Page 258 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 23: Hilbert Spaces: The Definition and Some Simple Properties

Notes
Now x = Limx n
n
= Lim x n ( norm is continuous mapping)
n
= .
Uniqueness of x.
Let us suppose that y E, y x and y = .

x y
Convexity of E E
2

x y
… (3)
2
Also by parallelogram law, we have

2 2 2
2
x y x y x y
=
2 2 2 2 2
2 2 2 2
x y 2 x y
=
2 2 2 2
< .
2
So that

x y 2
< , a result contrary to (3).
2
Hence we must have y = x.
This completes the proof of the theorem.

Example: Give an example of a Banach space which is not an Hilbert space.

Solution: C [a, b] is a Banach space with supremum norm, i.e. if x C [a, b] then
x = Sup {|x(t)| : t [a, b]}.
Then this norm does not satisfy parallelogram law as shown below:

t a
Let x(t) = 1 and y (t) = . Then x = 1, y = 1
b a
t a
Now x (t) + y (t) = 1 + so that x + y = 2
b a

t a
x (t) – y (t) = 1 – so that x – y = 1
b a
2
Hence 2 ( x – y ) = 4, and x + y + x – y = 5
2
2
2
2
2
So that x + y + x – y 2 x + 2 y .
2
2
C [a, b] is not a Hilbert space.
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