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I was told that the formal adjoint of the gradient is the negative divergence.

Let $A : H\to H$ be a bounded, linear operator, The adjoint of $A$, i.e. $A^*: H\to H$ satisfies \begin{equation*} (Au,v)=(u,A^*v) \end{equation*} for all $u, v\in H$. Moreover, $A$ is symmetric if $A^*=A$.

Could anyone help me to verify the claim:

what would be the adjoint of the gradient in the space $L^2(\mathbb{R}^d)$?

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  • $\begingroup$ Do you mean $L^2(\Bbb R^n)$? $\endgroup$
    – Berci
    Aug 31, 2015 at 10:08
  • $\begingroup$ I think there should be answers in both both $L^(\mathbb{R})$ and $L^(\mathbb{R}^n)$ cases. $\endgroup$
    – math101
    Aug 31, 2015 at 13:42
  • $\begingroup$ This is a new question to the other 3. $\endgroup$
    – math101
    Sep 3, 2015 at 3:33

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