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Several Important Theorems

by

Francis J. Narcowich

November, 2014

1The Projection Theorem

LetHbe a Hilbert space. WhenVis a nite dimensional subspace of Handf2 H, we can always nd a uniquep2Vsuch thatkfpk= min v2Vkfvk. This fact is the foundation of least-squares approximation. What happens when we allowVto be innite dimensional? We will see that the minimization problem can be solved if and only ifVis closed. Theorem 1.1(The Projection Theorem).LetHbe a Hilbert space and let Vbe a subspace ofH. For everyf2 Hthere is a uniquep2Vsuch that kfpk= minv2Vkfvkif and only ifVis a closed subspace ofH.

To prove this, we need the following lemma.

Lemma 1.2(Polarization Identity).LetHbe a Hilbert space. For every pairf;g2 H, we have kf+gk2+kfgk2= 2(kfk2+kgk2): Proof.Adding theidentitieskfgk2=kfk2hf;gihg;fi+kgk2yields the result.The polarization identity is an easy consequence of having an inner prod- uct. It is surprising that if anormsatises the polarization identity, then the normcomesfrom an inner product1. Proof.(Projection Theorem) Showing that the existence of minimizer im- plies thatVis closed is left as an exercise. So we assume thatVis closed. Forf2 H, let:= infv2Vkvfk. It is a little easier to work with this in an equivalent form,2= infv2Vkvfk2. Thus, for every" >0 there is a v "2Vsuch that2 kv"fk2< 2+". By choosing"= 1=n, wheren is a positive integer, we can nd a sequencefvng1n=1inVsuch that

0 kvnfk22<1n

(1.1)1 Jordan, P. ; Von Neumann, J. On inner products in linear, metric spaces. Ann. of

Math. (2) 36 (1935), no. 3, 719{723.

1 Of course, the same inequality holds for a possibly dierent integerm, 0 kvmfk22<1m . Adding the two yields this:

0 kvnfk2+kvmfk222<1n

+1m :(1.2) By polarization identity and a simple manipulation, we have kvnvmk2+ 4kfvn+vm2 k2= 2kfvnk2+kfvmk2: We can subtract 42from both sides and use (1.2) to get kvnvmk2+4(kfvn+vm2 k22) = 2(kfvnk2+kfvmk222)<2n +2m

Because

12 (vn+vm)2V,kfvn+vm2 k2infv2Vkvfk2=2. It follows that the second term on the left is nonnegative. Dropping it makes the left side smaller: kvnvmk2<2n +2m :(1.3) Asn;m! 1, we see thatkvnvmk !0. Thusfvng1n=1is a Cauchy sequence inHand is therefore convergent to a vectorp2 H. SinceVis closed,p2V. Furthermore, taking limits in (1.1) implies thatkpfk= inf

v2Vkvfk. The uniqueness ofpis left as an exercise.There are two important corollaries to this theorem; they follow from

problem 4 of Assignment 1, 2021, and Theorem 1.1. We list them below. Corollary 1.3.LetVbe a subspace ofH. There exists an orthogonal projectionP:H !Vfor whichkfPfk= minv2Vkfvkif and only if

Vis closed.

Corollary 1.4.LetVbe a closed subspace ofH. Then,H=VV?and (V?)?=V.

2The Riesz Representation Theorem

LetVbe a Banach space. A bounded linear transformation that mapsV intoRorCis called alinear functionalonV. The linear functionals form a Banach spaceV, called thedual spaceofV, with norm dened by kkV:= sup v6=0j(v)jkvkV: 2

2.1 The linear functionals on Hilbert space

Theorem 2.1(The Riesz Representation Theorem).LetHbe a Hilbert space and let :H !C(orR)be a bounded linear functional onH. Then, there is a uniqueg2 Hsuch that, for allf2 H,(f) =hf;gi. Proof.The functional is a bounded operator that mapsHinto the scalars. It follows from our discussion of bounded operators that the null space of ,N(), is closed. IfN() =H, then (f) = 0 for allf2 H, hence = 0. Thus we may takeg= 0. IfN()6=H, then, sinceN() is closed, we have thatH=N()N()?. In addition, sinceN()6=H, there exists a nonzero vectorh2N()?. Moreover, (h)6= 0, becausehis not in the null spaceN(). Next, note that forf2 H, the vectorw:= (h)f(f)h is inN(). To see this, observe that (w) = (h)f(f)h= (h)(f)(f)(h) = 0: Becausew= (h)f(f)h2N(), it is orthogonal toh2N()?, we have that

0 =h(h)f(f)h;hi= (h)hf;hi (f)hh;hi|{z}

khk2:

