I Vector spaces and their topology, Basic definitions 1 Norm and seminorm on vector spaces real or complex A norm. defines a Hausdorff topology on a vector space in which the algebraic operations are con. tinuous resulting in a normed linear space If it is complete it is called a Banach space. 2 Inner product and semi inner product In the real case an inner product is a positive. definite symmetric bilinear form on X X R In the complex case it is positive definite. Hermitian symmetric sesquilinear form X X C An semi inner product gives rise. to a semi norm An inner product space is thus a special case of a normed linear space. A complete inner product space is a Hilbert space a special case of a Banach space. The polarization identity expresses the norm of an inner product space in terms of the. inner product For real inner product spaces it is, x y kx yk2 kx yk2. For complex spaces it is, x y kx yk2 ikx iyk2 kx yk2 ikx iyk2. In inner product spaces we also have the parallelogram law. kx yk2 kx yk2 2 kxk2 kyk2, This gives a criterion for a normed space to be an inner product space Any norm coming. from an inner product satisfies the parallelogram law and conversely if a norm satisfies the. parallelogram law we can show but not so easily that the polarization identity defines. an inner product which gives rise to the norm, 3 A topological vector space is a vector space endowed with a Hausdorff topology such. that the algebraic operations are continuous Note that we can extend the notion of Cauchy. sequence and therefore of completeness to a TVS a sequence xn in a TVS is Cauchy if. for every neighborhood U of 0 there exists N such that xm xn U for all m n N. A normed linear space is a TVS but there is another more general operation involving. norms which endows a vector space with a topology Let X be a vector space and suppose. T a family k k A of seminorms on X is given which are sufficient in the sense that. kxk 0 0 Then the topology generated by the sets kxk r A r 0. makes X a TVS A sequence or net xn converges to x iff kxn xk 0 for all Note. that a fortiori kxn k kxk 0 showing that each seminorm is continuous. If the number of seminorms is finite we may add them to get a norm generating the. same topology If the number is countable we may define a metric. so the topology is metrizable, Examples 0 On Rn or Cn we may put the lp norm 1 p or the weighted. lp norm with some arbitrary positive weight All of these norms are equivalent indeed. all norms on a finite dimensional space are equivalent and generate the same Banach. topology Only for p 2 is it a Hilbert space, 2 If is a subset of Rn or more generally any Hausdorff space we may define the. space Cb of bounded continuous functions with the supremum norm It is a Banach. space If X is compact this is simply the space C of continuous functions on. 3 For simplicity consider the unit interval and define C n 0 1 and C n 0 1. n N 0 1 Both are Banach spaces with the natural norms C 0 1 is the space of. Lipschitz functions C 0 1 C 0 C 0 C 1 0 1 if 0 1, 4 For 1 p and an open or closed subspace of Rn or more generally a finite. measure space we have the space Lp of equivalence classes of measurable p th power. integrable functions with equivalence being equality off a set of measure zero and for. p equivalence classes of essentially bounded functions bounded after modification. on a set of measure zero For 1 p the triangle inequality is not obvious it is. Minkowski s inequality Since we modded out the functions with Lp seminorm zero this. is a normed linear space and the Riesz Fischer theorem asserts that it is a Banach space. L2 is a Hilbert space If meas then Lp Lq if 1 q p, 5 The sequence space lp 1 p is an example of 4 in the case where the. measure space is N with the counting measure Each is a Banach space l2 is a Hilbert. space lp lq if 1 p q note the inequality is reversed from the previous example. The subspace c0 of sequences tending to 0 is a closed subspace of l. 6 If is an open set in Rn or any Hausdorff space we can equip C with the. norms f 7 f x indexed by x This makes it a TVS with the topology being that. of pointwise convergence It is not complete pointwise limit of continuous functions may. not be continuous, 7 If is an open set in Rn we can equip C with the norms f 7 kf kL K indexed. by compact subsets of thus defining the topology of uniform convergence on compact. subsets We get the same toplogy by using only the countably many compact sets. Kn x x n dist x 1 n, The topology