How is the acting of $H^{-1}$ on $H^1_0$ defined?

Solution 1:

As you defined, $H^{-1}$ is an abstract space, consisting of continuous linear functionals on $H^1_0$. So let us take an element $\phi\in H^{-1}$. What is $\phi$? This is a linear functional on $H^1_0$, so it can act on any function $u\in H^1_0$ and spit out a number. We denote this number by $\phi(u)$ or $\langle \phi,u\rangle$. The latter notation has the advantage of appearing more symmetric, and cleaner when you have complicated expressions instead of $\phi$ or $u$, as in $\langle \alpha\phi+\beta \exp(F),uv+\xi\rangle$.

Now $H^{-1}$ contains many familiar operations on functions. Remember that the elements of $H^{-1}$ are indeed operations, that do something on functions to get numbers. Evaluating a function at a given point $x$, and integrating a function against another given function are examples of such operations, which are therefore potential elements of $H^{-1}$. Let us take a function $v\in L^2$. Then we can define a linear operation $\phi$ by $$ \phi(u) = \int vu = (v,u)_{L^2}.\qquad\qquad(*) $$ Is $\phi\in H^{-1}$? We have to check two things: linearity, and continuity. Obviously $\phi$ is linear: $$ \phi(\alpha u+\beta w)=\int v(\alpha u+\beta w)=\alpha\int vu + \beta\int vw =\alpha\phi(u)+\beta\phi(w). $$ Continuity can be checked by using the Cauchy-Bunyakowsky-Schwarz inequality: $$ |\phi(u)|=|(v,u)_{L^2}|\leq \|v\|_{L^2}\|u\|_{L^2}\leq \|v\|_{L^2}\|u\|_{H^1}. $$ So $\phi\in H^{-1}$. In other words, if we define a mapping $T:v\mapsto \phi$ by ($*$), then $T(L^2)\subset H^{-1}$. Is $T:L^2\to H^{-1}$ injective? In other words, is it possible that two different functions $v_1$ and $v_2$ to give rise to the same functional $\phi$? This would mean that $$ (v_1,u)_{L^2} = (v_2,u)_{L^2}, \qquad\textrm{or}\qquad (v_1-v_2,u)_{L^2} = 0, $$ for all $u\in H^1_0$. In particular, the preceding is true for all compactly supported smooth functions $u$, which implies that $v_1$ and $v_2$ must agree with each other almost everywhere. This means $v_1=v_2$ in $L^2$. So $T:L^2\to H^{-1}$ is injective. What all this means is that we can think the image $T(L^2)$ as being $L^2$ itself, so we can think of $L^2$ being a subset of $H^{-1}$. Now we write simply $v$ instead of $\phi=T(v)$. Hence $$ \langle v,u\rangle = (v,u)_{L^2}. $$