Proposition 17

The space $\mathscr{L}_1(E; E)$ coincides with the space $\mathscr{L}_1(E)$ of endomorphisms of finite trace of E, and one has $|$Tr($u$)$|\leqslant \|u\|_1$ for every endomorphism $u$ of E of finite trace.

We may suppose that $K =\mathbf{C}$. The space $\mathscr{L}_1(E; E)$ contains by definition the set of positive endomorphisms of finite trace, and therefore $\mathscr{L}_1(E)\subset \mathscr{L}_1(E; E)$ (EVT, p. 50, def. 8). Conversely, let us prove that $\mathscr{L}_1(E; E)$ is contained in $\mathscr{L}_1(E)$.

Let $u\in \mathscr{L}_1(E; E)$. We have $u^*\in \mathscr{L}_1(E; E)$ (lemma 8); it is therefore enough to prove that the Hermitian elements of $\mathscr{L}_1(E; E)$ are of finite trace (lemma 2 of I, p. 96).

Let $u$ be such an endomorphism. It is compact (remark 2, p. 169). Let B be an orthonormal basis of E such that $u$ is diagonal in the basis B (Theorem 1 of IV, p. 149), and let $\lambda$ be the family of eigenvalues of $u$ relative to B. The endomorphism $|u|$ is diagonalizable in the basis B and the family of its eigenvalues is $|\lambda |$ (prop. 1, d) of IV, p. 147). Since $|u|$ is of finite trace, the latter family is summable (prop. 12 of IV, p. 164), hence the family $\lambda$ is summable. Consequently, $u$ is of finite trace (loc. cit.), and we have $|$Tr($u$)$|\leqslant$ Tr($|u|$) $=\|u\|_1$.

Consequently, the space $\mathscr{L}_1(E)$ is a self-adjoint two-sided ideal of the star-algebra $\mathscr{L}(E)$. If E is of infinite dimension, this ideal is not closed in $\mathscr{L}(E)$.

Propositions 17 and 16 prove that the bilinear form $b$ on $\mathscr{L}_1(E; F)\times \mathscr{L}(F; E)$ with values in $\mathbf{C}$ defined by $b(u, v) =$ Tr($vu$) is a continuous bilinear form which puts the spaces $\mathscr{L}_1(E; F)$ and $\mathscr{L}(F; E)$ in duality.