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Locally convex topological vector space

Fréchet spaces are locally convex spaces that are completely metrizable (with a choice of complete metric). They are generalizations of Banach spaces, which are complete vector spaces with respect to a metric generated by a norm.

History[edit]

Metrizable topologies on vector spaces have been studied since their introduction in Maurice Fréchet's 1902 PhD thesis Sur quelques points du calcul fonctionnel (wherein the notion of a metric was first introduced). After the notion of a general topological space was defined by Felix Hausdorff in 1914,^[1] although locally convex topologies were implicitly used by some mathematicians, up to 1934 only John von Neumann would seem to have explicitly defined the weak topology on Hilbert spaces and strong operator topology on operators on Hilbert spaces.^[2][3] Finally, in 1935 von Neumann introduced the general definition of a locally convex space (called a convex space by him).^[4][5]

A notable example of a result which had to wait for the development and dissemination of general locally convex spaces (amongst other notions and results, like nets, the product topology and Tychonoff's theorem) to be proven in its full generality, is the Banach–Alaoglu theorem which Stefan Banach first established in 1932 by an elementary diagonal argument for the case of separable normed spaces^[6] (in which case the unit ball of the dual is metrizable).

Definition[edit]

Suppose ${\displaystyle X}$ is a vector space over ${\displaystyle \mathbb {K} ,}$asubfield of the complex numbers (normally ${\displaystyle \mathbb {C} }$ itself or ${\displaystyle \mathbb {R} }$). A locally convex space is defined either in terms of convex sets, or equivalently in terms of seminorms.

Definition via convex sets[edit]

Atopological vector space (TVS) is called locally convex if it has a neighborhood basis (that is, a local base) at the origin consisting of balanced, convex sets.^[7] The term locally convex topological vector space is sometimes shortened to locally convex spaceorLCTVS.

A subset ${\displaystyle C}$in${\displaystyle X}$ is called

1. Convex if for all ${\displaystyle x,y\in C,}$ and ${\displaystyle 0\leq t\leq 1,}$ ${\displaystyle tx+(1-t)y\in C.}$ In other words, ${\displaystyle C}$ contains all line segments between points in ${\displaystyle C.}$
2. Circled if for all ${\displaystyle x\in C}$ and scalars ${\displaystyle s,}$if${\displaystyle |s|=1}$ then ${\displaystyle sx\in C.}$If${\displaystyle \mathbb {K} =\mathbb {R} ,}$ this means that ${\displaystyle C}$ is equal to its reflection through the origin. For ${\displaystyle \mathbb {K} =\mathbb {C} ,}$ it means for any ${\displaystyle x\in C,}$ ${\displaystyle C}$ contains the circle through ${\displaystyle x,}$ centred on the origin, in the one-dimensional complex subspace generated by ${\displaystyle x.}$
3. Balanced if for all ${\displaystyle x\in C}$ and scalars ${\displaystyle s,}$if${\displaystyle |s|\leq 1}$ then ${\displaystyle sx\in C.}$If${\displaystyle \mathbb {K} =\mathbb {R} ,}$ this means that if ${\displaystyle x\in C,}$ then ${\displaystyle C}$ contains the line segment between ${\displaystyle x}$ and ${\displaystyle -x.}$ For ${\displaystyle \mathbb {K} =\mathbb {C} ,}$ it means for any ${\displaystyle x\in C,}$ ${\displaystyle C}$ contains the disk with ${\displaystyle x}$ on its boundary, centred on the origin, in the one-dimensional complex subspace generated by ${\displaystyle x.}$ Equivalently, a balanced set is a circled cone (in the TVS ${\textstyle \mathbb {R} ^{2}}$, ${\textstyle C={}}$ball centered at the origin of radius ${\textstyle {\sqrt {2}}}$, ${\textstyle x=(1,1)}$ belongs, ${\textstyle s=2}$, ${\textstyle sx=(2,2)}$ does not belong, C is not a cone but C is balanced, sx is in C, for all x belonging to C and scalar s for which ${\textstyle |s|<1}$).
4. Acone (when the underlying field is ordered) if for all ${\displaystyle x\in C}$ and ${\displaystyle t\geq 0,}$ ${\displaystyle tx\in C.}$
5. Absorbent or absorbing if for every ${\displaystyle x\in X,}$ there exists ${\displaystyle r>0}$ such that ${\displaystyle x\in tC}$ for all ${\displaystyle t\in \mathbb {K} }$ satisfying ${\displaystyle |t|>r.}$ The set ${\displaystyle C}$ can be scaled out by any "large" value to absorb every point in the space.
   • In any TVS, every neighborhood of the origin is absorbent.^[7]
6. Absolutely convex or a disk if it is both balanced and convex. This is equivalent to it being closed under linear combinations whose coefficients absolutely sum to ${\displaystyle \leq 1}$; such a set is absorbent if it spans all of ${\displaystyle X.}$

