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Several Important Theorems
byFrancis 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 that0 kvnfk22<1n
(1.1)1 Jordan, P. ; Von Neumann, J. On inner products in linear, metric spaces. Ann. ofMath. (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 +2mBecause
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= infv2Vkvfk. 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 ifVis 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: 22.1 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 that0 =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=