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Banach Spaces Ck[A, B] 1. Banach Spaces C K[A, B]

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Banach Spaces Ck[A, B] 1. Banach Spaces C K[A, B] (November 13, 2016) Banach spaces Ck[a; b] Paul Garrett [email protected] http:=/www.math.umn.edu/egarrett/ [This document is http://www.math.umn.edu/~garrett/m/real/notes 2016-17/02d spaces Ck.pdf] 1. Banach spaces Ck[a; b] 2. Non-Banach limit C1[a; b] of Banach spaces Ck[a; b] We specify natural topologies, in which differentiation or other natural operators are continuous, and so that the space is complete. Many familiar and useful spaces of continuous or differentiable functions, such as Ck[a; b], have natural metric structures, and are complete. In these cases, the metric d(; ) comes from a norm j · j, on the functions, giving Banach spaces. Other natural function spaces, such as C1[a; b], are not Banach, but still do have a metric topology and are complete: these are Fr´echetspaces, appearing as (projective) limits of Banach spaces, as below. These lack some of the conveniences of Banach spaces, but their expressions as limits of Banach spaces is often sufficient. 1. Banach spaces Ck[a; b] We give the vector space Ck[a; b] of k-times continuously differentiable functions on an interval [a; b] a metric which makes it complete. Mere pointwise limits of continuous functions easily fail to be continuous. First recall the standard [1.0.1] Claim: The set Co(K) of complex-valued continuous functions on a compact set K is complete with o the metric jf − gjCo , with the C -norm jfjCo = supx2K jf(x)j. Proof: This is a typical three-epsilon argument. To show that a Cauchy sequence ffig of continuous functions has a pointwise limit which is a continuous function, first argue that fi has a pointwise limit at every x 2 K. Given " > 0, choose N large enough such that jfi − fjj < " for all i; j ≥ N. Then jfi(x) − fj(x)j < " for any x in K. Thus, the sequence of values fi(x) is a Cauchy sequence of complex numbers, so has a limit 0 0 f(x). Further, given " > 0 choose j ≥ N sufficiently large such that jfj(x) − f(x)j < " . For i ≥ N 0 jfi(x) − f(x)j ≤ jfi(x) − fj(x)j + jfj(x) − f(x)j < " + " 0 This is true for every positive " , so jfi(x) − f(x)j ≤ " for every x in K. That is, the pointwise limit is approached uniformly in x 2 [a; b]. To prove that f(x) is continuous, for " > 0, take N be large enough so that jfi − fjj < " for all i; j ≥ N. From the previous paragraph jfi(x) − f(x)j ≤ " for every x and for i ≥ N. Fix i ≥ N and x 2 K, and choose a small enough neigborhood U of x such that jfi(x) − fi(y)j < " for any y in U. Then jf(x) − f(y)j ≤ jf(x) − fi(x)j + jfi(x) − fi(y)j + jf(y) − fi(y)j ≤ " + jfi(x) − fi(y)j + " < " + " + " Thus, the pointwise limit f is continuous at every x in U. === Unsurprisingly, but significantly: [1.0.2] Claim: For x 2 [a; b], the evaluation map f ! f(x) is a continuous linear functional on Co[a; b]. Proof: For jf − gjCo < ", we have jf(x) − g(x)j ≤ jf − gjCo < " 1 Paul Garrett: Banach spaces Ck[a; b] (November 13, 2016) proving the continuity. === As usual, a real-valued or complex-valued function f on a closed interval [a; b] ⊂ R is continuously differentiable when it has a derivative which is itself a continuous function. That is, the limit f(x + h) − f(x) f 0(x) = lim h!0 h exists for all x 2 [a; b], and the function f 0(x) is in Co[a; b]. Let Ck[a; b] be the collection of k-times continuously differentiable functions on [a; b], with the Ck-norm X (i) X (i) jfjCk = sup jf (x)j = jf j1 0≤i≤k x2[a;b] 0≤i≤k (i) th k where f is the i derivative of f. The associated metric on C [a; b] is jf − gjCk . Similar to the assertion about evaluation on Co[a; b], [1.0.3] Claim: For x 2 [a; b] and 0 ≤ j ≤ k, the evaluation map f ! f (j)(x) is a continuous linear functional on Ck[a; b]. Proof: For jf − gjCk < ", (j) (j) jf (x) − g (x)j ≤ jf − gjCk < " proving the continuity. === We see that Ck[a; b] is a Banach space: [1.0.4] Theorem: The normed metric space Ck[a; b] is complete. k (j) Proof: For a Cauchy sequence ffig in C [a; b], all the pointwise limits limi fi (x) of j-fold derivatives exist (j) for 0 ≤ j ≤ k, and are uniformly continuous. The issue is to show that limi f is differentiable, with (j+1) 1 derivative limi f . It suffices to show that, for a Cauchy sequence fn in C [a; b], with pointwise limits 0 0 f(x) = limn fn(x) and g(x) = limn fn(x) we have g = f . By the fundamental theorem of calculus, for any index i, Z x 0 fi(x) − fi(a) = fi (t) dt a 0 0 Since the fi uniformly approach g, given " > 0 there is io such that jfi (t) − g(t)j < " for i ≥ io and for all t in the interval, so for such i Z x Z x Z x 0 0 fi (t) dt − g(t) dt ≤ jfi (t) − g(t)j dt ≤ " · jx − aj −! 