Page 282 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 26: The Conjugate Space H*

Notes
y y 1
But fy ,y y,y y ...(5)
y y y
Using (5) in (4) we obtain

fy y
From (2) and (6) it follows that

fy y

This completes the proof of the theorem.
Theorem 2: (Riesz-representation Theorem for Continuous Linear Functional on a Hilbert Space):
Let H be a Hilbert space and let f be an arbitrary functional on H . Then there exists a unique
*
vector y in H such that
f = fy, i.e. f(x) = (x,y) for every vector x H and f y .

Proof: We prove the following three steps to prove the theorem.
*
Step 1: Here we show that any f H has the representation f = fy.
If f = 0 we take y = 0 so that result follows trivially.
So let us take f 0.

We note the following properties of y in representation if it exists. First of all y 0, since
otherwise f = 0.

Further (x,y) = 0 x for which f(x) = 0. This means that if x belongs to the null space N(f) of f, then
(x,y) = 0.

y N f .
So let us consider the null space N(f) of f. Since f is continuous, we know that N(f) is a proper
closed subspace and since f 0,N f H and so N f 0 .

Hence by the orthogonal decomposition theorem, ay 0 in N f . Let us define any
0
arbitrary x H.
z f x y 0 f y x
0
Now f z f x f y f y f x 0
0 0
z N f .

Since y 0 N f , we get

0 z,y f x y f y x,y
0 0 0 0
f x y ,y f y x,y
0 0 0 0
Hence we get

f x y ,y 0 f y 0 x,y 0 0 ...(3)
0

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