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Recently I found this statement -- the gradient operator is the adjoint of the minus divergence operator -- in one of my lecture notes. Knowing only a little about functional analysis, I'm looking for an intuitive interpretation.

I already found this topic which has a few good answers, but I'd like to view it from a somewhat different angle.

So say there are two vector spaces $X$ and $Y$, and that $L$ is a linear operator that maps elements from $X$ to $Y$, i.e. $L:X\to Y$. Furthermore, the set of all bounded/continuous linear functionals on $X$ (i.e. $f\;| f:X\to\mathbb{R}$) is called the dual (space) of $X$, marked $X'$. Mutatis mutandis for $Y'$.

Then, apparently if we take a look at our $L$, there is an operator (called the adjoint operator) $L'$ that maps elements from $Y'$ to $X'$, right? So $L':Y'\to X'$.

To make this a little less abstract, let's take the gradient operator $\nabla$. Now, I don't know how to write down appropriate spaces $X$ and $Y$ such that $\nabla:X\to Y$. I thought about $C^1[a,b]$, the space of continuous differentiable functions on the interval $[a,b]$, which is then mapped to $C^0[a,b]$. But these are only functions of a single variable, right? How to denote the space of functions that depend on two or three variables, say $(x,y)$ resp. $(x,y,z)$?

If I know the above mentioned spaces, then it should be possible to think of some operators in their duals, $X'$ and $Y'$. If I understand it correctly, the divergence operator should then be the operator to map operators from $Y'$ to $X'$.

So my actual question: how to denote the mentioned spaces, come up with a few operators in their duals $X'$ and $Y'$ (so some examples), and then show that the divergence indeed maps the operators from $Y'$ to $X'$?

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    There are some subtleties here. The gradient operator acting on an appropriate subspace of the space of square-integrable functions $\mathbb{R}^n \to \mathbb{C}$ (say) is unbounded, so cannot be continuously extended to the whole space. One must talk about densely-defined operators instead and then talking about adjoints of such things rigorously is subtle.2012-07-16

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