Past exam of the analysis course of the University of Oxford / 2010 / B / 5 / vi

Table of contents
- Solution vi

Solution

By the Cauchy-Schwarz inequality for a positive operator, for every

y

,

∣ ⟨ T x, y ⟩ ∣^{2} \leq ⟨ T x, x ⟩ ⟨ T y, y ⟩ = 0.

Taking

y = T x

gives

T x = 0

, and

(S + T) x = 0

then gives

S x = 0

. Thus

ker (S + T) = ker S \cap ker T .

The reverse inclusion is immediate. This is the kernel of a sum of positive operators identity.

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