and use continuity of inversion.
Our proof that the spectrum of a bounded operator is always non-empty uses Liouville's Theorem and the Hahn-Banach Theorem, specifically the consequence of the latter asserting that the intersection of the kernels of all the bounded linear functionals on a normed linear space is
The spectrum is nonempty for every
has empty spectrum. That is,
It follows from Proposition II.40 and the continuity of
that is analytic on (with derivative
by Proposition II.40 — so is bounded on .
Then, by Liouville's Theorem, must be constant — and so
This means for all
and for all
It follows from the Hahn-Banach Theorem that
for every , which is impossible.