Page 266 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 24: Orthogonal Complements

Since M M = {0} and Notes
H = M + M ,
we have from the definition of the direct sum of subspaces,
H = M M .

This completes the proof of the theorem.
Theorem 6: Let M be a proper closed linear sub space of a Hilbert space H. Then there exists a
non-zero vector z in H such that z M.
o o
Proof: Since M is a proper subspace of H, there exists a vector x in H which is not in M.
Let d = d (x, M) = inf { x – y } : y M}.
Since x M, we have d > 0.
Also M is a proper closed subspace of H, then by theorem: “Let M be a closed linear subspace of
a Hilbert space H. Let x be a vector not in M and let d = d (x, m) (or d is the distance from x to M).
Then there exists a unique vector y in M such that x – y = d.”
o o
There exists a vector y in M such that
o
x – y = d.
o
Let z = x – y . We then here
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z = x – y = d > 0.
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z is a non-zero vector.
o
Now we claim that Z M.
o
Let y be an arbitrary vector in M. We shall show that z y. For any scalar , we have
o
z – y = x – y – y = x – (y + y).
o o o
since M is a subspace of H and y , y M,
o
y + M M.
o
Then by definition of d, we have
x – (y + y) d
o
Now z – y = x – (y + y) d = z
o o o
z – y 2 z 2
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or (z – y, z – y) – (z , z ) 0
o o o o
or (z , z ) – (z , y) – (y, z ) + (y, y) – (z , z ) 0
o o o o o o
or (z ,y) (z ,y) (y,y) 0 … (1)
o o
The above result is true for all scalars .
Let us take (z ,y) .
o
Putting the value of , in (1), we get

(z ,y)(z ,y) (z ,y)(z ,y) 2 (z ,y)(z ,y) y 2 0
o o o o o o
2
2
2
or –2 |(z , y)| + |(z , y)| y 0
2
o o
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