Lemma 6

Let E be a normed real vector space, let F be a real vector space, and let $u: E\rightarrow F$ be a non-injective affine mapping. Let B be a ball in E and S the corresponding sphere. Then $u(S) =$ $u(B)$, and in particular, $u(S)$ is convex.

We reduce to the case where $u$ is linear and where B is the unit ball of E. We have $u(S)\subset u(B)$. Conversely, let $x\in B$ and let $y$ be a nonzero element of Ker($u$). The image of the continuous mapping $t\mapsto \|x+ty\|$ of $\mathbf{R}$ into $\mathbf{R}_+$ is an unbounded interval containing the real number $\|x\|\leqslant 1$. There therefore exists $t\in \mathbf{R}$ such that $\|x+ty\|= 1$. Then $x+ty\in S$ and $u(x+ty) =u(x)$, hence $u(x)\in u(S)$.

Let us prove prop. 7. Let $x$ and $y$ be elements of the unit sphere of E; let us prove that the segment with endpoints $\langle x|u(x)\rangle$ and $\langle y|u(y)\rangle$ is contained in the numerical image of $u$.

Let F be the subspace of E generated by $x$ and $y$. If dim(F) = 1, we have $\langle x|u(x)\rangle =\langle y|u(y)\rangle$, whence the assertion. Otherwise, dim(F) = 2; let $p$ then be the orthogonal projector of E with image F, and let us denote by $u_F$ the endomorphism of F given by $x\mapsto p(u(x))$. Since $p$ is hermitian (lemma 3 of I, p. 133), we have $\langle z|u_F(z)\rangle =\langle z|u(z)\rangle$ for every $z\in F$, so that $\iota (u_F)\subset \iota (u)$. We may therefore suppose that E = F.

For every element $z$ of E, let $v_z$ be the hermitian endomorphism of E defined by $t\mapsto \langle z|t\rangle z$; one has $\langle z|u(z)\rangle =$ Tr($u\circ v_z$). When $z$ runs through the unit sphere of E$,v_z$ ranges over the set of orthoprojectors of rank 1 of E, which is a sphere S in the real affine subspace V of $\mathscr{L}(E)$ formed by the hermitian endomorphisms of E of trace 1 (lemma 5). The numerical image of E is therefore the set of Tr($u\circ v$), for $v\in S$. The mapping $v\mapsto$ Tr($u\circ v$) of V into $\mathbf{C}$ is linear. Since dim$_{\mathbf{R}}(V) = 3>$ dim$_{\mathbf{R}}(\mathbf{C})$, it is not injective; it therefore follows from lemma 6 that $\iota (u)$ is convex.