Auxiliary normed space

Let ${\displaystyle X}$ will be a real or complex vector space. For any subset ${\displaystyle D}$ of ${\displaystyle X,}$ the Minkowski functional of ${\displaystyle D}$ defined by:

• If ${\displaystyle D=\varnothing }$ then define ${\displaystyle p_{\varnothing }(x):\{0\}\to [0,\infty )}$ to be the trivial map ${\displaystyle p_{\varnothing }=0}$[2] and it will be assumed that ${\displaystyle \operatorname {span} \varnothing =\{0\}.}$[note 1]
• If ${\displaystyle D\neq \varnothing }$ and if ${\displaystyle D}$ is absorbing in ${\displaystyle \operatorname {span} D}$ then denote the Minkowski functional of ${\displaystyle D}$ in ${\displaystyle \operatorname {span} D}$ by
  $${\displaystyle p_{D}:\operatorname {span} D\to [0,\infty )}$$

  
  where for all ${\displaystyle x\in \operatorname {span} D,}$ this is defined by
  $${\displaystyle p_{D}(x):=\inf _{}\{r:x\in rD,r>0\}.}$$

  

Let ${\displaystyle X}$ will be a real or complex vector space. For any subset ${\displaystyle D}$ of ${\displaystyle X}$ such that the Minkowski functional ${\displaystyle p_{D}}$is a seminorm on ${\displaystyle \operatorname {span} D,}$ let ${\displaystyle X_{D}}$ denote

$${\displaystyle X_{D}:=\left(\operatorname {span} D,p_{D}\right)}$$

which is called the seminormed space induced by ${\displaystyle D,}$ where if ${\displaystyle p_{D}}$ is a norm then it is called the normed space induced by ${\displaystyle D.}$

Assumption (Topology): ${\displaystyle X_{D}=\operatorname {span} D}$ is endowed with the seminorm topology induced by ${\displaystyle p_{D},}$ which will be denoted by ${\displaystyle \tau _{D}}$ or ${\displaystyle \tau _{p_{D}}}$

Importantly, this topology stems entirely from the set ${\displaystyle D,}$ the algebraic structure of ${\displaystyle X,}$ and the usual topology on ${\displaystyle \mathbb {R} }$ (since ${\displaystyle p_{D}}$is defined using only the set ${\displaystyle D}$ and scalar multiplication). This justifies the study of Banach disks and is part of the reason why they play an important role in the theory of nuclear operators and nuclear spaces.

The inclusion map ${\displaystyle \operatorname {In} _{D}:X_{D}\to X}$ is called the canonical map.[1]

Suppose that ${\displaystyle D}$ is a disk. Then ${\textstyle \operatorname {span} D=\bigcup _{n=1}^{\infty }nD}$ so that ${\displaystyle D}$ is absorbing in ${\displaystyle \operatorname {span} D,}$ the linear span of ${\displaystyle D.}$ The set ${\displaystyle \{rD:r>0\}}$ of all positive scalar multiples of ${\displaystyle D}$ forms a basis of neighborhoods at the origin for a locally convex topological vector space topology ${\displaystyle \tau _{D}}$ on ${\displaystyle \operatorname {span} D.}$ The Minkowski functional of the disk ${\displaystyle D}$ in ${\displaystyle \operatorname {span} D}$ guarantees that ${\displaystyle p_{D}}$is well-defined and forms a seminorm on ${\displaystyle \operatorname {span} D.}$[3] The locally convex topology induced by this seminorm is the topology ${\displaystyle \tau _{D}}$ that was defined before.

A bounded disk ${\displaystyle D}$ in a topological vector space ${\displaystyle X}$ such that ${\displaystyle \left(X_{D},p_{D}\right)}$ is a Banach space is called a Banach disk, infracomplete, or a bounded completant in ${\displaystyle X.}$

If its shown that ${\displaystyle \left(\operatorname {span} D,p_{D}\right)}$ is a Banach space then ${\displaystyle D}$ will be a Banach disk in any TVS that contains ${\displaystyle D}$ as a bounded subset.

