From: mccarthy@artsci.wustl.edu (John McCarthy) Newsgroups: sci.math.research Subject: Re: metric on continuous maps on R Date: 16 Nov 1994 17:44:03 GMT In article <1994Nov15.131744.1@brest>, wrote: >Question : >it is so easy to turn the set of continuous maps from [0,1] or [a,b] to R in a >normed vect.space. But I cannot find a way to define a norm on the set of >continuous maps on R at large (barring conditions on the maps to infinity). >Lingering in the distant memories of my mind there is a notion somebody might >have proved this to be impossible. More generally, >IS IT IMPOSSIBLE TO EMBBED THE SET OF ALL CONTINUOUS MAPS ON the set of real >numbers IN A METRIC SPACE ? >Thanks for a comprehensive answer or reference. > >Emmanuel Amiot, Perpignan, France > How about this: let d_N(f,g) = \sup_{-N \leq x \leq N} |f(x) - g(x)|. Let p(f,g) = \sum_{N=1}^\infty 2^{-N} \frac{d_N(f,g)}{1 + d_N(f,g)} Then p(.,.) is a complete metric on the space of continuous maps from R to R. John McCarthy From: mccarthy@artsci.wustl.edu (John McCarthy) Newsgroups: sci.math.research Subject: Re: metric on continuous maps on R Date: 16 Nov 1994 17:44:03 GMT In article <1994Nov15.131744.1@brest>, wrote: >Question : >it is so easy to turn the set of continuous maps from [0,1] or [a,b] to R in a >normed vect.space. But I cannot find a way to define a norm on the set of >continuous maps on R at large (barring conditions on the maps to infinity). >Lingering in the distant memories of my mind there is a notion somebody might >have proved this to be impossible. More generally, >IS IT IMPOSSIBLE TO EMBBED THE SET OF ALL CONTINUOUS MAPS ON the set of real >numbers IN A METRIC SPACE ? >Thanks for a comprehensive answer or reference. > >Emmanuel Amiot, Perpignan, France > How about this: let d_N(f,g) = \sup_{-N \leq x \leq N} |f(x) - g(x)|. Let p(f,g) = \sum_{N=1}^\infty 2^{-N} \frac{d_N(f,g)}{1 + d_N(f,g)} Then p(.,.) is a complete metric on the space of continuous maps from R to R. John McCarthy ============================================================================== From: schectex@math.vanderbilt.edu (Eric Schechter) Newsgroups: sci.math.research Subject: Re: metric on continuous maps on R Date: Mon, 21 Nov 94 10:57:08 -0600 Here are additional answers and comments on Amiot's question, plus (at the very end) a related question of my own. In an article dated Nov. 15, Emmanueal Amiot asked whether one can put a norm or metric on the set of all continuous maps from R into R. In an article dated Nov. 16, John McCarthy gave a complete metric on the set of all continuous maps from R into R. I will repeat that metric here: >let d_N(f,g) = \sup_{-N \leq x \leq N} |f(x) - g(x)|. >Let p(f,g) = \sum_{N=1}^\infty 2^{-N} \frac{d_N(f,g)}{1 + d_N(f,g)} >Then p(.,.) is a complete metric on the space of continuous maps >from R to R. I will now add a few other answers and comments: (i) The metric given by McCarthy yields the topology of uniform convergence on compact sets. It is translation-invariant and has other nice properties too. It is a simple example of what some mathematicians call an "F-norm" (although the terminology varies a bit in the literature). Some introductory material about F-norms and about metrizable, not-necessarily-locally-convex topological vector spaces can be found in the books ``An F-Space Sampler'', by Kalton, Peck, and Roberts, and ``Metric Linear Spaces'' by Rolewicz. A larger amount of more elementary material on F-norms can be found in the book I'm writing, which will probably be published about 2/3 of a year from now. All of the results below are also discussed in my book. (ii) The ``usual'' topology on C(R,R) --- i.e., the one which has the nicest properties and is used most often --- is the topology given by the F-norm described above. But that F-norm is not equivalent to a norm --- i.e., it is NOT POSSIBLE to give a norm for the usual topology on C(R,R). Indeed, a theorem of Kolmogorov shows that a topological vector space is normable if and only if it is Hausdorff, locally convex, and locally