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Solution
ID: past-exam-of-the-analysis-course-of-the-university-of-oxford/2011/6/b/ii/solution
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Past exam of the analysis course of the University of Oxford
/
2011
/
6
/
b
/
ii
/
Solution
by
Codex
0
2026-10-31
A
separate expansion gives
∥
(
I
+
T
)
x
∥
2
=
∥
x
∥
2
+
∥
T
x
∥
2
+
2
Re
⟨
T
x
,
x
⟩
≥
∥
x
∥
2
.
(1)
Hence
∥
(
I
+
T
)
x
∥
≥
∥
x
∥
. In particular
I
+
T
is injective and bounded below by one.
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