Solving this equation for (f) yields (f) =hf;(h)khk2hi. The vectorg:=(h)khk2hthen satises (f) =hf;gi. To show uniqueness, supposeg1;g22 H

satisfy (f) =hf;g1iand (f) =hf;g2i. Subtracting these two gives hf;g2g1i= 0 for allf2 H. Lettingf=g2g1results inhg2g1;g2g1i=

0. Consequently,g2=g1.2.2 Adjoints of bounded linear operators

We now turn the problem of showing that an adjoint for a bounded operator always exists. This is just a corollary of the Riesz Representation Theorem. Corollary 2.2.LetL:H ! Hbe a bounded linear operator. Then there exists a bounded linear operatorL:H ! H, called the adjoint ofL, such thathLf;hi=hf;Lhi, for allf;h2 H. Proof.Fixh2 Hand dene the linear functional h(f) =hLf;hi. Us- ing the boundedness ofLand Schwarz's inequality, we havejh(f)j kLkkfkkhk=Kkfk, and so his bounded. By Theorem 2.1, there is a 3 unique vectorginHfor which h(f) =hf;gi. The vectorgis uniquely determined by h; thusg=gha function ofh. We claim thatghis a linear function ofh. Considerh=ah1+bh2. Note that h(f) =hLf;ah1+bh2i= ah1(f)+bh2(f). Since h1(f) =hf;g1iand h2(f) =hf;g2i, we see that h(f) =hf;ghi= ah2(f) +bh2(f) =hf;agh1+bgh2i: It follows thatgh=agh1+bgh2and thatghis a linear function ofh. It is also bounded. Iff=gh, then h(gh) =kghk2. From the boundjh(f)j kLkkfkkhk, we havekghk2 jLkkghkkhk. Dividing bykghkthen yields kghk kLkkhk. Thus the correspondenceh!ghis a bounded linear function onH. Denote this function byL. SincehLf;hi=hf;ghi, we have thathLf;hi=hf;Lhi.Corollary 2.3.kLk=kLk. Proof.By problem 7 in Assignment 7, 2021,kLk= supf;hjhLf;hij, where khk=kfk= 1. On the other hand,kLk= supf;hjhLh;fij. Since hLh;fi=hf;Lhi, we have that supf;hjhLh;fij= supf;hjhLf;hij. It immediately follows thatkLk=kLk.Example 2.4.LetR= [0;1][0;1]and suppose thatkis a Hilbert-Schmidt kernel. IfLu(x) =R1

0k(x;y)u(y)dy, thenLv(x) =R1

0k(y;x)v(y)dy.

Proof.We will uses;tas the integration variables and switch back, to avoid confusion. We begin withhLu;vi=R1 0 R1

0k(s;t)u(t)dtv(s)ds. By Fubini's

theorem, we may switch the variables of integration to get this: Z 1 0 Z1 0 k(s;t)u(t)dyv(s)ds=Z 1 0 Z1 0 k(s;t)v(s)ds u(t)dt Z 1 0Z 1

0k(s;t)v(s)ds

|{z} L v u(t)dt: =hu;Lvi The result follows by changing variables fromt;stox;yin the second equa- tion above .3The Fredholm Alternative Theorem 3.1(The Fredholm Alternative).LetL:H ! Hbe a bounded linear operator whose range,R(L), is closed. Then, the equationLf=g 4 and be solved if and only ifhg;vi= 0for allv2N(L). Equivalently,

R(L) =N(L)?.

Proof.Letg2R(L), so that there is anh2 Hsuch thatg=Lh. If v2N(L), thenhg;vi=hLh;vi=hh;Lvi= 0. Consequently,R(L) N(L)?. Letf2N(L)?. SinceR(L) is closed, the projection theorem, Theorem 1.1, and Corollary 1.3, imply that there exists an orthogonal pro- jectionPontoR(L) such thatPf2R(L) andf0=fPf2R(L)?. More- over, sincefandPfare both inN(L)?, we have thatf02R(L)?\N(L)?. Hence,hLh;f0i= 0 =hh;Lf0i, for allh2 H. Settingh=Lf0then yields L f0= 0, sof02N(L). Butf02N(L)?and is thus orthogonal to itself; hence,f0= 0 andf=Pf2R(L). It immediately follows that N(L)?R(L). Since we already know thatR(L)N(L)?, we have R(L) =N(L)?.We want to point out thatR(L) being closed is crucial for the theorem to be true. If it is not closed, then the projectionPwill not exist and the proof breaks down. In that case, one actually hasR(L) =N(L)?, butnot

R(L) =N(L)?.

The theorem is stated in a variety of ways. The form that emphasizes the \alternative" is given in the result below, which follows immediately from Theorem 3.1. Corollary 3.2.LetL:H ! Hbe a bounded linear operator whose range, R(L), is closed. Then, either the equationLf=ghas a solution or there exists a vectorv2N(L)such thathg;vi 6= 0.

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