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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 $\chi_v$?

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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.

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