Page 220 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 18: The Natural Imbedding of N in N**

Let f, g N* and , be scalars. Then Notes
F ( f + g) = ( f + g) x = f (x) + g (x)
x
= F (f) + F (g)
x x
F is linear
x
F is bounded.
x
For any f N*, we have
|F (f)| = |f (x)|
x
f x … (1)

Thus the constant x is bounded (in the sense of a bounded linear functional) for F . Hence F is
x x
a functional on N*.
We now prove F = x
x
We have F = sup {|F (f)| : f 1}
x x
sup { F x : f 1 } (Using (1))
x
Hence F x … (2)
x
To prove the reverse inequality we consider the case when x = 0. In this case (2) gives
F = 0 = 0.
o
But F = 0 always. Hence F = 0 i.e. F = x for x = 0.
x o x
Not let x 0 be a vector in N. Then by theorem (If N is a normal linear space and x N, x 0,
o o
then there exists a functional f N* such that
o
f (x ) = x and f = 1.)
o o o o
a functional f N* such that
f (x) = x and f = 1.
But F = sup {|F (f)| : f 1}
x x
= sup { f (x) : f = 1}

and since x = f (x) sup {|f (x)| : f = 1}
we conclude that F x … (3)
x
[Note that since f (x) = x 0 we have f (x) = |f (x)|]
From (2) and (3); we have

F = x … (4)
x
Finally, we show that J is an isometric isomorphism of N into N**. For any x, y N and scalar.
F (f) = f (x + y) = f (x) + f (y)
x+y
= F (f) + F (f)
x y
F (f) = (F + F ) f … (5)
x+y x y
F = F + F … (6)
x+y x y
Further, F (f) = f ( x) = f (x) = ( F ) (f)
x x

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