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If
First, note that if
If
Thus, upon choosing
is well-defined on
WLOG, suppose
Once
Since
where
So we can choose an
Using the same process for the imaginary part of
To see this, we can verify that since the bound holds on both the real and imaginary components of
This defines our function
on all of
We use this lemma in our proof for the Hahn-Banach theorem
File | Last Modified |
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Functional Analysis Lecture 5 | 2025-07-01 |
Hahn-Banach theorem | 2025-07-01 |
Created 2025-07-01 Last Modified 2025-07-01