Page 254 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 23: Hilbert Spaces: The Definition and Some Simple Properties

Theorem 2: If x and y are any two vectors in an inner product space then Notes
|(x, y)| x y … (1)
Proof: If y = 0, we get y = 0 and also theorem 1 implies that |(x, y)| = 0 so that (1) holds.
Now, let y 0, then for any scalar C we have

0 x – y 2 = (x – y,x – y)
But
(x – y, x – y) = (x, x) – (x, y) – ( y, x) + ( y, y)

= (x, x) – (x, y) – (y, x) + (y, y)
= x – (y, x) – (x, y) + | | y 2
2
2
2
2
2
x – (y, x) – (x, y) +| | y = (x, y, x – y)
2
= x – y 0 … (2)
(x,y)
Choose = 2 , y 0, y 0 .
y
We get from (2)
2
2 (x,y)(x y) (x, y) (x,y) 2
x 2 2 (x,y) 4 y 0
y y y
2
2 |(x,y)| 2 |(x,y)| 2 (x,y)
x 2 2 2 0
y y y
2 |(x,y)| 2
x 2 0
y
x y |(x, y)| 2
2
2
or |(x, y)| x y .
This completes the proof of the theorem.

Theorem 3: If X is an inner product space, then (x, x) has the properties of a norm, i.e.

x = (x, x) is a norm on X.
Proof: We shall show that satisfies the condition of a norm.

(i) x = (x, x) x = (x, x) 0 and x = 0 x = 0.
2
(ii) Let x, y X, then
x + y 2 = (x + y, x + y)

= (x, x) + (x, y) + (y, x) + (y, y) … (1)
= x + (x, y) + (x, y) + y 2
2
= x + 2Re (x, y) + y 2 [ (x,y) (x, y) 2Re(x,y)]
2
x + 2|(x, y)| + y 2 [ Re (x, y) |(x, y)|]
2

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