In fact, every locally convex TVS has a neighborhood basis of the origin consisting of absolutely convex sets (that is, disks), where this neighborhood basis can further be chosen to also consist entirely of open sets or entirely of closed sets.^[8] Every TVS has a neighborhood basis at the origin consisting of balanced sets, but only a locally convex TVS has a neighborhood basis at the origin consisting of sets that are both balanced and convex. It is possible for a TVS to have some neighborhoods of the origin that are convex and yet not be locally convex because it has no neighborhood basis at the origin consisting entirely of convex sets (that is, every neighborhood basis at the origin contains some non-convex set); for example, every non-locally convex TVS ${\displaystyle X}$ has itself (that is, ${\displaystyle X}$) has a convex neighborhood of the origin.

Because translation is continuous (by definition of topological vector space), all translations are homeomorphisms, so every base for the neighborhoods of the origin can be translated to a base for the neighborhoods of any given vector.

Definition via seminorms[edit]

Aseminormon${\displaystyle X}$ is a map ${\displaystyle p:X\to \mathbb {R} }$ such that

1. ${\displaystyle p}$ is nonnegative or positive semidefinite: ${\displaystyle p($x)\geq 0} ;
2. ${\displaystyle p}$ is positive homogeneous or positive scalable: ${\displaystyle p($sx)=|s|p(x)} for every scalar ${\displaystyle s.}$ So, in particular, ${\displaystyle p(0)=0}$;
3. ${\displaystyle p}$ is subadditive. It satisfies the triangle inequality: ${\displaystyle p(x+y)\leq p($x)+p(y).}

If${\displaystyle p}$ satisfies positive definiteness, which states that if ${\displaystyle p($x)=0} then ${\displaystyle x=0,}$ then ${\displaystyle p}$ is a norm. While in general seminorms need not be norms, there is an analogue of this criterion for families of seminorms, separatedness, defined below.

If${\displaystyle X}$ is a vector space and ${\displaystyle {\mathcal {P}}}$ is a family of seminorms on ${\displaystyle X}$ then a subset ${\displaystyle {\mathcal {Q}}}$of${\displaystyle {\mathcal {P}}}$ is called a base of seminorms for ${\displaystyle {\mathcal {P}}}$ if for all ${\displaystyle p\in {\mathcal {P}}}$ there exists a ${\displaystyle q\in {\mathcal {Q}}}$ and a real ${\displaystyle r>0}$ such that ${\displaystyle p\leq rq.}$^[9]

Definition (second version): A locally convex space is defined to be a vector space ${\displaystyle X}$ along with a family ${\displaystyle {\mathcal {P}}}$ of seminorms on ${\displaystyle X.}$

Seminorm topology[edit]

Suppose that ${\displaystyle X}$ is a vector space over ${\displaystyle \mathbb {K} ,}$ where ${\displaystyle \mathbb {K} }$ is either the real or complex numbers. A family of seminorms ${\displaystyle {\mathcal {P}}}$ on the vector space ${\displaystyle X}$ induces a canonical vector space topology on ${\displaystyle X}$, called the initial topology induced by the seminorms, making it into a topological vector space (TVS). By definition, it is the coarsest topology on ${\displaystyle X}$ for which all maps in ${\displaystyle {\mathcal {P}}}$ are continuous.