This is because the Minkowski functional ${\displaystyle p_{D}}$is defined in purely algebraic terms. Consequently, the question of whether or not ${\displaystyle \left(X_{D},p_{D}\right)}$ forms a Banach space is dependent only on the disk ${\displaystyle D}$ and the Minkowski functional ${\displaystyle p_{D},}$ and not on any particular TVS topology that ${\displaystyle X}$ may carry. Thus the requirement that a Banach disk in a TVS ${\displaystyle X}$ be a bounded subset of ${\displaystyle X}$ is the only property that ties a Banach disk's topology to the topology of its containing TVS ${\displaystyle X.}$

Bounded disks

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

Hausdorffness

The space ${\displaystyle \left(X_{D},p_{D}\right)}$ is Hausdorff if and only if ${\displaystyle p_{D}}$is a norm, which happens if and only if ${\displaystyle D}$ does not contain any non-trivial vector subspace.[6] In particular, if there exists a Hausdorff TVS topology on ${\displaystyle X}$ such that ${\displaystyle D}$ is bounded in ${\displaystyle X}$ then ${\displaystyle p_{D}}$is a norm. An example where ${\displaystyle X_{D}}$ is not Hausdorff is obtained by letting ${\displaystyle X=\mathbb {R} ^{2}}$ and letting ${\displaystyle D}$ be the ${\displaystyle x}$-axis.

Convergence of nets

Suppose that ${\displaystyle D}$ is a disk in ${\displaystyle X}$ such that ${\displaystyle X_{D}}$ is Hausdorff and let ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}}$ be a net in ${\displaystyle X_{D}.}$ Then ${\displaystyle x_{\bullet }\to 0}$ in ${\displaystyle X_{D}}$ if and only if there exists a net ${\displaystyle r_{\bullet }=\left(r_{i}\right)_{i\in I}}$ of real numbers such that ${\displaystyle r_{\bullet }\to 0}$ and ${\displaystyle x_{i}\in r_{i}D}$ for all ${\displaystyle i}$; moreover, in this case it will be assumed without loss of generality that ${\displaystyle r_{i}\geq 0}$ for all ${\displaystyle i.}$

Relationship between disk-induced spaces

If ${\displaystyle C\subseteq D\subseteq X}$then ${\displaystyle \operatorname {span} C\subseteq \operatorname {span} D}$ and ${\displaystyle p_{D}\leq p_{C}}$ on ${\displaystyle \operatorname {span} C,}$ so define the following continuous[5] linear map:

If ${\displaystyle C}$ and ${\displaystyle D}$ are disks in ${\displaystyle X}$ with ${\displaystyle C\subseteq D}$ then call the inclusion map ${\displaystyle \operatorname {In} _{C}^{D}:X_{C}\to X_{D}}$ the canonical inclusion of ${\displaystyle X_{C}}$ into ${\displaystyle X_{D}.}$

In particular, the subspace topology that ${\displaystyle \operatorname {span} C}$ inherits from ${\displaystyle \left(X_{D},p_{D}\right)}$ is weaker than ${\displaystyle \left(X_{C},p_{C}\right)}$'s seminorm topology.[5]

The disk as the closed unit ball

The disk ${\displaystyle D}$ is a closed subset of ${\displaystyle \left(X_{D},p_{D}\right)}$ if and only if ${\displaystyle D}$ is the closed unit ball of the seminorm ${\displaystyle p_{D}}$; that is, ${\displaystyle D=\left\{x\in \operatorname {span} D:p_{D}(x)\leq 1\right\}.}$

If ${\displaystyle D}$ is a disk in a vector space ${\displaystyle X}$ and if there exists a TVS topology ${\displaystyle \tau }$ on ${\displaystyle \operatorname {span} D}$ such that ${\displaystyle D}$ is a closed and bounded subset of ${\displaystyle \left(\operatorname {span} D,\tau \right),}$ then ${\displaystyle D}$ is the closed unit ball of ${\displaystyle \left(X_{D},p_{D}\right)}$ (that is, ${\displaystyle D=\left\{x\in \operatorname {span} D:p_{D}(x)\leq 1\right\}}$ ) (see footnote for proof).[note 2]

The following theorem may be used to establish that ${\displaystyle \left(X_{D},p_{D}\right)}$ is a Banach space. Once this is established, ${\displaystyle D}$ will be a Banach disk in any TVS in which ${\displaystyle D}$ is bounded.

Proof

Assume without loss of generality that ${\displaystyle X=\operatorname {span} D}$ and let ${\displaystyle p:=p_{D}}$ be the Minkowski functional of ${\displaystyle D.}$ Since ${\displaystyle D}$ is a bounded subset of a Hausdorff TVS, ${\displaystyle D}$ do not contain any non-trivial vector subspace, which implies that ${\displaystyle p}$ is a norm. Let ${\displaystyle \tau _{D}}$ denote the norm topology on ${\displaystyle X}$ induced by ${\displaystyle p}$ where since ${\displaystyle D}$ is a bounded subset of ${\displaystyle (X,\tau ),}$ ${\displaystyle \tau _{D}}$ is finer than ${\displaystyle \tau .}$