is complete, 8 In the previous example in the case is a region in C and we take complex valued. functions we may consider the subspace H of holomorphic functions By Weierstrass s. theorem it is a closed subspace hence itself a complete TVS. 9 If f g L1 I I 0 1 and, f x x dx g x 0 x dx, for all infinitely differentiable with support contained in I so is identically zero near. 0 and 1 then we say that f is weakly differentiable and that f 0 g We can then define. the Sobolev space Wp1 I f Lp I f 0 Lp I with the norm. Z 1 Z 1 1 p, kf kWp1 I f x dx f x dx, but still incorporates first order differentiability of f. This is a larger space than C 1 I, The case p 2 is particularly useful because it allows us to deal with differentiability. in a Hilbert space context Sobolev spaces can be extended to measure any degree of. differentiability even fractional and can be defined on arbitrary domains in Rn. Subspaces and quotient spaces, If X is a vector space and S a subspace we may define the vector space X S of cosets. If X is normed we may define, kukX S inf kxkX or equivalently kx kX S inf kx skX. This is a seminorm and is a norm iff S is closed, Theorem If X is a Banach space and S is a closed subspace then S is a Banach space. and X S is a Banach space, Sketch Suppose xn is a sequence of elements of X for which the cosets x n are Cauchy. We can take a subsequence with kx n x n 1 kX S 2 n 1 n 1 2 Set s1 0 define. s2 S such that kx1 x2 s2 kX 1 2 define s3 S such that k x2 s2 x3 s3 kX. 1 4 Then xn sn is Cauchy in X, A converse is true as well and easily proved. Theorem If X is a normed linear space and S is a closed subspace such that S is a. Banach space and X S is a Banach space then X is a Banach space. Finite dimensional subspaces are always closed they re complete More generally. Theorem If S is a closed subspace of a Banach space and V is a finite dimensional. subspace then S V is closed, Sketch We easily pass to the case V is one dimensional and V S 0 We then have that. S V is algebraically a direct sum and it is enough to show that the projections S V S. and S V V are continuous since then a Cauchy sequence in S V will lead to a. Cauchy sequence in each of the closed subspaces and so to a convergent subsequence. Now the projection X X S restricts to a 1 1 map on V so an isomorphism of V onto. its image V Let V V be the continuous inverse Since S V V we may form. the composition S V S V V and it is continuous But it is just the projection. onto V The projection onto S is id so it is also continuous. Note The sum of closed subspaces of a Banach space need not be closed For a coun. terexample in a separable Hilbert space let S1 be the vector space of all real sequences. n 1 for which xn 0 if n is odd and S2 be the sequences for which x2n nx2n 1. n 1 2 Clearly X1 l2 S1 and X2 l2 S2 are closed subspaces of l2 the space. of square integrable sequences they are defined as the intersection of the null spaces of. continuous linear functionals Obviously every sequence can be written in a unique way. as sum of elements of S1 and S2, x1 x2 0 x2 x1 0 x4 2x3 0 x6 3x5 x1 x1 x3 2x3 x5 3x5. If a sequence has all but finitely many terms zero so do the two summands Thus all. such sequences belong to X1 X2 showing that X1 X2 is dense in l2 Now consider the. sequence 1 0 1 2 0 1 3 l2 Its only decomposition as elements of S1 and S2 is. 1 0 1 2 0 1 3 0 0 1 0 1 0 1 1 1 1 2 1 1 3 1, and so it does not belong to X1 X2 Thus X1 X2 is not closed in l2. Basic properties of Hilbert spaces, An essential property of Hilbert space is that the distance of a point to a closed convex. set is alway attained, Projection Theorem Let X be a Hilbert space K a closed convex subset and x X. Then there exists a unique x K such that, kx x k inf kx yk. Proof Translating we may assume that x 0 and so we must show that there is a unique. element of K of minimal norm Let d inf y K kyk and chose xn K with kxn k d. Then the parallelogram law gives, xn xm 2 xn xm 2, kxn k kxm k 2 1 kxn k2 1 kxm k2 d2. 