Page 214 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 17: The Hahn-Banach Theorem

Let g = Re (f) then g (x) |f (x)| p (x). Notes
So by the generalised Hahn-Banach Theorem for Real Linear space, can be extended to a linear
functional g on L into R such that g = g on M and g (x) p (x) x L.
o o o
Define f (x) = g (x) – i g (i x) for x L as in the Hahn-Banach Theorem, f is linear functional on
o o o o
L such that f = f on M.
o
To complete the proof we have to prove that
|f (x)| p (x) x L.
o
i
Let x L and f (x) = r e , r > o and real. Then
o
|f (x)| = r = e re = e f (x)
–
–i
i
o o
= f (e x).
–i
o
–i
Since r = f (e x), f is real so that we can take
o o
–i
|f (x)| = r = f (e x) = g (e x) … (1)
–i
o o o
–i
–i
Since g (x) p (x), g (e x) p (e x) for x L.
o o
–i
–i
But p (e x) = |e | p (x) so that g (e x) p (x) … (2)
–i
o
It follows from (1) and (2) that
|f (x)| p (x)
o
This completes the proof of the theorem.
Corollary 1: Deduce the Hahn-Banach theorem for normed linear spaces from the generalised
Hahn-Banach theorem.
Proof: Let p (x) = f x for x N.
We first note that p (x) 0 for all x N.
Then for any x, y N, we have
p (x + y) = f x + y
f ( x + y )
= f x + f y
= p (x) + p (y)
p (x + y) p (x) + p (y)

Also p ( x) = f x = | | f x = | | p (x).
Hence p satisfies all the conditions of the generalized Hahn-Banach Theorem for
Complex Linear space. Therefore a functional f defined on all of N such that f = f on M and
o o
|f (x)| p (x) = f x x N.
o
f f … (3)
o
Since f is the extension of f from a subspace M, we get
o
f f … (4)
o
From (3) and (4) it follows that
f = f
o

LOVELY PROFESSIONAL UNIVERSITY 207