0 a a a Thus, Z x Z x 0 lim fi(x) − fi(a) = lim fi (t) dt = g(t) dt i i a a from which f 0 = g. === By design, we have d k k−1 [1.0.5] Theorem: The map dx : C [a; b] ! C [a; b] is continuous. Proof: As usual, for a linear map T : V ! W , by linearity T v − T v0 = T (v − v0) it suffices to check continuity at 0. For Banach spaces the homogeneity jσ · vjV = jαj · jvjV shows that continuity is equivalent to existence of a constant B such that jT vjW ≤ B · jvjV for v 2 V . Then d X df (i) X (i) j fj k−1 = sup j( ) (x)j = sup jf (x)j ≤ 1 · jfj k dx C dx C 0≤i≤k−1 x2[a;b] 1≤i≤k x2[a;b] 2 Paul Garrett: Banach spaces Ck[a; b] (November 13, 2016) as desired. === 2. Non-Banach limit C1[a; b] of Banach spaces Ck[a; b] The space C1[a; b] of infinitely differentiable complex-valued functions on a (finite) interval [a; b] in R is not a Banach space. [1] Nevertheless, the topology is completely determined by its relation to the Banach spaces Ck[a; b]. That is, there is a unique reasonable topology on C1[a; b]. After explaining and proving this uniqueness, we also show that this topology is complete metric. This function space can be presented as \ C1[a; b] = Ck[a; b] k≥0 and we reasonably require that whatever topology C1[a; b] should have, each inclusion C1[a; b] −! Ck[a; b] is continuous. At the same time, given a family of continuous linear maps Z ! Ck[a; b] from a vector space Z in some reasonable class, with the compatibility condition of giving commutative diagrams ⊂ Ck[a; b] / Ck−1[a; b] fMM O MMM MMM MMM Z the image of Z actually lies in the intersection C1[a; b]. Thus, diagrammatically, for every family of compatible maps Z ! Ck[a; b], there is a unique Z ! C1[a; b] fitting into a commutative diagram * ( C1[a; b] ::: / C1[a; b] / Co[a; b] c w; j j4 w 8 j j 9! w j j w j j Z jw j We require that this induced map Z ! C1[a; b] is continuous. When we know that these conditions are met, we would say that C1[a; b] is the (projective) limit of the spaces Ck[a; b], written C1[a; b] = lim Ck[a; b] k with implicit reference to the inclusions Ck+1[a; b] ! Ck[a; b] and C1[a; b] ! Ck[a; b]. [2.0.1] Claim: Up to unique isomorphism, there exists at most one topology on C1[a; b] such that to every compatible family of continuous linear maps Z ! Ck[a; b] from a topological vector space Z there is a unique continuous linear Z ! C1[a; b] fitting into a commutative diagram as just above. Proof: Let X; Y be C1[a; b] with two topologies fitting into such diagrams, and show X ≈ Y , and for a unique isomorphism. First, claim that the identity map idX : X ! X is the only map ' : X ! X fitting into a commutative diagram [1] It is not essential to prove that there is no reasonable Banach space structure on C1[a; b], but this can be readily proven in a suitable context. 3 Paul Garrett: Banach spaces Ck[a; b] (November 13, 2016) * ' X ::: / C1[a; b] / Co[a; b] O ' X ::: / C1[a; b] / Co[a; b] 4 7 Indeed, given a compatible family of maps X ! Ck[a; b], there is unique ' fitting into * ' X ::: / C1[a; b] / Co[a; b] ` w; j j4 w 8 j j ' w j j w j j X jw j Since the identity map idX fits, necessarily ' = idX . Similarly, given the compatible family of inclusions Y ! Ck[a; b], there is unique f : Y ! X fitting into * ' X ::: / C1[a; b] / Co[a; b] ` ; j4 ww jjjj ww jjjj f ww jjj ww jjjj Y jwjj Similarly, given the compatible family of inclusions X ! Ck[a; b], there is unique g : X ! Y fitting into * ' Y ::: / C1[a; b] / Co[a; b] _ ; j4 ww jjj ww jjjj g ww jjjj ww jjjj X jwjj Then f ◦ g : X ! X fits into a diagram * ' X ::: / C1[a; b] / Co[a; b] ` ; j4 ww jjjj ww jjjj f◦g ww jjj ww jjjj X jwjj Therefore, f ◦ g = idX .
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