since then w e i 0 and w f i 0 for all i It follows that if we write R W W for

Since then w e i 0 and w f i 0 for all i it follows

This preview shows page 28 - 30 out of 34 pages.

since then(w0, ei) = 0 and (w0, fi) = 0 for alli.It follows that if we writeR:W←→Wforthe linear map on this finite dimensional space which is equal to Id-Tacting onit, thenR*is given by Id-T*acting onWand we use the Hilbert space structureonWinduced as a subspace ofH.So, what we have just shown is that(3.172)(Id-T)u= 0⇐⇒uWandRu= 0,(Id-T*)u= 0⇐⇒uWandR*u= 0.Thus we really are reduced to the finite-dimensional theorem(3.173)dim Nul(R) = dim Nul(R*) onW.You no doubt know this result. It follows by observing that in this case, every-thing now onW,Ran(W) = Nul(R*)and finite dimensions(3.174)dim Nul(R) + dim Ran(R) = dimW= dim Ran(W) + dim Nul(R*).21. Fredholm operatorsDefinition21.A bounded operatorF∈ B(H) on a Hilbert space is said to beFredholmif it has the three properties in (3.154) – its null space is finite dimensional,its range is closed and the orthocomplement of its range is finite dimensional.For general Fredholm operators the row-rank=colum-rank result (3.155) does nothold. Indeed the difference of these two integers(3.175)ind(F) = dim (null(Id-K))-dim(Ran(Id-K))is a very important number with lots of interesting properties and uses.Notice that the last two conditions in (3.154) are really independent since theorthocomplement of a subspace is the same as the orthocomplement of its closure.
Background image
22. KUIPER’S THEOREM – UNDER CONSTRUCTION95There is for instance a bounded operator on a separable Hilbert space with trivialnull space and dense range which is not closed.How could this be?Think forinstance of the operator onL2(0,1) which is multiplication by the functionx.This is assuredly bounded and an element of the null space would have to satisfyxu(x) = 0 almost everywhere, and hence vanish almost everywhere. Moreover thedensity of theL2functions vanishing inx <for some (non-fixed)>0 showsthat the range is dense. However it is clearly not invertible.Before proving this result let’s check that the third condition in (3.154) reallyfollows from the first. This is a general fact which I mentioned, at least, earlier butlet me pause to prove it.Proposition36.IfB∈ B(H)is a bounded operator on a Hilbert space andB*is its adjoint then(3.176) Ran(B)= (Ran(B))={v∈ H; (v, w) = 0wRan(B)}= Nul(B*).Proof.The definition of the orthocomplement of Ran(B) shows immediatelythat(3.177)v(Ran(B))⇐⇒(v, w) = 0wRan(B)←→(v, Bu) = 0u∈ H⇐⇒(B*v, u) = 0u∈ H ⇐⇒B*v= 0⇐⇒vNul(B*).On the other hand we have already observed thatV= (V)for any subspace –since the right side is certainly contained in the left and (u, v) = 0 for allvVimplies that (u, w) = 0 for allwVby using the continuity of the inner productto pass to the limit of a sequencevnw.
Background image
Image of page 30

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture