Auxiliary normed space

In functional analysis, two methods of constructing normed spaces from disks were systematically employed by Alexander Grothendieck to define nuclear operators and nuclear spaces.[1] One method is used if the disk D is bounded: in this case, the auxiliary normed space is span D with norm pD :=infxrD, r>0 r. The other method is used if the disk D is absorbing: in this case, the auxiliary normed space is the quotient space X / p-1
D
(0)
. If the disk is both bounded and absorbing then the two auxiliary normed spaces are canonically isomorphic (as topological vector spaces and as normed spaces).

Preliminaries

Definition: A subset of a vector space is called a disk and is said to be disked, absolutely convex, or convex balanced if it is convex and balanced.
Definition: If C and D are subsets of a vector space X then we say that D absorbs C if there exists a real r > 0 such that CaD for any scalar a satisfying |a|r. We say that D is absorbing in X if D absorbs { x } for every xX.
Definition: A subset B of a topological vector space (TVS) X is said to be bounded in X if every neighborhood of the origin in X absorbs B.
Definition:[2] A subset of a TVS X is called bornivorous if it absorbs all bounded subsets of X.

Induced by a bounded disk – Banach disks

Seminormed space induced by a disk

Henceforth, X will be a real or complex vector space (not necessarily a TVS, yet) and let D will be a disk in X.

If D = ∅ then set XD = { 0 } and give XD the indiscrete topology.[3] Henceforth assume that D ≠ ∅ (alternatively, if D = ∅ then one may instead replace D with { 0}).

Since D is a disk, span D =
n=1
nD
so that D is absorbing in span D, the linear span of D. Note that the set { rD : r > 0 } of all positive scalar multiples of D forms a basis of neighborhoods at 0 for a locally convex topological vector space topology 𝜏D on span D. The Minkowski functional of D in span D, which we denote and define by

pD(x) := inf { r : xrD, r > 0}

is well-defined and forms a seminorm on span D.[4] The locally convex topology topology induced by this seminorm is the topology 𝜏D that was defined before.

Definition and notation: When we give span D the vector topology induced by this seminorm then we denote the resulting seminormed topological vector space by (XD, pD) or simply XD and we denote its topology by 𝜏D. We call this space the (seminormed) space induced by D and say that it is a disk induced space.
Definition: The natural inclusion InD : XDX is called the canonical map.[1]

Banach disk definition

Definition: A bounded disk D in a TVS X such that (XD, pD) is a Banach space is called a Banach disk, infracomplete, or a bounded completant in X.

Observe that if its shown that (span D, pD) is a Banach space then D will be a Banach disk in any TVS that contains D as a bounded subset.

This is because the Minkowski functional pD is defined in purely algebraic terms. Consequently, the question of whether or not (XD, pD) forms a Banach space is dependent only on the disk D and the Minkowski functional pD, and not on any particular TVS topology that X may carry. Thus, the requirement that a Banach disk in a TVS X be a bounded subset of X is the only property that ties a Banach disk to the topology of its containing TVS.

Related definitions
Definition:[2] A disk in a TVS is called infrabornivorous if it absorbs all Banach disks.
Definition:[2] A linear map between two TVSs is called infrabounded if it maps Banach disks to bounded disks.

Properties of disk induced seminormed spaces

Bounded disks

The following result explains why Banach disks are required to be bounded.

Theorem[5][2][1]  If D is a disk in a topological vector space (TVS) X, then D is bounded in X if and only if the natural inclusion InD : XDX is continuous.

Proof 

If the disk D is bounded in the TVS X then for all neighborhoods U of 0 in X, there exists some r > 0 such that rDUXD. It follows that in this case the topology of (XD, pD) is finer than the subspace topology that XD inherits from X, which implies that the natural inclusion InD : XDX is continuous. Conversely, if X has a TVS topology such that InD : XDX is continuous, then for every neighborhood U of 0 in X there exists some r > 0 such that rDUXD, which shows that D is bounded in X.

Hausdorffness

The space (XD, pD) is Hausdorff if and only if pD is a norm, which happens if and only if D does not contain any non-trivial vector subspace.[6] In particular, if there exists a Hausdorff TVS topology on X such that D is bounded in X then pD is a norm. An example where XD is not Hausdorff is obtained by letting X = ℝ2 and letting D be the x-axis.

Convergence of nets

Suppose that D is a disk in X such that XD is Hausdorff and let x = (xi)iI be a net in XD. Then x → 0 in XD if and only if there exists a net r = (ri)iI of real numbers such that r → 0 and xiriD for all i; moreover, in this case we can assume without loss of generality that ri ≥ 0 for all i.

Relationship between disk-induced spaces

Note that if CDX then span C ⊆ span D and pDpC on span C, so we can define the following continuous[2] linear map:

Definition: If C and D are disks in X with CD then call the natural inclusion InD
C
: XCXD
the canonical inclusion of XC into XD.

In particular, the subspace topology that span C inherits from (XD, pD) is weaker than (XC, pC)'s seminorm topology.[2]

D as the closed unit ball

Clearly, the disk D is a closed subset of (XD, pD) if and only if D is the closed unit ball of the seminorm pD i.e. D = { x ∈ span D : pD(x) ≤ 1}.

  • If D is a disk in a vector space X and if there exists a TVS topology 𝜏 on span D such that D is a closed and bounded subset of (span D, 𝜏), then D is the closed unit ball of (XD, pD) (i.e. D = { x ∈ Span D : pD(x) ≤ 1}) (see footnote for proof).[7]

Examples and sufficient conditions

The following theorem may be used to establish that (XD, pD) is a Banach space. Once this is established, D will be a Banach disk in any TVS in which D is bounded.

Theorem[8]  Let D be a disk in a vector space X. If there exists a Hausdorff TVS topology 𝜏 on span D such that D is a bounded sequentially complete subset of (span D, 𝜏), then (XD, pD) is a Banach space.

Proof 

Assume without loss of generality that X = span D and let p := pD be the Minkowski functional of D. Since D is a bounded subset of a Hausdorff TVS, D do not contain any non-trivial vector subspace, which implies that p is a norm. Let 𝜏D denote the norm topology on X induced by p where note that since D is a bounded subset of (X, 𝜏), 𝜏D is finer than 𝜏.

Since D is convex and balanced, for any 0 < m < n we have

2-(n+1)D + ⋅⋅⋅ + 2-(m+2)D = 2-(m+1)(1-2m-n) D ⊆ 2-(m+2)D.

Let x = (xi)
i=1
be a Cauchy sequence in (XD, p). By replacing x with a subsequence, we may assume without loss of generality that for all i,

xi+1 - xi1/2i+2 D.

This implies that for any 0 < m < n,

xn - xm = (xn - xn-1) + ⋅⋅⋅ + (xm+1 - xm) ∈ 2-(n+1)D + ⋅⋅⋅ + 2-(m+2)D ⊆ 2-(m+2)D

so that in particular, by taking m = 1 we see that x is contained in x1 + 2-3D. Since 𝜏D is finer than 𝜏, x is a Cauchy sequence in (X, 𝜏). Note that for all m > 0, 2-(m+2)D is a Hausdorff sequentially complete subspace of subset of (X, 𝜏). In particular, this is true for x1 + 2-3D so there exists some xx1 + 2-3D such that xx in (X, 𝜏).

Since xn - xm ∈ 2-(m+2)D for all 0 < m < n, by fixing m and taking the limit (in (X, 𝜏)) as n → ∞, we see that x - xm ∈ 2-(m+2)D for each m > 0. This implies that p(x - xm) → 0 as m → ∞, which says exactly that xx in (XD, p). Thus (XD, p) is complete.

This assumption is allowed because x is a Cauchy sequence in a metric space (so the limits of all subsequences are equal) and a sequence in a metric space converges if and only if every subsequence has a sub-subsequence that converges.

Note that even if D is not a bounded and sequentially complete subset of any Hausdorff TVS, one might still be able to conclude that (XD, pD) is a Banach space by applying this theorem to some disk K satisfying { x ∈ span D : pD(x) < 1 } ⊆ K ⊆ { x ∈ span D : pD(x) ≤ 1 } (since pD = pK).

The following are consequences of the above theorem:

  • A sequentially complete bounded disk in a Hausdorff TVS is a Banach disk.[2]
  • Any disk in a Hausdorff TVS that is complete and bounded (e.g. compact) is a Banach disk.[9]
  • The closed unit ball in a Fréchet space is sequentially complete and thus a Banach disk.[2]

Suppose that D is a bounded disk in a TVS X.

  • If L : XY is a continuous linear map and BX is a Banach disk, then L(B) is a Banach disk and L|XB : XBL(XB) induces an isometric TVS-isomorphism YL(B)XB / (XB ker L).

Properties of Banach disks

Let X be a TVS and let D be a bounded disk in X.

  • If D is a bounded Banach disk in a Hausdorff locally convex space X and if T is a barrel in X then T absorbs D (i.e. there is a number r > 0 such that DrT).[5]
  • If U is a convex balanced closed neighborhood of 0 in X then the collection of all neighborhoods rU, where r > 0 ranges over the positive real numbers, induces a topological vector space topology on X. When X has this topology, it is denoted by XU. Since this topology is not necessarily Hausdorff nor complete, the completion of the Hausdorff space X / p-1
    U
    (0)
    is denoted by XU so that XU is a complete Hausdorff space and pU :=infxrU, r>0 r is a norm on this space making XU into a Banach space. The polar of U, U°, is a weakly compact bounded equicontinuous disk in X' and so is infracomplete.

