Question about the action of $H^{-1}$ After reading the post on How is the acting of $H^{-1}$ on $H^1_0$ defined?, I have the following question.
For each $v \in H^1_0$, then I can associate the functional $\phi_v:=f \to (v,f)_{L^2}$ for $f \in H^1_0$, as is done in the linked answer above. But, it seems to me that I can define $\chi_v = f \to (v,f)_{H^1}$.
But then, if I use the mapping $\chi$, I wouldn't get the following result

  
*
  
*If $v,u\in H^1_0$, 
  $$(u,v)_{L^2} = \langle u,v\rangle$$
  where the latter should denote again the dual pairing of $H^1_0$ and $H^{-1}$. 
  

because it would mean that $(u,v)_{L^2} = (u,v)_{H^1}$
What is wrong with using functional $\chi_v$?
 A: This is a reasonable question... and nothing is "wrong" with using the $H^1$ pairing, it's just that it doesn't accomplish what we want, in terms of having that nice Gelfand triple $H^1_0\to L^2\to H^{-1}$ (and variations thereupon) with $H^{-1}$ the/a dual of the left-hand side object.
It is understandable to ask the question, certainly, because many of us have been led to believe that we should automatically invoke the Riesz-Frechet theorem about identification (anyway, up to complex conjugation) of a Hilbert space with its own dual. These Gelfand triple situations seem incompatible with the Riesz-Frechet result. Indeed, the "problem" (which is not really a problem, just a psychological hazard) is that very few maps among Hilbert spaces behave "functorially" with respect to the Riesz-Frechet map. This fails already for one-dimensional real vector spaces where the map is multiplication by something other than $\pm 1$. 
So, you certainly can make your map, but, as it turns out, it is unreasonable to expect compatibility with the Gelfand triple set-up.
