Page 263 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes (d) Let S S , we have
1 2
x S x is orthogonal to every vector in S
2 2
x is orthogonal to every vector in S because S S .
1 1 2
x S
1
S S
2 1
(e) Let x S. Then (x, y) = 0 y S .

by definition of (S ) , x (S ) .
Thus x S x S .
S S .
This completes the proof of the theorem.

Theorem 2: If S is a non-empty subset of a Hilbert space H, then S is a closed linear subspace of
H and hence a Hilbert space.
Proof: We have

S = {x H : (x, y) = 0 y S} by definition. Since (0, y) = 0 y S, therefore at least 0 S and
thus S is non-empty.

Now let x , x S and , be scalars. Then (x , y) = 0, (x , y) = 0 for every y S.
1 2 1 2
For every y S, we have
( x + x , y) = (x , y) + (x , y)
1 2 1 2
= (0) + (0)

= 0
x + x S
1 2
S is a subspace of H.
Next we shall show that S is a closed subset of H.

Let (x ) S and x x in H.
n n
Then we have to show that x S .
For this we have to prove (x, y) = 0 for every y S.
Since x S , (x , y) = 0 for every y S and for n = 1, 2, 3, …
n n
Since the inner product is a continuous function, we get
(x , y) (x, y) as n
n
Since (x , y) = 0 n, (x, y) = 0
n
x S .
Hence S is a closed subset of H.

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