Proposition 14

Let $u\in \mathscr{L}_2(E; F)$. Then

$\|u\|_1=$ sup$_{v\in\mathscr{L}_2(E;F)}|\langle v|u\rangle |$,

$\|v\|\leqslant 1$

the least upper bound being taken in $\mathbf{R}_+\cup \{+\infty \}$.

Let $B = (e_i)_{i\in I}$ be an orthonormal basis of the initial space Ker($u$)$^{\circ}$ of $u, C = (f_i)_{i\in I}$ an orthonormal family in F and $(\alpha_i)_{i\in I}$ a family in $(\mathbf{R}^*_+)^I$ such that $u(e_i) =\alpha_if_i$ for every $i\in I$ (Cor. 2 of IV, p. 150). The family $(\alpha_i)_{i\in I}$ is square-summable since $u$ belongs to $\mathscr{L}_2(E; F)$ (Cor. 4 of IV, p. 165). Let $(e_j)_{j\in J}$ be an orthonormal basis of E extending $(e_i)_{i\in I}$.

Let L be a finite subset of I. Let $v_L$ be the continuous finite-rank linear mapping of E into F defined by

for every $x$ in E. We have $\|v_L\|\leqslant 1$ and $v_L\in \mathscr{L}_2(E; F)$. Moreover, for every $j\in J$, we have $v_L(e_j) = 0$ if $j \notin L$ and $v_L(e_j) =f_j$ if $j\in L$. Thus

$|\langle v_L|u\rangle |=|$Tr($v^*_Lu$)$|=|\sum_{j\in J}\langle v_L(e_j)|u(e_j)\rangle |=\sum_{j\in L}\alpha_j$.

From loc. cit. we deduce that

(10) $\|u\|_1=$ sup$_L\sum_{j\in L}\alpha_j\leqslant$ sup$_{v\in\|\mathscr{L}v^2\|(E;F)\leqslant 1}|\langle v|u\rangle |$

in $\mathbf{R}_+\cup \{+\infty \}$.

This implies the equality of the proposition when $\|u\|_1= +\infty$.

Suppose that $\|u\|_1$ is finite. For every $i\in I$, let $p_i$ be the linear mapping of E into F such that $p_i(e_i) =f_i$ and $p_i(x) = 0$ for every $x\in e^{\circ}_i$. For all $j$ and $k$ in I, one has

$\langle p_j|p_k\rangle =$ Tr($p^*_jp_k$) $=\sum_{i\in J}\langle p_j(e_i)|p_k(e_i)\rangle$,

which is zero unless $j=k$, in which case this quantity is $\|f_k\|^2= 1$. The family $(p_i)_{i\in I}$ is therefore orthonormal in $\mathscr{L}_2(E; F)$. Consequently, the family $(\alpha_ip_i)_{i\in I}$ is summable in $\mathscr{L}_2(E; F)$; its sum is equal to $u$ since these two continuous linear mappings coincide on the elements $e_i$ for every $i\in J$.

Let $v\in \mathscr{L}_2(E; F)$. For every $i\in I$, one has

$\langle v|p_i\rangle =$ Tr($v^*p_i$) $=\sum_{j\in J}\langle v(e_j)|p_i(e_j)\rangle =\langle v(e_i)|f_i\rangle$.

If $\|v\|\leqslant 1$, one then obtains the estimate

$=\sum_{i\in I}\alpha_i|\langle v(e_i)|f_i\rangle |\leqslant \sum_{i\in I}\alpha_i=$ Tr($|u|$),

whence the inequality

sup$_{v\in\|\mathscr{L}v^2\|(E;F)\leqslant 1}|\langle v|u\rangle |\leqslant \|u\|_1$,

which, combined with formula (10), completes the proof of the proposition.

It follows from this proposition that the set $\mathscr{L}_1(E; F)$ of continuous linear mappings $u$ of E into F such that $\|u\|_1$ is finite is a vector subspace of $\mathscr{L}_2(E; F)$ and that the mapping $u\mapsto \|u\|_1$ is a semi-norm on $\mathscr{L}_1(E; F)$; inequality (9) shows that it is a norm.