bounded. The space C(R,R) is not locally bounded --- i.e., 0 does not have a neighborhood which is bounded --- so C(R,R) cannot be normed. (Here ``bounded'' refers to the topological vector space structure, not to metric boundedness.) Local boundedness is discussed in some textbooks on functional analysis. (iii) If we are merely interested embedding C(R,R) --- with its usual topology --- in a larger space, then we CAN use a norm. Any metric space (M,d) can be embedded isometrically in the sup-normed Banach space of all bounded maps from M into R, by fixing some particular member of M --- call it 0 --- and then sending each m in M to the function d(m,.)-d(0,.). However, this does not make C(R,R) into linear subspace of that Banach space --- indeed, it cannot, since we have noted in the previous paragraph that C(R,R) is not normable. (iv) If we do not insist on using the usual topology, and we permit the use of an entirely bizarre topology, then C(R,R) can be normed. Indeed, C(R,R) is separable, and so is (for instance) the Hilbert space L2[0,1]. Hence, by the Axiom of Choice, each has a vector basis (i.e., maximal linearly independent set) with cardinality equal to the continuum (if we assume the Continuum Hypothesis). Since C(R,R) and L2[0,1] have vector bases with the same cardinality, there exists a linear bijection between those two spaces. Use that linear bijection to copy the Hilbert space norm of L2[0,1] onto C(R,R). Thus we can make C(R,R) into a Hilbert space, by keeping its linear structure but completely altering its topology. (v) Are there any other interesting, useful, non-bizarre topologies on C(R,R)? I'm not sure how to give a complete answer to that, but I can give some partial answers: If the new topology on C(R,R) makes it into a metrizable topological vector space, then that topology can also be given by an F-norm. We probably want the metric on C(R,R) to be complete. Then the old and new topologies on C(R,R) are the same if we add any of the following three hypotheses: - If the two metrics are comparable --- i.e., if one is stronger than the other --- then they are equal. This is a corollary of the Closed Graph Theorem (which is true for F-norms, not just for norms). - If we want the new topology to have a nice order structure (as the old topology does), then the topologies have to be the same. Any two complete Riesz F-norms on a vector space are equivalent; this can be found in books on ordered topological vector spaces. - If we insist on explicitly constructible objects (which is often the case in applied mathematics --- no Axiom of Choice permitted here!), then the two topologies must be identical. Indeed, Wright proved (in a paper published in 1977) this variant of the Closed Graph Theorem: if we replace conventional ZF + AC with the alternative ZF + DC + BP, then any linear map from an F-space into a Hausdorff topological vector space is continuous. Thus inequivalent complete F-norms on a vector space cannot be proved to exist using ZF + DC + BP; thus they certainly cannot be proved to exist using just ZF + DC --- which is a superset of the reasoning system used by most constructivists. This explains why, in applied mathematics, we often speak of ``the usual norm'' (or more generally, ``the usual F-norm'') for a vector space. (Wright's 1977 paper was motivated by Solovay's famous 1970 model, but a better motivation has subsequently developed: Solovay's system, ZF+DC+BP+LM, has turned out to be of greater consistency strength than conventional set theory, but in 1984 Shelah proved that ZF+DC+BP is equiconsistent with conventional set theory. For Closed Graph Theorems, LM isn't really relevant.) (vi) This leaves unanswered the following question: Is C(R,R) (not with its usual topology, but with a different topology that isn't too bizarre) equal to a linear subspace of a Banach space? I don't know how to answer this, or even how to make the question precise. Any comments or suggestions? ==============================================================================