2 2 2 2 2 2, where we have used convexity to infer that xn xm 2 K Thus xn is a Cauchy. sequence and so has a limit x which must belong to K since K is closed Since the norm. is continuous kx k limn kxn k d, For uniqueness note that if kx k kx k d then k x x 2k d and the parallelogram. kx x k2 2kx k2 2kx k2 kx x k2 2d2 2d2 4d2 0, The unique nearest element to x in K is often denoted PK x and referred to as the. projection of x onto K It satisfies PK PK PK the definition of a projection This. terminology is especially used when K is a closed linear subspace of X in which case PK. is a linear projection operator, Projection and orthogonality If S is any subset of a Hilbert space X let. S x X hx si 0 for all s S, Then S is a closed subspace of X We obviously have S S 0 and S S. Claim If S is a closed subspace of X x X and PS x the projection of x onto S then. x PS x S Indeed if s S is arbitrary and t R then, kx PS xk2 kx PS x tsk2 kx PS xk2 2t x PS x s t2 ksk2. so the quadratic polynomial on the right hand side has a minimum at t 0 Setting the. derivative there to 0 gives x PS x s 0, Thus we can write any x X as s s with s S and s S namely s PS x. s x PS x Such a decomposition is certainly unique if s s were another one we. would have s s s s S S 0 We clearly have kxk2 ksk2 ks k2. An immediate corollary is that S S for S a closed subspace since if x S we. can write it as s s whence s S S 0 i e x S We thus see that the. decomposition, x I PS x PS x, is the unique decomposition of x into elements of S and S Thus PS I PS For. any subset S of X S is the smallest closed subspace containing S. Orthonormal sets and bases in Hilbert space, Let e1 e2 P. eN be orthonormal elements, P of a Hilbert space, P X and let S be their. span Then n hx en ien S and x n hx en ien S so n hx en ien PS x But. k n hx en ien k2 n 1 hx en i2 so, hx en i2 kxk2, Bessel s inequality Now let E be an orthonormal set of arbitrary cardinality It follows. from Bessel s inequality that for 0 and x X e E hx ei is finite and. hence that e E hx ei 0 is countable We can thus extend Bessel s inequality to. an arbitrary orthonormal set X, hx ei2 kxk2, where the sum is just a countable sum of positive terms. It is useful to extend the notion of sums over sets of arbitrary cardinality If E is an. arbitary set and f E X a function mapping into a Hilbert space or any normed linear. space or even TVS we say, if the net e F f e indexed by the finite subsets F of E converges to x In other words. P if for any neighborhood U of the origin there is a finite set F0 E such that. x e F f e U whenever F is a finite subset of E containing F0P In the case E N. this is equivalent to absolute convergence of a series Note that if e E f e converges. then P is a finite F0 such that if F1 and F2 are finite supersets of F0 then. P for all there, k e F1 f e e F2 f e k It follows easily that each of the sets e E kf e k. 1 n is finite and hence f e 0 for all but countably many e E. Lemma If E is an orthonormal subset of a Hilbert space X and x X then. Proof We may order the elements e1 e2 of E for which hx ei. 6 0 Note that, k hx en ien k2 hx en i 2 kxk2, This shows that the partial P sums sN n 1 hx en ien form a Cauchy sequence and so. converge to anP element n 1 hx P en ien of X As an exercise in applying the definition. we show that e E hx eie n 1 hx en ien Given 0 pick N large enough that. n N 1 hx en i If M N and F is a finite subset of E containing e1 eN. k hx en ien hx eiek2, Letting M tend to infinity, k hx en ien hx eiek2. as required, Recall the proof that every vector space has a basis We consider the set of all linearly. independent subsets of the vector space ordered by inclusions and note that if we have a. totally ordered subset of this set then the union is a linearly independent subset containing. all its members Therefore Zorn s lemma implies that there exists a maximal linearly. independent set It follows directly from the maximality that this set also spans i e is a. basis In an inner product space we can use the same argument to establish the existence. of an orthonormal basis, In fact while bases exist for all vector spaces for infinite dimensional spaces they are. difficult or impossible to construct and almost never used Another notion of basis is much. more useful namely one that uses the topology to allow infinite linear combinations To. distinguish ordinary bases from such notions an ordinary basis is called a Hamel basis. Here we describe an orthonormal Hilbert space basis By definition this is a maximal. orthonormal set By Zorn s lemma any orthonormal set in a Hilbert space can be extended. to a basis and so orthonormal bases exist If E is such an orthonormal basis and x X. Indeed we know that the sum on the right exists in PX and it is easy to check that its inner. on functional analysis at the beginning graduate level at Penn State not meant to replace a good text on the n is a sequence of elements of X for which the

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