Fréchet algebra

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In mathematics, especially functional analysis, a Fréchet algebra, named after Maurice René Fréchet, is an associative algebra A over the real or complex numbers that at the same time is also a (locally convex) Fréchet space. The multiplication operation (a,b)a*b for a,bA is required to be jointly continuous. If {n}n=0 is an increasing family[lower-alpha 1] of seminorms for the topology of A, the joint continuity of multiplication is equivalent to there being a constant Cn>0 and integer mn for each n such that abnCnambm for all a,bA.[lower-alpha 2] Fréchet algebras are also called B0-algebras.[1] A Fréchet algebra is m-convex if there exists such a family of semi-norms for which m=n. In that case, by rescaling the seminorms, we may also take Cn=1 for each n and the seminorms are said to be submultiplicative: abnanbn for all a,bA.[lower-alpha 3] m-convex Fréchet algebras may also be called Fréchet algebras.[2] A Fréchet algebra may or may not have an identity element 1A. If A is unital, we do not require that 1An=1, as is often done for Banach algebras.

Properties

  • Continuity of multiplication. Multiplication is separately continuous if akbab and bakba for every a,bA and sequence aka converging in the Fréchet topology of A. Multiplication is jointly continuous if aka and bkb imply akbkab. Joint continuity of multiplication is part of the definition of a Fréchet algebra. For a Fréchet space with an algebra structure, if the multiplication is separately continuous, then it is automatically jointly continuous.[3]
  • Group of invertible elements. If invA is the set of invertible elements of A, then the inverse map {invAinvAuu1 is continuous if and only if invA is a Gδ set.[4] Unlike for Banach algebras, invA may not be an open set. If invA is open, then A is called a Q-algebra. (If A happens to be non-unital, then we may adjoin a unit to A[lower-alpha 4] and work with invA+, or the set of quasi invertibles[lower-alpha 5] may take the place of invA.)
  • Conditions for m-convexity. A Fréchet algebra is m-convex if and only if for every, if and only if for one, increasing family {n}n=0 of seminorms which topologize A, for each m there exists pm and Cm>0 such that a1a2anmCmna1pa2panp, for all a1,a2,,anA and n.[5] A commutative Fréchet Q-algebra is m-convex,[6] but there exist examples of non-commutative Fréchet Q-algebras which are not m-convex.[7]
  • Properties of m-convex Fréchet algebras. A Fréchet algebra is m-convex if and only if it is a countable projective limit of Banach algebras.[8] An element of A is invertible if and only if its image in each Banach algebra of the projective limit is invertible.[lower-alpha 6][9][10]

Examples

  • Zero multiplication. If E is any Fréchet space, we can make a Fréchet algebra structure by setting e*f=0 for all e,fE.
  • Smooth functions on the circle. Let S1 be the 1-sphere. This is a 1-dimensional compact differentiable manifold, with no boundary. Let A=C(S1) be the set of infinitely differentiable complex-valued functions on S1. This is clearly an algebra over the complex numbers, for pointwise multiplication. (Use the product rule for differentiation.) It is commutative, and the constant function 1 acts as an identity. Define a countable set of seminorms on A by φn=φ(n),φA, where φ(n)=supxS1|φ(n)(x)| denotes the supremum of the absolute value of the nth derivative φ(n).[lower-alpha 7] Then, by the product rule for differentiation, we have φψn=i=0n(ni)φ(i)ψ(ni)i=0n(ni)φiψnii=0n(ni)φ'nψ'n=2nφ'nψ'n, where (ni)=n!i!(ni)!, denotes the binomial coefficient and 'n=maxknk. The primed seminorms are submultiplicative after re-scaling by Cn=2n.
  • Sequences on . Let be the space of complex-valued sequences on the natural numbers . Define an increasing family of seminorms on by φn=maxkn|φ(k)|. With pointwise multiplication, is a commutative Fréchet algebra. In fact, each seminorm is submultiplicative φψnφnψn for φ,ψA. This m-convex Fréchet algebra is unital, since the constant sequence 1(k)=1,k is in A.
  • Equipped with the topology of uniform convergence on compact sets, and pointwise multiplication, C(), the algebra of all continuous functions on the complex plane , or to the algebra Hol() of holomorphic functions on .
  • Convolution algebra of rapidly vanishing functions on a finitely generated discrete group. Let G be a finitely generated group, with the discrete topology. This means that there exists a set of finitely many elements U={g1,,gn}G such that: n=0Un=G. Without loss of generality, we may also assume that the identity element e of G is contained in U. Define a function :G[0,) by (g)=min{ngUn}. Then (gh)(g)+(h), and (e)=0, since we define U0={e}.[lower-alpha 8] Let A be the -vector space S(G)={φ:G|φd<,d=0,1,2,}, where the seminorms d are defined by φd=dφ1=gG(g)d|φ(g)|.[lower-alpha 9] A is an m-convex Fréchet algebra for the convolution multiplication φ*ψ(g)=hGφ(h)ψ(h1g),[lower-alpha 10] A is unital because G is discrete, and A is commutative if and only if G is Abelian.
  • Non m-convex Fréchet algebras. The Aren's algebra A=Lω[0,1]=p1Lp[0,1] is an example of a commutative non-m-convex Fréchet algebra with discontinuous inversion. The topology is given by Lp norms fp=(01|f(t)|pdt)1/p,fA, and multiplication is given by convolution of functions with respect to Lebesgue measure on [0,1].[11]