It is possible for a locally convex topology on a space ${\displaystyle X}$ to be induced by a family of norms but for ${\displaystyle X}$tonotbenormable (that is, to have its topology be induced by a single norm).

Basis and subbases[edit]

Let ${\displaystyle B_{<r}}$ denote the open ball of radius ${\displaystyle r>0}$in${\displaystyle \mathbb {K} }$. The family of sets ${\displaystyle p^{-1}\left(B_{<r}\right)=\{x\in X:p($x)<r\}} as${\displaystyle p}$ ranges over a family of seminorms ${\displaystyle {\mathcal {P}}}$ and ${\displaystyle r}$ ranges over the positive real numbers is a subbasis at the origin for the topology induced by ${\displaystyle {\mathcal {P}}}$. These sets are convex, as follows from properties 2 and 3 of seminorms. Intersections of finitely many such sets are then also convex, and since the collection of all such finite intersections is a basis at the origin it follows that the topology is locally convex in the sense of the first definition given above.

Recall that the topology of a TVS is translation invariant, meaning that if ${\displaystyle S}$ is any subset of ${\displaystyle X}$ containing the origin then for any ${\displaystyle x\in X,}$ ${\displaystyle S}$ is a neighborhood of the origin if and only if ${\displaystyle x+S}$ is a neighborhood of ${\displaystyle x}$; thus it suffices to define the topology at the origin. A base of neighborhoods of ${\displaystyle y}$ for this topology is obtained in the following way: for every finite subset ${\displaystyle F}$of${\displaystyle {\mathcal {P}}}$ and every ${\displaystyle r>0,}$ let $${\displaystyle U_{F,r}($$y):=\{x\in X:p(x-y)<r\ {\text{ for all }}p\in F\}.}

Bases of seminorms and saturated families[edit]

If${\displaystyle X}$ is a locally convex space and if ${\displaystyle {\mathcal {P}}}$ is a collection of continuous seminorms on ${\displaystyle X}$, then ${\displaystyle {\mathcal {P}}}$ is called a base of continuous seminorms if it is a base of seminorms for the collection of all continuous seminorms on ${\displaystyle X}$.^[9] Explicitly, this means that for all continuous seminorms ${\displaystyle p}$on${\displaystyle X}$, there exists a ${\displaystyle q\in {\mathcal {P}}}$ and a real ${\displaystyle r>0}$ such that ${\displaystyle p\leq rq.}$^[9]If${\displaystyle {\mathcal {P}}}$ is a base of continuous seminorms for a locally convex TVS ${\displaystyle X}$ then the family of all sets of the form ${\displaystyle \{x\in X:q($x)<r\}} as${\displaystyle q}$ varies over ${\displaystyle {\mathcal {P}}}$ and ${\displaystyle r}$ varies over the positive real numbers, is a base of neighborhoods of the origin in ${\displaystyle X}$ (not just a subbasis, so there is no need to take finite intersections of such sets).^{[9][proof 1]}

A family ${\displaystyle {\mathcal {P}}}$ of seminorms on a vector space ${\displaystyle X}$ is called saturated if for any ${\displaystyle p}$ and ${\displaystyle q}$in${\displaystyle {\mathcal {P}},}$ the seminorm defined by ${\displaystyle x\mapsto \max\{p($x),q(x)\}} belongs to ${\displaystyle {\mathcal {P}}.}$

If${\displaystyle {\mathcal {P}}}$ is a saturated family of continuous seminorms that induces the topology on ${\displaystyle X}$ then the collection of all sets of the form ${\displaystyle \{x\in X:p($x)<r\}} as${\displaystyle p}$ ranges over ${\displaystyle {\mathcal {P}}}$ and ${\displaystyle r}$ ranges over all positive real numbers, forms a neighborhood basis at the origin consisting of convex open sets;^[9] This forms a basis at the origin rather than merely a subbasis so that in particular, there is no need to take finite intersections of such sets.^[9]

Basis of norms[edit]