Because ${\displaystyle D}$ is convex and balanced, for any ${\displaystyle 0<m<n}$

$${\displaystyle 2^{-(n+1)}D+\cdots +2^{-(m+2)}D=2^{-(m+1)}\left(1-2^{m-n}\right)D\subseteq 2^{-(m+2)}D.}$$

Let ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }}$ be a Cauchy sequence in ${\displaystyle \left(X_{D},p\right).}$ By replacing ${\displaystyle x_{\bullet }}$ with a subsequence, we may assume without loss of generality^† that for all ${\displaystyle i,}$

$${\displaystyle x_{i+1}-x_{i}\in {\frac {1}{2^{i+2}}}D.}$$

This implies that for any ${\displaystyle 0<m<n,}$

$${\displaystyle x_{n}-x_{m}=\left(x_{n}-x_{n-1}\right)+\left(x_{m+1}-x_{m}\right)\in 2^{-(n+1)}D+\cdots +2^{-(m+2)}D\subseteq 2^{-(m+2)}D}$$

so that in particular, by taking ${\displaystyle m=1}$ it follows that ${\displaystyle x_{\bullet }}$ is contained in ${\displaystyle x_{1}+2^{-3}D.}$ Since ${\displaystyle \tau _{D}}$ is finer than ${\displaystyle \tau ,}$ ${\displaystyle x_{\bullet }}$ is a Cauchy sequence in ${\displaystyle (X,\tau ).}$ For all ${\displaystyle m>0,}$ ${\displaystyle 2^{-(m+2)}D}$ is a Hausdorff sequentially complete subset of ${\displaystyle (X,\tau ).}$ In particular, this is true for ${\displaystyle x_{1}+2^{-3}D}$ so there exists some ${\displaystyle x\in x_{1}+2^{-3}D}$ such that ${\displaystyle x_{\bullet }\to x}$ in ${\displaystyle (X,\tau ).}$

Since ${\displaystyle x_{n}-x_{m}\in 2^{-(m+2)}D}$ for all ${\displaystyle 0<m<n,}$ by fixing ${\displaystyle m}$ and taking the limit (in ${\displaystyle (X,\tau )}$) as ${\displaystyle n\to \infty ,}$ it follows that ${\displaystyle x-x_{m}\in 2^{-(m+2)}D}$ for each ${\displaystyle m>0.}$ This implies that ${\displaystyle p\left(x-x_{m}\right)\to 0}$ as ${\displaystyle m\to \infty ,}$ which says exactly that ${\displaystyle x_{\bullet }\to x}$ in ${\displaystyle \left(X_{D},p\right).}$ This shows that ${\displaystyle \left(X_{D},p\right)}$ is complete.

^†This assumption is allowed because ${\displaystyle x_{\bullet }}$ 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 ${\displaystyle D}$ is not a bounded and sequentially complete subset of any Hausdorff TVS, one might still be able to conclude that ${\displaystyle \left(X_{D},p_{D}\right)}$ is a Banach space by applying this theorem to some disk ${\displaystyle K}$ satisfying

$${\displaystyle \left\{x\in \operatorname {span} D:p_{D}(x)<1\right\}\subseteq K\subseteq \left\{x\in \operatorname {span} D:p_{D}(x)\leq 1\right\}}$$

because ${\displaystyle p_{D}=p_{K}.}$

The following are consequences of the above theorem:

• A sequentially complete bounded disk in a Hausdorff TVS is a Banach disk.[5]
• Any disk in a Hausdorff TVS that is complete and bounded (e.g. compact) is a Banach disk.[8]
• The closed unit ball in a Fréchet space is sequentially complete and thus a Banach disk.[5]

Suppose that ${\displaystyle D}$ is a bounded disk in a TVS ${\displaystyle X.}$

• If ${\displaystyle L:X\to Y}$ is a continuous linear map and ${\displaystyle B\subseteq X}$ is a Banach disk, then ${\displaystyle L(B)}$ is a Banach disk and ${\displaystyle L{\big \vert }_{X_{B}}:X_{B}\to L\left(X_{B}\right)}$ induces an isometric TVS-isomorphism ${\displaystyle Y_{L(B)}\cong X_{B}/\left(X_{B}\cap \operatorname {ker} L\right).}$

Let ${\displaystyle X}$ be a TVS and let ${\displaystyle D}$ be a bounded disk in ${\displaystyle X.}$