Generalizations

We can drop the requirement for the algebra to be locally convex, but still a complete metric space. In this case, the underlying space may be called a Fréchet space[12] or an F-space.[13] If the requirement that the number of seminorms be countable is dropped, the algebra becomes locally convex (LC) or locally multiplicatively convex (LMC).[14] A complete LMC algebra is called an Arens-Michael algebra.[15]

Michael's Conjecture

The question of whether all linear multiplicative functionals on an m-convex Frechet algebra are continuous is known as Michael's Conjecture.[16] For a long time, this conjecture was perhaps the most famous open problem in the theory of topological algebras. Michael's Conjecture was solved completely and affirmatively in 2022.[17]

Notes

  1. An increasing family means that for each aA,
    a0a1an.
  2. Joint continuity of multiplication means that for every absolutely convex neighborhood V of zero, there is an absolutely convex neighborhood U of zero for which U2V, from which the seminorm inequality follows. Conversely,
    akbkabn=akbkabk+abkabnakbkabkn+abkabnCn(akambkm+ambkbm)Cn(akambm+akambkbm+ambkbm).
  3. In other words, an m-convex Fréchet algebra is a topological algebra, in which the topology is given by a countable family of submultiplicative seminorms: p(fg)p(f)p(g), and the algebra is complete.
  4. If A is an algebra over a field k, the unitization A+ of A is the direct sum Ak1, with multiplication defined as (a+μ1)(b+λ1)=ab+μb+λa+μλ1.
  5. If aA, then bA is a quasi-inverse for a if a+bab=0.
  6. If A is non-unital, replace invertible with quasi-invertible.
  7. To see the completeness, let φk be a Cauchy sequence. Then each derivative φk(l) is a Cauchy sequence in the sup norm on S1, and hence converges uniformly to a continuous function ψl on S1. It suffices to check that ψl is the lth derivative of ψ0. But, using the fundamental theorem of calculus, and taking the limit inside the integral (using uniform convergence), we have
    ψl(x)ψl(x0)=limk(φk(l)(x)φk(l)(x0))=limkx0xφk(l+1)(t)dt=x0xψl+1(t)dt.
  8. We can replace the generating set U with UU1, so that U=U1. Then satisfies the additional property (g1)=(g), and is a length function on G.
  9. To see that A is Fréchet space, let φn be a Cauchy sequence. Then for each gG, φn(g) is a Cauchy sequence in . Define φ(g) to be the limit. Then
    gS(g)d|φn(g)φ(g)|gS(g)d|φn(g)φm(g)|+gS(g)d|φm(g)φ(g)|φnφmd+gS(g)d|φm(g)φ(g)|,
    where the sum ranges over any finite subset S of G. Let ϵ>0, and let Kϵ>0 be such that φnφmd<ϵ for m,nKϵ. By letting m run, we have
    gS(g)d|φn(g)φ(g)|<ϵ
    for nKϵ. Summing over all of G, we therefore have φnφd<ϵ for nKϵ. By the estimate
    gS(g)d|φ(g)|gS(g)d|φn(g)φ(g)|+gS(g)d|φn(g)|φnφd+φnd,
    we obtain φd<. Since this holds for each d, we have φA and φnφ in the Fréchet topology, so A is complete.
  10. φ*ψdgG(hG(g)d|φ(h)||ψ(h1g)|)g,hG((h)+(h1g))d|φ(h)||ψ(h1g)|=i=0d(di)(g,hG|iφ(h)||diψ(h1g)|)=i=0d(di)(hG|iφ(h)|)(gG|diψ(g)|)=i=0d(di)φiψdi2dφ'dψ'd

Citations

  1. Mitiagin, Rolewicz & Żelazko 1962; Żelazko 2001.
  2. Husain 1991; Żelazko 2001.
  3. Waelbroeck 1971, Chapter VII, Proposition 1; Palmer 1994, § 2.9.
  4. Waelbroeck 1971, Chapter VII, Proposition 2.
  5. Mitiagin, Rolewicz & Żelazko 1962, Lemma 1.2.
  6. Żelazko 1965, Theorem 13.17.
  7. Żelazko 1994, pp. 283–290.
  8. Michael 1952, Theorem 5.1.
  9. Michael 1952, Theorem 5.2.
  10. See also Palmer 1994, Theorem 2.9.6.
  11. Fragoulopoulou 2005, Example 6.13 (2).
  12. Waelbroeck 1971.
  13. Rudin 1973, 1.8(e).
  14. Michael 1952; Husain 1991.
  15. Fragoulopoulou 2005, Chapter 1.
  16. Michael 1952, § 12, Question 1; Palmer 1994, § 3.1.
  17. Patel, S. R. (2022-06-28). "On affirmative solution to Michael's acclaimed problem in the theory of Fréchet algebras, with applications to automatic continuity theory". arXiv:2006.11134 [math.FA].

Sources