form $A^*f=\varphi f$ (and I'm not sure it's the case), we should have $(f(\Delta f−f),\varphi−\phi)_{L^2(\mathbb R^n)}=0$ for all $f$ smooth with compact support. –  Davide Giraudo Sep 16 '11 at 15:55
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