Page 285 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes 1
Using the definition of f we have f(x) = .
n
Suppose y y n M satisfying the condition of the theorem, then

f x x,y x y y as x e .
n n n n
1
Thus Riesz representation theorem is valid if and only if y n 0 for every n.
n
Hence y y M.
n
no y M such that f(x,y) = (x,y) for every x H.

the completeness assumption cannot be left out from the Riesz-representation theorem.

*
*
Theorem 3: The mapping :H H defined by :H H defined by y fy where
fy(x) = (x,y) for every x H is an (i) additive, (ii) one-to-one, (iii) onto, (iv) symmetry, (v) not
linear.
Proof:
(i) Let us show that is additive, i.e.,

y y y y for y ,y H.
1 2 1 2 1 2
Now from the definition y y f
1 2 y 1 y 2
Hence for every x H, we get

f y y x x,y 1 y 2 x,y 1 x,y 2
1 2
f y x f y x
1 2
f y y f f y y
y y 1 2 y y 1 2
1 2 1 2
(i) is one-to-one. Let y ,y 2 H
1
Then y =f and y =f . Then
1
y
1 2 y 2
y = (y ) f = f y
1
y
2
1 2
f x = f x x H. ...(1)
y 1 y 2
f x = x,y and f x x,y
y 1 y 2
1 2
from (1), we get
x,y 1 x,y 2 x,y 1 x,y 2 0
x,y y 0 x H ...(2)
1 2
Choose x = y – y then from (2) if follows that (y – y ,y – y ) = 0
1 2 1 2 1 2

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