Proposition 9

Tag 081E, permanent. Cite it as [Bourbaki, Tag 081E].

Topological Vector Spaces, chapter V, § 4. Some classes of operators in hilbertian spaces

For every hermitian endomorphism u of E, we have

\|u\| = \sup_{\|x\| \leqslant 1} |\langle x|u(x)\rangle|.

Put $\Phi = \Phi_u$ and $c = \sup_{\|x\| \leqslant 1} |\Phi(x, x)|$ , then evidently $c \leqslant \|u\|$ . Let $x, y$ be in E such that $\|x\| \leqslant 1, \|y\| \leqslant 1$ . Then

\Phi(x + y, x + y) = \Phi(x, x) + \Phi(y, y) + 2\Re\Phi(x, y),

hence

4\Re\Phi(x, y) = \Phi(x + y, x + y) - \Phi(x - y, x - y);

but

|\Phi(t, t)| \leqslant c\|t\|^2

for all

t \in E

, thus

4|\Re\Phi(x, y)| \leqslant c(\|x + y\|^2 + \|x - y\|^2) = 2c(\|x\|^2 + \|y\|^2) \leqslant 4c.

Let

a = \Phi(x, y)

; there exists a complex number

\lambda

with absolute value 1 such that

λa = |a|. Replacing y by λy in the preceding inequality, we get $|\Phi(x, y)| \leq c$ . By (15), $\|u\| \leq c$ and the proposition follows. Q.E.D.

Evidently every hermitian endomorphism is normal. Conversely :

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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