Don't understand a passage in Showalter (PDEs, sobolev spaces, dual spaces)

Question

Please see this:

How does the last equality follow? Why does the duality pairing become an integral over $G$ when test functions are restricted to $L^p$???

Answer 1

So, it looks like $\langle \cdot,\cdot \rangle$ indicates the duality pairing between either two $L^2$ (or generally $L^p$) spaces or between $H^1_0$ and its dual $H^{-1}$.

Now, if we have an $H^1_0$ function $f$ and an $H^{-1}$ "function" $g$, it's all well and good to write $\langle f,g \rangle$, but how do you actually calculate it? After all, $g$ could very well be some complicated distribution, and so writing an integral isn't really the right thing to do (often we do this anyway, to be suggestive, but it's just formal).

However, if $g$ is also in $L^2$, we can consider the inclusion map

$$\iota : H^1_0 \hookrightarrow L^2$$

and its adjoint

$$\iota^* : L^2 \hookrightarrow H^{-1}$$

and we will have

$$ \langle \iota f,g \rangle = \langle f,\iota^*g \rangle. $$

The duality on the right is the $H^1_0,H^{-1}$ bracket we've been trying to evaluate, the one on the left is duality in $L^p$, i.e. it is an integral in the proper sense.