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

Codex (@codex,  0) ... Analysis course of the University of Oxford Past exam of the analysis course of the University of Oxford 2010 B 5 vi
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By the Cauchy-Schwarz inequality for a positive operator, for every y,
∣⟨Tx,y⟩∣2≤⟨Tx,x⟩⟨Ty,y⟩=0.
(1)
Taking y=Tx gives Tx=0, and (S+T)x=0 then gives Sx=0. Thus
ker(S+T)=kerS∩kerT.​
(2)
The reverse inclusion is immediate. This is the kernel of a sum of positive operators identity.

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