If ${\displaystyle D}$ is a bounded Banach disk in a Hausdorff locally convex space ${\displaystyle X}$ and if ${\displaystyle T}$ is a barrel in ${\displaystyle X}$ then ${\displaystyle T}$ absorbs ${\displaystyle D}$ (that is, there is a number ${\displaystyle r>0}$ such that ${\displaystyle D\subseteq rT.}$[4]

If ${\displaystyle U}$ is a convex balanced closed neighborhood of the origin in ${\displaystyle X}$ then the collection of all neighborhoods ${\displaystyle rU,}$ where ${\displaystyle r>0}$ ranges over the positive real numbers, induces a topological vector space topology on ${\displaystyle X.}$ When ${\displaystyle X}$ has this topology, it is denoted by ${\displaystyle X_{U}.}$ Since this topology is not necessarily Hausdorff nor complete, the completion of the Hausdorff space ${\displaystyle X/p_{U}^{-1}(0)}$ is denoted by ${\displaystyle {\overline {X_{U}}}}$ so that ${\displaystyle {\overline {X_{U}}}}$ is a complete Hausdorff space and ${\displaystyle p_{U}(x):=\inf _{x\in rU,r>0}r}$ is a norm on this space making ${\displaystyle {\overline {X_{U}}}}$ into a Banach space. The polar of ${\displaystyle U,}$ ${\displaystyle U^{\circ },}$ is a weakly compact bounded equicontinuous disk in ${\displaystyle X^{\prime }}$ and so is infracomplete.

If ${\displaystyle X}$ is a metrizable locally convex TVS then for every bounded subset ${\displaystyle B}$ of ${\displaystyle X,}$ there exists a bounded disk ${\displaystyle D}$ in ${\displaystyle X}$ such that ${\displaystyle B\subseteq X_{D},}$ and both ${\displaystyle X}$ and ${\displaystyle X_{D}}$ induce the same subspace topology on ${\displaystyle B.}$[5]

The canonical map is the quotient map ${\displaystyle q_{V}:X\to X_{V}=X/p_{V}^{-1}(0),}$ which is continuous when ${\displaystyle X_{V}}$ has either the norm topology or the quotient topology.[1]

If ${\displaystyle U}$ and ${\displaystyle V}$ are radial disks such that ${\displaystyle U\subseteq V}$then ${\displaystyle p_{U}^{-1}(0)\subseteq p_{V}^{-1}(0)}$ so there is a continuous linear surjective canonical map ${\displaystyle q_{V,U}:X/p_{U}^{-1}(0)\to X/p_{V}^{-1}(0)=X_{V}}$ defined by sending ${\displaystyle x+p_{U}^{-1}(0)\in X_{U}=X/p_{U}^{-1}(0)}$ to the equivalence class ${\displaystyle x+p_{V}^{-1}(0),}$ where one may verify that the definition does not depend on the representative of the equivalence class ${\displaystyle x+p_{U}^{-1}(0)}$ that is chosen.[1] This canonical map has norm ${\displaystyle \,\leq 1}$[1] and it has a unique continuous linear canonical extension to ${\displaystyle {\overline {X_{U}}}}$ that is denoted by ${\displaystyle {\overline {g_{V,U}}}:{\overline {X_{U}}}\to {\overline {X_{V}}}.}$

Suppose that in addition ${\displaystyle B\neq \varnothing }$ and ${\displaystyle C}$ are bounded disks in ${\displaystyle X}$ with ${\displaystyle B\subseteq C}$ so that ${\displaystyle X_{B}\subseteq X_{C}}$ and the inclusion ${\displaystyle \operatorname {In} _{B}^{C}:X_{B}\to X_{C}}$ is a continuous linear map. Let ${\displaystyle \operatorname {In} _{B}:X_{B}\to X,}$ ${\displaystyle \operatorname {In} _{C}:X_{C}\to X,}$ and ${\displaystyle \operatorname {In} _{B}^{C}:X_{B}\to X_{C}}$ be the canonical maps. Then ${\displaystyle \operatorname {In} _{C}=\operatorname {In} _{B}^{C}\circ \operatorname {In} _{C}:X_{B}\to X_{C}}$ and ${\displaystyle q_{V}=q_{V,U}\circ q_{U}.}$[1]

A sequence ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }}$ in a TVS ${\displaystyle X}$ is said to be fast convergent[5] to a point ${\displaystyle x\in X}$ if there exists a Banach disk ${\displaystyle D}$ such that both ${\displaystyle x}$ and the sequence is (eventually) contained in ${\displaystyle \operatorname {span} D}$ and ${\displaystyle x_{\bullet }\to x}$ in ${\displaystyle \left(X_{D},p_{D}\right).}$

Every fast convergent sequence is Mackey convergent.[5]