The following theorem implies that if ${\displaystyle X}$ is a locally convex space then the topology of ${\displaystyle X}$ can be a defined by a family of continuous normson${\displaystyle X}$ (anorm is a seminorm ${\displaystyle s}$ where ${\displaystyle s($x)=0} implies ${\displaystyle x=0}$) if and only if there exists at least one continuous normon${\displaystyle X}$.^[10] This is because the sum of a norm and a seminorm is a norm so if a locally convex space is defined by some family ${\displaystyle {\mathcal {P}}}$ of seminorms (each of which is necessarily continuous) then the family ${\displaystyle {\mathcal {P}}+n:=\{p+n:p\in {\mathcal {P}}\}}$ of (also continuous) norms obtained by adding some given continuous norm ${\displaystyle n}$ to each element, will necessarily be a family of norms that defines this same locally convex topology. If there exists a continuous norm on a topological vector space ${\displaystyle X}$ then ${\displaystyle X}$ is necessarily Hausdorff but the converse is not in general true (not even for locally convex spaces or Fréchet spaces).

1. ${\displaystyle X}$ does not admit a continuous norm (that is, any continuous seminorm on ${\displaystyle X}$ can not be a norm).
2. ${\displaystyle X}$ contains a vector subspace that is TVS-isomorphic to ${\displaystyle \mathbb {K} ^{\mathbb {N} }.}$
3. ${\displaystyle X}$ contains a complemented vector subspace that is TVS-isomorphic to ${\displaystyle \mathbb {K} ^{\mathbb {N} }.}$

Nets[edit]

Suppose that the topology of a locally convex space ${\displaystyle X}$ is induced by a family ${\displaystyle {\mathcal {P}}}$ of continuous seminorms on ${\displaystyle X}$. If ${\displaystyle x\in X}$ and if ${\displaystyle x_{\bullet }=\left(x_{i}\right)_{i\in I}}$ is a netin${\displaystyle X}$, then ${\displaystyle x_{\bullet }\to x}$in${\displaystyle X}$ if and only if for all ${\displaystyle p\in {\mathcal {P}},}$ ${\displaystyle p\left(x_{\bullet }-x\right)=\left(p\left(x_{i}\right)-x\right)_{i\in I}\to 0.}$^[12] Moreover, if ${\displaystyle x_{\bullet }}$ is Cauchy in ${\displaystyle X}$, then so is ${\displaystyle p\left(x_{\bullet }\right)=\left(p\left(x_{i}\right)\right)_{i\in I}}$ for every ${\displaystyle p\in {\mathcal {P}}.}$^[12]

Equivalence of definitions[edit]

Although the definition in terms of a neighborhood base gives a better geometric picture, the definition in terms of seminorms is easier to work with in practice. The equivalence of the two definitions follows from a construction known as the Minkowski functional or Minkowski gauge. The key feature of seminorms which ensures the convexity of their ${\displaystyle \varepsilon }$-balls is the triangle inequality.

For an absorbing set ${\displaystyle C}$ such that if ${\displaystyle x\in C,}$ then ${\displaystyle tx\in C}$ whenever ${\displaystyle 0\leq t\leq 1,}$ define the Minkowski functional of ${\displaystyle C}$ to be $${\displaystyle \mu _{C}($$x)=\inf\{r>0:x\in rC\}.}

From this definition it follows that ${\displaystyle \mu _{C}}$ is a seminorm if ${\displaystyle C}$ is balanced and convex (it is also absorbent by assumption). Conversely, given a family of seminorms, the sets $${\displaystyle \left\{x:p_{\alpha _{1}}($$x)<\varepsilon _{1},\ldots ,p_{\alpha _{n}}(x)<\varepsilon _{n}\right\}} form a base of convex absorbent balanced sets.

