What is the difference between the algebraic and the topological dual of a topological vector space, such as for example the Euclidean space $\mathbb{H}$?

I am interested in intuitive as well as in detailled technical answers.

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    $\begingroup$ If the vector space is finite dimensional, there is no difference at all. $\endgroup$
    – Crostul
    Sep 11, 2016 at 13:09
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    $\begingroup$ The algebraic dual is all linear maps from the vector space to the scalar field. The topological dual is all continuous linear maps from the vector space to the scalar field. For finite-dimensional vector spaces (over $\mathbf R$ or $\mathbf C$) all linear maps between them and the scalar field are continuous. $\endgroup$
    – KCd
    Sep 11, 2016 at 13:24
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    $\begingroup$ @KCd how do you define continuity without a topology and simply on a vector space? $\endgroup$
    – user288972
    Jun 27, 2019 at 19:52
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    $\begingroup$ @Andrew you don't. The question was about a topological vector space, so when I wrote "the vector space" in my previous comment I was referring to a topological vector space. Finite-dimensional vector spaces over $\mathbf R$ and $\mathbf C$ (or over any other field complete with respect to an absolute value, such as the $p$-adic numbers $\mathbf Q_p$ for a prime $p$) have exactly one nondiscrete Hausdorff topology, namely the topology coming from the sup-norm for any choice of basis. That's the topology we always use on such spaces. $\endgroup$
    – KCd
    Jun 27, 2019 at 21:43
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    $\begingroup$ For vector spaces that have no natural topology, you might use the discrete topology on them if you must use a topology at all, in which case the algebraic and topological dual are exactly the same thing. $\endgroup$
    – KCd
    Jun 27, 2019 at 21:44

2 Answers 2


Let $E$ a vector space on $\mathbb{K}$ with topology $\mathcal{T}$. $(E, \mathcal{T})$ is said a vector topological space and $\mathcal{T}$ a vector topology if the functions \begin{align*} \displaystyle (x,y) \in E \times E \longmapsto x+y \in E \\ \displaystyle (\lambda,x) \in \mathbb{K} \times E \longmapsto \lambda x \in E \end{align*} are continuous. Now basically, the vector topology is invariant to translations, so that $U$ is a neighborhood of $0 \in E$ if and only if $x_0+U$ is a neighborhood of $x_0 \in E$, and $\mathcal{U}$ is a basis of neighborhood of $0 \in E$ if and only if $x_0+\mathcal{U}$ is a local basis of $x_0 \in E$.

We can consider the following facts

Now, if $\mathcal{U}$ and $\mathcal{V}$ are two bases of origin of the topological vector space $E$ and $F$ respectively, then a linear map $T:E \longrightarrow F$ is continuous if and only if $\forall V \in \mathcal{V}$ there is $U \in \mathcal{U}$ such that $T(U) \subset V$. This result is not difficult to verify.

For definition, if $E$ is a topological vector space, a subset $F \subset E$ is said bounded topologically if $\forall U \in \mathcal{U}$ there is $\lambda > 0$ such that $F \subset \lambda U$. Now a map $T:E \longrightarrow F$ is said bounded if it sends topologically bounded sets in topologically bounded sets, and there is the following result:

$Theorem$. Let $T:E \longrightarrow F$ a linear operator, consider the following properties

(a) $T$ is continuos

(b) $T$ is bounded

(c) If $x_k \rightarrow 0$ in $E$, then $\lbrace T(x_k) \rbrace$ is topologically bounded in $F$

(d) If $x_k \rightarrow 0$ in $E$, then $T(x_k) \rightarrow 0$ in $F$.

We have that $(a) \Longrightarrow (b) \Longrightarrow (c)$, and if $E$ is metrizable all properties are equivalent.

In case of metrizability we can define the dual space $\mathcal{L}(E, \mathbb{K})$ that consists of all linear functional continuous of type $T:E \longrightarrow \mathbb{K}$. The difference with the dual algebraic $E'$ of $E$ is that it's consists simply of all linear functional of the type $T:E \longrightarrow \mathbb{K}$.

Even in the case of not metrizable space $E$, you can be considered a locally convex space $E$ (which they are also topological vector spaces) defined by a separable family of seminorm, and it can be shown the following result

$Theorem$. If $\mathcal{P}$ and $\mathcal{Q}$ are two separable families seminorm which define a vector topology Hausdorff so that $E$ and $F$ are LCS, then a linear application $T:E \longrightarrow F$ is continuous if and only if $\forall q \in \mathcal{Q}$ there are $p_j \in \mathcal{P}$ with $j \in J$ ($J$ finite) and there is a constant $C > 0$ such that $q(Tx) \leq C \max_{j \in J} p_j(x)$.

Therefore in this case we can define the dual $E'=\mathcal{L}(E,\mathbb{K})$ of the LCS $E$, that is the set of continuous linear functional $T:E \rightarrow \mathbb{K}$ according to the previous characterization of continuity.

Note that this characterization is a generalization of what is happening for normed spaces $E$ and $F$ so that $T:E \longrightarrow F$ is continuous if and only if $||Tx||_F \leq C ||x||_E$ for some $C \geq 0$.


In finite dimensions, there is no difference; see the post Let $X$ be a finite dimensional normed space. Does the algebraic dual space $X^*$ and the dual space $X'$ coincide?

In infinite dimensions, the difference is huge. For example, the topological dual (the space of all continuous linear functionals) of a Hilbert space is the Hilbert space itself, by the Riesz representation theorem, while the algebraic dual (the space of all linear functionals) is vastly bigger since there are lots of non-continuous linear functionals.

For example, one can use Zorn's lemma to extend an orthonormal basis (in the Hilbert space sense) to an algebraic basis by adding inductively elements that cannot be written as finite linear combinations of basis elements already chosen. Then one can map each basis element to an arbitrary number and extend by linearity to a linear functional. Most of these are not continuous since the continuous ones are already fully determined by the image of the orthonormal basis.

This only works in infinite dimensions since an orthonormal basis is not already an algebraic basis iff the dimension is infinite.


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