(November 28, 2016) Topological vector spaces Paul Garrett [email protected] http:=/www.math.umn.edu/egarrett/ [This document is http://www.math.umn.edu/~garrett/m/fun/notes 2016-17/tvss.pdf] 1. Banach spaces Ck[a; b] 2. Non-Banach limit C1[a; b] of Banach spaces Ck[a; b] 3. Sufficient notion of topological vector space 4. Unique vectorspace topology on Cn 5. Non-Fr´echet colimit C1 of Cn, quasi-completeness 6. Seminorms and locally convex topologies 7. Quasi-completeness theorem 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 [1.1] 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: [1.2] 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 < " proving the continuity. === 1 Paul Garrett: Topological vector spaces (November 28, 2016) 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], [1.3] 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: [1.4] 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 [1.5] 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: Topological vector spaces (November 28, 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 (specified in the next section), 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]. [2.1] 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: Topological vector spaces (November 28, 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 .