Page 284 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 26: The Conjugate Space H*

Let us choose x to be y – y so that Notes
1 2
2
y y ,y y y y 0
1 2 1 2 1 2
y y 0
1 2
y 1 y 2

y is unique in the representation of f(x) = (x,y)
This completes the proof of the theorem.

Note The above Riesz representation theorem does not hold in an inner product space
which is not complete as shown by the example given below. In other words the
completeness assumption cannot be dropped in the above theorem.

Example: Let us consider the subspace M of l consisting of all finite sequences. This is the
2
set of all scalar sequence whose terms are zero after a finite stage. It is an incomplete inner
product space with inner product

x,y x y x,y M
n n
n 1
Now let us define

x
f x n as x x M.
n n
n 1
Linearity of f together with Hölder’s inequality yields

2
2 x n 1 2
f x x n
n n 2
n 1 n 1 n 1
2 2
2
x,x x ,
6 6
1 2
since .
n 2 6
n 1
f is a continuous linear functional on M.

We now prove that there is no y M such that

f x x,y x M.
Let us take x = e = (0,0,....,1,0,0,.....) where 1 is in n place.
th
n

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