Ways of defining a locally convex topology[edit]

1. Every ${\displaystyle B\in {\mathcal {B}}}$isconvex, balanced, and absorbing;
2. For every ${\displaystyle B\in {\mathcal {B}}}$ there exists some real ${\displaystyle r}$ satisfying ${\displaystyle 0<r\leq 1/2}$ such that ${\displaystyle rB\in {\mathcal {B}}.}$

Then ${\displaystyle {\mathcal {B}}}$ is a neighborhood base at 0 for a locally convex TVS topology on ${\displaystyle X.}$

Example: auxiliary normed spaces

If${\displaystyle W}$isconvex and absorbingin${\displaystyle X}$ then the symmetric set ${\displaystyle D:=\bigcap _{|u|=1}uW}$ will be convex and balanced (also known as an absolutely convex set or a disk) in addition to being absorbing in ${\displaystyle X.}$ This guarantees that the Minkowski functional ${\displaystyle p_{D}:X\to \mathbb {R} }$of${\displaystyle D}$ will be a seminormon${\displaystyle X,}$ thereby making ${\displaystyle \left(X,p_{D}\right)}$ into a seminormed space that carries its canonical pseudometrizable topology. The set of scalar multiples ${\displaystyle rD}$as${\displaystyle r}$ ranges over ${\displaystyle \left\{{\tfrac {1}{2}},{\tfrac {1}{3}},{\tfrac {1}{4}},\ldots \right\}}$ (or over any other set of non-zero scalars having ${\displaystyle 0}$ as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If ${\displaystyle X}$ is a topological vector space and if this convex absorbing subset ${\displaystyle W}$ is also a bounded subsetof${\displaystyle X,}$ then the absorbing disk ${\displaystyle D:=\bigcap _{|u|=1}uW}$ will also be bounded, in which case ${\displaystyle p_{D}}$ will be a norm and ${\displaystyle \left(X,p_{D}\right)}$ will form what is known as an auxiliary normed space. If this normed space is a Banach space then ${\displaystyle D}$ is called a Banach disk.

Further definitions[edit]

• A family of seminorms ${\displaystyle \left(p_{\alpha }\right)_{\alpha }}$ is called totalorseparated or is said to separate points if whenever ${\displaystyle p_{\alpha }($x)=0} holds for every ${\displaystyle \alpha }$ then ${\displaystyle x}$ is necessarily ${\displaystyle 0.}$ A locally convex space is Hausdorff if and only if it has a separated family of seminorms. Many authors take the Hausdorff criterion in the definition.
• Apseudometric is a generalization of a metric which does not satisfy the condition that ${\displaystyle d(x,y)=0}$ only when ${\displaystyle x=y.}$ A locally convex space is pseudometrizable, meaning that its topology arises from a pseudometric, if and only if it has a countable family of seminorms. Indeed, a pseudometric inducing the same topology is then given by $${\displaystyle d(x,y)=\sum _{n}^{\infty }{\frac {1}{2^{n}}}{\frac {p_{n}(x-y)}{1+p_{n}(x-y)}}}$$ (where the ${\displaystyle 1/2^{n}}$ can be replaced by any positive summable sequence ${\displaystyle a_{n}}$). This pseudometric is translation-invariant, but not homogeneous, meaning ${\displaystyle d(kx,ky)\neq |k|d(x,y),}$ and therefore does not define a (pseudo)norm. The pseudometric is an honest metric if and only if the family of seminorms is separated, since this is the case if and only if the space is Hausdorff. If furthermore the space is complete, the space is called a Fréchet space.
• As with any topological vector space, a locally convex space is also a uniform space. Thus one may speak of uniform continuity, uniform convergence, and Cauchy sequences.
• ACauchy net in a locally convex space is a net ${\displaystyle \left(x_{a}\right)_{a\in A}}$ such that for every ${\displaystyle r>0}$ and every seminorm ${\displaystyle p_{\alpha },}$ there exists some index ${\displaystyle c\in A}$ such that for all indices ${\displaystyle a,b\geq c,}$ ${\displaystyle p_{\alpha }\left(x_{a}-x_{b}\right)<r.}$ In other words, the net must be Cauchy in all the seminorms simultaneously. The definition of completeness is given here in terms of nets instead of the more familiar sequences because unlike Fréchet spaces which are metrizable, general spaces may be defined by an uncountable family of pseudometrics. Sequences, which are countable by definition, cannot suffice to characterize convergence in such spaces. A locally convex space is complete if and only if every Cauchy net converges.
• A family of seminorms becomes a preordered set under the relation ${\displaystyle p_{\alpha }\leq p_{\beta }}$ if and only if there exists an ${\displaystyle M>0}$ such that for all ${\displaystyle x,}$ ${\displaystyle p_{\alpha }($x)\leq Mp_{\beta }(x).} One says it is a directed family of seminorms if the family is a directed set with addition as the join, in other words if for every ${\displaystyle \alpha }$ and ${\displaystyle \beta ,}$ there is a ${\displaystyle \gamma }$ such that ${\displaystyle p_{\alpha }+p_{\beta }\leq p_{\gamma }.}$ Every family of seminorms has an equivalent directed family, meaning one which defines the same topology. Indeed, given a family ${\displaystyle \left(p_{\alpha }($x)\right)_{\alpha \in I},} let ${\displaystyle \Phi }$ be the set of finite subsets of ${\displaystyle I}$ and then for every ${\displaystyle F\in \Phi }$ define $${\displaystyle q_{F}=\sum _{\alpha \in F}p_{\alpha }.}$$ One may check that ${\displaystyle \left(q_{F}\right)_{F\in \Phi }}$ is an equivalent directed family.
• If the topology of the space is induced from a single seminorm, then the space is seminormable. Any locally convex space with a finite family of seminorms is seminormable. Moreover, if the space is Hausdorff (the family is separated), then the space is normable, with norm given by the sum of the seminorms. In terms of the open sets, a locally convex topological vector space is seminormable if and only if the origin has a bounded neighborhood.

