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.
[Math] Difference between the algebraic and topological dual of a topological vector space
functional-analysislinear algebratopological-vector-spacesvector-spaces
Related Solutions
Recall that if a vector space has a finite basis it is said to be finite dimensional, and the dimension is defined to be the number of vectors that make up this basis. Basis are (possibly finite) sets of vectors that span the vector space and are linearly independent. One can prove that every vector in said vector space can be written in one and only one way as a linear combination of these basis vectors. Say $V$ is a $K$ vector space with basis $B=\{v_1,\ldots,v_n\}$. Then if we have $$v=\alpha_1v_1+\cdots+\alpha_nv_n$$
we write $(v)_B=(\alpha_1,\ldots,\alpha_n)$ and say $v$ has coordinates $(\alpha_1,\ldots,\alpha_n)$ in the basis $B$. This immediately gives a mapping $V\to F^n$ given by $$v\mapsto (v)_B$$
This is the same as mapping each basis vector $v_i$ to $$(0,0,\ldots,\underbrace{1}_i,\ldots,0)$$ which entirely determines the transformation.
Note that $0\mapsto (0,0,\ldots,0)$; that $(v+w)_B=(v)_B+(w)_B$ and $(\lambda v)_B=\lambda (v)_B$ so this is a linear transformation, which gives an isomorphism between $V$ and $F^n$. This means $V$ and $F^n$ are essentially the same as vector spaces, that is, "there is only one vector space of dimension $n$ over a field $F$ up to isomorphism."
A vector space over a field $\mathbb{F}$ is a set $V$ with operations $+$ and $\cdot$ satisfying the vector space axioms. Given a vector space $V$, it's dual space $V^\star$ is defined as $\mathrm{Hom}(V,\mathbb{F})$, i.e. the set of all linear maps(functionals) between the vector space and its underlying field(considered as an own vector space in this case).
Normally, the Dirac notation $\langle v|w\rangle$ is a representation of a scalar product and the Bra and Ket correspond on a low level to vectors input in this scalar product. Note, that for the complex scalar product, order of the arguments is important due to its hermitian nature and its semi-bilinearity, i.e. that it is conjugate linear w.r.t. to one argument. A classical way is to define the standard complex scalar product to be conjugate linear in the second argument. In Bra-Ket-notation, you usually reverse the order of the arguments of the complex scalar product, i.e. $\langle x,y\rangle=\langle y|x\rangle$ resulting in conjugate linearity in the first argument.
The key thing is that there is a strong correspondence between scalar products and members of the dual space, i.e. linear functionals.
There is a way that any functional corresponds in a one-to-one fashion to a representation using the scalar product(of the associated vector space). This is known as Riesz representation theorem.
More precisely, you may look at a vector $v\in V$ and suppose that $\langle\cdot,\cdot\rangle$ is an associated scalar product(turning $V$ into a euclidean/unitary space in the real/complex case). Then the map $\varphi_v:w\mapsto\langle w,v\rangle$ is a member of the dual space $V^\star$ and the theorem says that any linear functional $\psi\in V^\star$ can be written as such a $\varphi_v$ uniquely.
Thus, you may convert a scalar product between two vectors into an application of a linear functional to another vector, pulling the problem statement into the realm of dual spaces, where you have other mathematical possibilities to tackle various questions.
EDIT: Note, that the definition of $\varphi_v$ of course depends also on the argument which is assumed to be conjugate linear, in this case the second, as linearity in the first is needed to make $\varphi_v$ a linear map(check this).
Best Answer
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 $C \geq 0$.