Proposition 7

Let $u\in \mathscr{L}(E)$.

a) If $u$ is normal, then $u$ is a Riesz endomorphism if and only if $u$ is a Fredholm endomorphism of index 0.

b) If $u$ is hermitian, then $u$ is a Fredholm endomorphism if and only if $u$ is a Riesz endomorphism.

Every Riesz endomorphism is a Fredholm endomorphism of index 0 (proposition 5 of III, p. 46). Conversely, suppose that $u$ is a Fredholm endomorphism of index 0, and that $u$ is normal. Its nilspace then coincides with its kernel (EVT, V, p. 43, prop. 8), and is therefore of finite dimension. It then results from lemma 2 of III, p. 45 and from the definition that $u$ is a Riesz endomorphism.

Let us prove b). By a), it is enough to verify that the index of a hermitian Fredholm endomorphism $u$ is zero. But the orthogonal complement of the image of $u$ (which is closed) is then equal to the kernel of $u$ (EVT, V, p. 41, prop. 4), whence the assertion.