# Completion of pre-Hilbert space in H. Brezis' Functional Analysis

I'm trying to solve the problem 5.12 of Harim Brezis' Functional Analysis, Sobolev Spaces and Partial Differential Equations; but I'm stucked understanding the statement which comes as follows:

Let $E$ be a vector space equipped with the scalar product $( , )$. One does not assume that $E$ is complete for the norm $|u| = (u, u)^{1/2}$ ($E$ is said to be a pre-Hilbert space).
Recall that the dual space $E^*$, equipped with the dual norm $||f||_E^*$, is complete. Let $T : E \rightarrow E^*$ be the map defined by $$\langle Tu, v\rangle_{E^*,E} = (u, v) \hspace{5mm} \forall u, v \in E.$$ Check that $T$ is a linear isometry. Is $T$ surjective? Our purpose is to show that $R(T)$ is dense in $E^*$ and that $||\hspace{2mm}||_{E^*}$ is a Hilbert norm.

I don't quite understand how does $T$ work. Does it take as arguments both $u$ and $v$ or just $v$? Is it using the Riesz-Fréchet representation theorem implicitly? Is $T \in E^*$?

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## 2 Answers

No, the confusing notation is that $\langle Tu,v\rangle_{E^*,E}$ is an alternative notation for $(Tu)(v)$.

$T$ is a mapping $E\to E^*$, and for an $u\in E$, it maps $Tu$ which maps $v\mapsto (u,v)$.

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Berci already clarified the notation issue. The author is definitely not using the Riesz Representation Theorem, which precisely the statement that, when $E$ is complete, $T$ is onto.

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