Sufficient conditions[edit]

Hahn–Banach extension property[edit]

Let ${\displaystyle X}$ be a TVS. Say that a vector subspace ${\displaystyle M}$of${\displaystyle X}$ has the extension property if any continuous linear functional on ${\displaystyle M}$ can be extended to a continuous linear functional on ${\displaystyle X}$.^[13] Say that ${\displaystyle X}$ has the Hahn-Banach extension property (HBEP) if every vector subspace of ${\displaystyle X}$ has the extension property.^[13]

The Hahn-Banach theorem guarantees that every Hausdorff locally convex space has the HBEP. For complete metrizable TVSs there is a converse:

If a vector space ${\displaystyle X}$ has uncountable dimension and if we endow it with the finest vector topology then this is a TVS with the HBEP that is neither locally convex or metrizable.^[13]

Properties[edit]

Throughout, ${\displaystyle {\mathcal {P}}}$ is a family of continuous seminorms that generate the topology of ${\displaystyle X.}$

Topological closure

If${\displaystyle S\subseteq X}$ and ${\displaystyle x\in X,}$ then ${\displaystyle x\in \operatorname {cl} S}$ if and only if for every ${\displaystyle r>0}$ and every finite collection ${\displaystyle p_{1},\ldots ,p_{n}\in {\mathcal {P}}}$ there exists some ${\displaystyle s\in S}$ such that ${\displaystyle \sum _{i=1}^{n}p_{i}(x-s)<r.}$^[14] The closure of ${\displaystyle \{0\}}$in${\displaystyle X}$ is equal to ${\displaystyle \bigcap _{p\in {\mathcal {P}}}p^{-1}(0).}$^[15]

Topology of Hausdorff locally convex spaces

Every Hausdorff locally convex space is homeomorphic to a vector subspace of a product of Banach spaces.^[16] The Anderson–Kadec theorem states that every infinite–dimensional separable Fréchet spaceishomeomorphic to the product space ${\textstyle \prod _{i\in \mathbb {N} }\mathbb {R} }$ of countably many copies of ${\displaystyle \mathbb {R} }$ (this homeomorphism need not be a linear map).^[17]

Properties of convex subsets[edit]

