I'm taking a course on functional analysis at the moment and the lecturer has defined the dual space of a normed space $X$ to be the space of all bounded linear functionals and denoted it $X^*$. However if I look at other references (such as Wikipedia or Wolfram MathWorld) the definition of the dual space is given to be the space of all linear functionals (ie, they don't specify that they have to be bounded). Does this imply that any linear functional is necessarily bounded, or is my lecturer giving an unusual definition, or is the definition on Wikipedia wrong? I'm just a bit confused about this.

  • $\begingroup$ Both definitions are out there. The continuous dual as you are working with is more common in my experience. $\endgroup$ – Ian Apr 6 '17 at 1:04
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    $\begingroup$ Yeah, you will almost always be working with bounded functionals. $\endgroup$ – David Bowman Apr 6 '17 at 1:07
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    $\begingroup$ If $X$ is finite dimensional then every linear functional on $X$ is bounded. In this context people often do not mention the boundedness because it comes for free. $\endgroup$ – BindersFull Apr 6 '17 at 1:33

There are (at least) two somewhat conflicting definitions of a dual space.

In functional analysis, we start with a topological vector space $V$, usually over real or complex numbers, and then the (continuous) dual is the space of all continuous functionals. For normed (in particular, Banach) spaces, a functional is continuous if and only if it is bounded.

In abstract algebra, we deal with vector spaces over arbitrary fields. Frequently, the vector spaces are finite-dimensional, in which case the dual actually coincides with algebraic dual, but for infinite dimensional spaces (over real or complex numbers), the algebraic dual is usually larger by far than the continuous dual*.

More abstractly, the difference of definition is a matter of perspective. In terms of category theory, you can think of the dual of a $K$-vector space $V$ as the space $\operatorname{Hom}(V,K)$, i.e. the set of morphisms from $V$ to the base field $K$ in your category. If the category is that of vector spaces, you obtain the algebraic dual. If the category is that of topological vector spaces, you obtain the continuous dual.

*${}$a pure vector space can be regarded as a discrete topological vector space. In this case, it is fairly easy to see that the continuous dual and the algebraic dual coincide.

  • $\begingroup$ I see, thanks for clearing it up. $\endgroup$ – IAlreadyHaveAKey Apr 6 '17 at 2:05

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