Proposition 2

Tag 07X4, permanent. Cite it as [Bourbaki, Tag 07X4].

Topological Vector Spaces, chapter V, § 1. Prehilbertian spaces and hilbertian spaces

— If f is a positive hermitian form, we have

|f(x, y)|^2 \leq f(x, x) \ f(y, y)

for every x and y in E (Cauchy-Schwarz inequality).

First assume that we have $f(y, y) \neq 0$ . For every $\xi \in \mathbf{K}$ , we have

f(y, y) \ f(x + \xi y, x + \xi y) \geq 0

which can be written as

f(x, x) \ f(y, y) - |f(x, y)|^2 + (\xi f(y, y) + \overline{f(x, y)}) (\overline{\xi} f(y, y) + f(x, y)) \geq 0 .

Replacing $\xi$ by $-\overline{f(x, y)}/f(y, y)$ in this inequality, we get (7). If $f(x, x) \neq 0$ , we argue similarly.

Finally, if $f(x, x) = f(y, y) = 0$ , we have $f(x + \xi y, x + \xi y) \geq 0$ for all $\xi \in \mathbf{K}$ , which can be written as

\xi f(x, y) + \overline{\xi} f(x, y) \geq 0 .

Replacing $\xi$ by $-\overline{f(x, y)}$ in this inequality, we get $-2 \ |f(x, y)|^2 \geq 0$ , and therefore $f(x, y) = 0$ ; we again get (7) in this case.

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

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