Induced by a radial disk – quotient

Suppose that X is a topological vector space and V is a convex balanced and radial set. Then { 1/n V : n = 1, 2, ...} is a neighborhood basis at the origin for some locally convex topology 𝜏V on X. This TVS topology 𝜏V is given by the Minkowski functional formed by V, pV : X → ℝ, which is a seminorm on X defined by pV :=infxrV, r>0 r. The topology 𝜏V is Hausdorff if and only if pV is a norm, or equivalently, if and only if X / p-1
V
(0) = { 0
} or equivalently, for which it suffices that V be bounded in X. The topology 𝜏V need not be Hausdorff but X / p-1
V
(0)
is Hausdorff. A norm on X / p-1
V
(0)
is given by ||x + p-1
V
(0)
|| := pV(x)
, where this value is in fact independent of the representative of the equivalence class x + p-1
V
(0)
chosen. The normed space (X / p-1
V
(0), ||||)
> is denoted by XV and its completion is denoted by XV.

If in addition V is bounded in X then the seminorm pV : X → ℝ is a norm so in particular, p-1
V
(0) = { 0
}. In this case, we take XV to be the vector space X instead of X / { 0 } so that the notation XV is unambiguous (whether XV denotes the space induced by a radial disk or the space induced by a bounded disk).[1]

The quotient topology 𝜏Q on X / p-1
V
(0)
(inherited from X's original topology) is finer (in general, strictly finer) than the norm topology.

Canonical maps

The canonical map is the quotient map qV : XXV = X / p-1
V
(0)
, which is continuous when XV has either the norm topology or the quotient topology.[1]

If U and V are radial disks such that UV then p-1
U
(0)
p-1
V
(0) so there is a continuous linear surjective canonical map qV,U : X / p-1
U
(0) → X / p-1
V
(0) = XV
defined by sending x + p-1
U
(0) XU = X / p-1
U
(0)
to the equivalence class x + p-1
V
(0)
, where one may verify that the definition does not depend on the representative of the equivalence class x + p-1
U
(0)
that is chosen.[1] This canonical map has norm ≤ 1[1] and it has a unique continuous linear canonical extension to XU that is denoted by qV,U : XUXV.

Suppose that in addition B ≠ ∅ and C are bounded disks in X with BC so that XBXC and the natural inclusion InC
B
: XBXC
is a continuous linear map. Let InB : XBX, InC : XCX, and InC
B
: XBXC
be the canonical maps. Then InC = InC
B
∘ InB
and qV = qV,UqU.[1]

Induced by a bounded radial disk

Suppose that S is a bounded radial disk. Since S is a bounded disk, if we let D := S then we may create the auxiliary normed space XD = span D with norm pD :=infxrD, r>0 r; since S is radial, XS = X. Since S is a radial disk, if we let V := S then we may create the auxiliary seminormed space X / p-1
V
(0)
with the seminorm pV :=infxrV, r>0 r; because S is bounded, this seminorm is a norm and p-1
V
(0) = { 0
} so X / p-1
V
(0) = X / { 0 } = X
. Thus, in this case the two auxiliary normed spaces produced by these two different methods result in the same normed space.

Duality

Suppose that H is a weakly closed equicontinuous disk in X' (this implies that H is weakly compact) and let U := H° = { x X : |h(x)| ≤1 for all h H} be the polar of H. Since U° = H°° = H by the bipolar theorem, it follows that a continuous linear functional f belongs to X'H = span H if and only if f belongs to the continuous dual space of (X, pU), where pU is the Minkowski functional of U defined by pU :=infxrU, r>0 r.[10]

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See also

References

  1. Schaefer 1999, p. 97.
  2. Narici 2011, pp. 441-457.
  3. Schaefer 1999, p. 169.
  4. Treves 2006, p. 370.
  5. Treves 2006, pp. 370-373.
  6. Narici 2011, pp. 115-154.
  7. Assume WLOG that X = span D. Since D is closed in (X, 𝜏), it is also closed in (XD, pD) and since the seminorm pD is the Minkowski functional of D, which is continuous on (XD, pD), it followsNarici 2011, p. 119-120 that D is the closed unit ball in (XD, p).
  8. Narici 2011, pp. 441-442.
  9. Treves 2006, pp. 370–371.
  10. Treves 2006, p. 477.
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