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Abstract spectral theory 1
Hu xiyu
September 6, 2016

1

Riesz Representation theorem

H is a Hilbert space,linear functional l : H → R,∃!lv ∈ H,s
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proof:
assume A is the zero space of the functional l,i
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∀x ∈ A,l(x) = 0 and
H = A ⊕ A⊥
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if not,exist a ∈ A⊥ ⊂
H, a = 0, l(a) = 0(because we always can find an a vertical to A)
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E
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Lax-Milgram theorem

H is a Hilbert space, a(u, v) : H × H → R is a bilinear form satisfied ∃α, C
constant
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t ∀x ∈ H,l(X) =< wr , x >
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so we need to prove:
t −→ wt

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a(t, x) look as linear functional A(x),At (X) =< wt , x >,W : t −→ wt is a
bijective:
injective: if t1 = t2 ∈ H,t1 , t2 −→ wt ,
0 < ||t1 − t2 || =< t1 − t2 , t1 − t2 >= At1 (t1 − t2 ) − At2 (t1 − t2 )
this is a contradiction! injective of W is proved!
surjective:
suffice to prove dense and closed!
dense is trivial:consider space ImW ⊥ ,suffice to prove ImW ⊥ is a zero space
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is a Cauchy sequences of || · || ⇒ X1 , X2 ,
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(*)
if this is right,we have:
a(xi , v) =< wxi , v >
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xi −→ x in|| · ||
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E
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Remark:the inner product is nesseray conditon because we do not have approximation theorem with only bilinear form
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e:

∞

T i
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2

consider a series operator Sk =
we have:

k
i
i=1 T

Sk (1 − T ) = (1 − T )Sk = 1 − T k+1
take the limit,we have:
(limk→∞ Sk )(1 − T ) = (1 − T )(limk→∞ Sk ) = I
limk→∞ Sk = ∞ T i
i=1
so we have: 1 − T =
Q
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D
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