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Schwartz space is Fréchet

Statement

The Schwartz topology is locally convex, metrizable and complete. Set qj=maxα,βjpαβ. A complete translation-invariant metric defining it is d(f,g)=j=02j1min(1,qj(fg)). No choice axiom is required.

Facts & Assumptions

Given: The seminorm topology of Schwartz topology and convergence, already verified to be Hausdorff and locally convex.

[F1]

The uniform Cauchy criterion gives uniform limits of real functions (A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy).

[F2]

Uniform limits of continuous real functions are continuous (The uniform limit of continuous real-valued functions on a metric space is continuous).

[F4]

Continuous higher mixed partials commute (Continuous mixed partials of order k are invariant under permutations), applied separately to real and imaginary parts.

Proof

technique · direct
1.1

The series converges since its terms are bounded by 2j1. Symmetry and translation invariance follow termwise; min(1,a+b)min(1,a)+min(1,b) proves the triangle inequality. Vanishing distance forces q0(fg)=0, hence f=g. To make d(f,0)<ε, choose J with j>J2j1<ε/2 and require qJ(f)<ε/2. Conversely, to ensure qJ(f)<η, put η=min(η/2,1/2) and require d(f,0)<2J1η. Then min(1,qJ(f))<η<1, as required. Every finite seminorm neighbourhood contains such a qJ ball and each qJ ball is a finite seminorm neighbourhood, proving equality of topologies.

givenalgebra
2.1

Let (fk) be d-Cauchy. The converse estimate in step 1.1 makes it Cauchy in every qJ. Each βfk therefore has a unique uniform complex limit gβ, by applying [F1] to its real and imaginary parts, and this limit is continuous by [F2]. On any fixed coordinate segment x+tej, atb with a<b, [F4] gives derivative β+ejfk. Apply [F3] componentwise to these restricted functions: their values converge at a and their derivatives converge uniformly. Consequently jgβ=gβ+ej along every segment. Induction on the length of an ordered derivative now gives f=g0C and βf=gβ. Limits are unique specified values, so forming this family requires no choice.

step 1.1F1F2F3F4
3.1

Fix α,β and ε>0. The seminorm Cauchy property gives N with xαβ(fkfl)(x)<ε/2 for every x and k,lN. For fixed x,k, let l using step 2.1. Then pαβ(fkf)ε/2<ε. Taking k=N also shows pαβ(f)pαβ(fN)+ε/2<. Thus fS and fkf in every seminorm and hence in d. Combined with local convexity, this proves the claimed Fréchet property.

step 2.1step 1.1given

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