we now wish to show that an infinite complete orthonormal set is never a basis in a Hilbert space. To this end suppose X is a Hilbert space and 'A is an infinite complete ortho- normal set in this space. Since A is assumed to be infinite, we can certainly extract a denumerable sequence of distinct points of A, X1, X2, X3, ... . Consider now the series 8 1 Since (1/k4) converges we can apply Theorem 9.5(1) to assert that 8 1 Σ k=1 where k=1 K Xar must converge to some xe X. Suppose now that A were a basis for X. If so, we could write x as some finite linear combination of basis elements; that is, we could write k² x = Vaxa + Xk + vv vs Ya, ..., Y, F. ..., X₂ € A and Now let j be any other index value different from a, 1 } = ( x,, ², ² x ) = Xk) = (Xj, YaXx + k == 1 which is ridiculous. Thus A cannot be a basis. v and compute + ₂x₁) = 0, Request Explota

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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we now wish to show that an infinite complete orthonormal set is never a
basis in a Hilbert space.
To this end suppose X is a Hilbert space and 'A is an infinite complete ortho-
normal set in this space. Since A is assumed to be infinite, we can certainly extract a
denumerable sequence of distinct points of A,
X1, X2, X3, ....
Consider now the series
1
k24
k=1
Since 1(1/k4) converges we can apply Theorem 9.5(1) to assert that
22/1
where
must converge to some x e X. Suppose now that A were a basis for X. If so, we
could write x as some finite linear combination of basis elements; that is, we could
write
x = V₁x₂ + + vxv
Xar
,X,ΕΑ and
Now let j be any other index value different from a,
} = (ׂ‚ ² // × ) =
k=1
which is ridiculous. Thus A cannot be a basis.
Ya, ..., Y₂ € F.
1/2 V₂X₂
Xk) = (Xj, VxXx +
v and compute
+₂x₂) = 0₂
Request
explain
Transcribed Image Text:we now wish to show that an infinite complete orthonormal set is never a basis in a Hilbert space. To this end suppose X is a Hilbert space and 'A is an infinite complete ortho- normal set in this space. Since A is assumed to be infinite, we can certainly extract a denumerable sequence of distinct points of A, X1, X2, X3, .... Consider now the series 1 k24 k=1 Since 1(1/k4) converges we can apply Theorem 9.5(1) to assert that 22/1 where must converge to some x e X. Suppose now that A were a basis for X. If so, we could write x as some finite linear combination of basis elements; that is, we could write x = V₁x₂ + + vxv Xar ,X,ΕΑ and Now let j be any other index value different from a, } = (ׂ‚ ² // × ) = k=1 which is ridiculous. Thus A cannot be a basis. Ya, ..., Y₂ € F. 1/2 V₂X₂ Xk) = (Xj, VxXx + v and compute +₂x₂) = 0₂ Request explain
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