Algebraic properties of convex subsets

A subset ${\displaystyle C}$ is convex if and only if ${\displaystyle tC+(1-t)C\subseteq C}$ for all ${\displaystyle 0\leq t\leq 1}$^[18] or equivalently, if and only if ${\displaystyle (s+t)C=sC+tC}$ for all positive real ${\displaystyle s>0{\text{ and }}t>0,}$^[19] where because ${\displaystyle (s+t)C\subseteq sC+tC}$ always holds, the equals sign ${\displaystyle \,=\,}$ can be replaced with ${\displaystyle \,\supseteq .\,}$If${\displaystyle C}$ is a convex set that contains the origin then ${\displaystyle C}$isstar shaped at the origin and for all non-negative real ${\displaystyle s\geq 0{\text{ and }}t\geq 0,}$ ${\displaystyle ($sC)\cap (tC)=(\min _{}\{s,t\})C.}

The Minkowski sum of two convex sets is convex; furthermore, the scalar multiple of a convex set is again convex.^[20]

Topological properties of convex subsets

• Suppose that ${\displaystyle Y}$ is a TVS (not necessarily locally convex or Hausdorff) over the real or complex numbers. Then the open convex subsets of ${\displaystyle Y}$ are exactly those that are of the form ${\displaystyle z+\{y\in Y:p($y)<1\}=\{y\in Y:p(y-z)<1\}} for some ${\displaystyle z\in Y}$ and some positive continuous sublinear functional ${\displaystyle p}$on${\displaystyle Y.}$^[21]
• The interior and closure of a convex subset of a TVS is again convex.^[20]
• If${\displaystyle C}$ is a convex set with non-empty interior, then the closure of ${\displaystyle C}$ is equal to the closure of the interior of ${\displaystyle C}$; furthermore, the interior of ${\displaystyle C}$ is equal to the interior of the closure of ${\displaystyle C.}$^[20][22]
  • So if the interior of a convex set ${\displaystyle C}$ is non-empty then ${\displaystyle C}$ is a closed (respectively, open) set if and only if it is a regular closed (respectively, regular open) set.
• If${\displaystyle C}$ is convex and ${\displaystyle 0<t\leq 1,}$ then^[23] ${\displaystyle t\operatorname {Int} C+(1-t)\operatorname {cl} C~\subseteq ~\operatorname {Int} C.}$ Explicitly, this means that if ${\displaystyle C}$ is a convex subset of a TVS ${\displaystyle X}$ (not necessarily Hausdorff or locally convex), ${\displaystyle y}$ belongs to the closure of ${\displaystyle C,}$ and ${\displaystyle x}$ belongs to the interior of ${\displaystyle C,}$ then the open line segment joining ${\displaystyle x}$ and ${\displaystyle y}$ belongs to the interior of ${\displaystyle C;}$ that is, ${\displaystyle \{tx+(1-t)y:0<t<1\}\subseteq \operatorname {int} _{X}C.}$^{[22][24][proof 2]}
• If${\displaystyle M}$ is a closed vector subspace of a (not necessarily Hausdorff) locally convex space${\displaystyle X,}$ ${\displaystyle V}$ is a convex neighborhood of the origin in ${\displaystyle M,}$ and if ${\displaystyle z\in X}$ is a vector notin${\displaystyle V,}$ then there exists a convex neighborhood ${\displaystyle U}$ of the origin in ${\displaystyle X}$ such that ${\displaystyle V=U\cap M}$ and ${\displaystyle z\not \in U.}$^[20]
• The closure of a convex subset of a locally convex Hausdorff space ${\displaystyle X}$ is the same for all locally convex Hausdorff TVS topologies on ${\displaystyle X}$ that are compatible with duality between ${\displaystyle X}$ and its continuous dual space.^[25]
• In a locally convex space, the convex hull and the disked hull of a totally bounded set is totally bounded.^[7]
• In a complete locally convex space, the convex hull and the disked hull of a compact set are both compact.^[7]
  • More generally, if ${\displaystyle K}$ is a compact subset of a locally convex space, then the convex hull ${\displaystyle \operatorname {co} K}$ (respectively, the disked hull ${\displaystyle \operatorname {cobal} K}$) is compact if and only if it is complete.^[7]
• In a locally convex space, convex hulls of bounded sets are bounded. This is not true for TVSs in general.^[26]
• In a Fréchet space, the closed convex hull of a compact set is compact.^[27]
• In a locally convex space, any linear combination of totally bounded sets is totally bounded.^[26]