On page 9 in http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.639.5952&rep=rep1&type=pdf
it is being shown why the negative divergence is the adjoint of the gradient. $V: \mathbb R^n \to \mathbb R^n$ being a vector field and $f$ a smooth function with compact support, it starts with
\begin{align}V \cdot \nabla f &= \sum_i \int_{\mathbb R^n} V^i\frac{\partial f}{\partial x^i}\, dx \\
&= \,\,… \\ &=\int_{\mathbb R^n}-div \,\cdot f \, dx \\&=-div \, V \cdot f \end{align}
How to I get from the first step to the second and from the second last to the last step? Particularly, what role does the integral play? I dont understand where it comes from.
Best Answer
Here is my understanding. In general on $\mathbb{R}^n$ you have a standard inner product $\cdot$ by which you can define a pairing $\langle \cdot, \cdot \rangle$ between $\mathbb{R}^n$ and its dual $(\mathbb{R}^n)^*$. More explicitly, if $\alpha \in (\mathbb{R}^n)^*$ and $v \in \mathbb{R}^n$, then $$\langle v, \alpha\rangle := \int_{\mathbb{R}^n} \alpha^{\sharp}\cdot v \,\mathrm{vol} = \int_{\mathbb{R}^n}\alpha(v)\,\mathrm{vol},$$ since $\alpha^{\sharp}$ is by definition the vector such that $\alpha(v) = \alpha^{\sharp}\cdot v$. Note that $\langle \cdot, \cdot \rangle$ is bilinear, symmetric, positive definite and non-degenerate (because $\cdot$ is).
Given that, at each point $p$ on $\mathbb{R}^n$ the gradient $(\nabla f)_p$ sits in $(\mathbb{R}^n)^*$ (it is usually denoted by $df_p$), whereas $V_p$ is an element of $\mathbb{R}^n$. By definition, the evaluation $df(V)$ is exactly the directional derivative of $f$ with respect to $V$, so that $$df(V) = \sum_{i=1}^n V_i \frac{\partial f}{\partial x^i}.$$ Thus $$ \langle V, df\rangle := \int_{\mathbb{R}^n} df(V)\,\mathrm{vol} = \sum_{i=1}^n\int_{\mathbb{R}^n} V_i \frac{\partial f}{\partial x^i}\,\mathrm{vol}. $$ The last step works similarly, it is enough to notice that the inner product inside the integral is $\cdot$, whereas the other one is $\langle \cdot, \cdot \rangle$. Indeed on your notes the notation is consistent.