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In functional analysis there are many places where dual space is mentioned, but I still don't understand the real power of that concept. Why do we need the dual space?

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closed as too broad by Nick Peterson, Thomas Andrews, AlexR, Yiorgos S. Smyrlis, TMM Feb 10 '14 at 0:17

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Do you have another viable source of questions for homework and exams in functional analysis? :-P $\endgroup$ – Asaf Karagila Feb 9 '14 at 23:30
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More generally, if $X$ and $Y$ are sets (resp. vector spaces over a common field $k$), we can consider the set $Y^X$ of all mappings (resp. the vector space $\mathrm{Hom}(X,Y)$ of all linear mappings) from $X$ to $Y$. These are obviously interesting to study if you think in terms of structures and structure preserving maps; and dual spaces are simply a special case (namely $\mathrm{Hom}(X,k)$). Also $\mathrm{Hom}$ behaves very nicely with respect to short exact sequences, and is deeply related to the tensor product.

Now apparently you are interested in a functional analysis point of view: here we assume that the dual space is the topological dual space (i.e. continuous linear functionals); the dual space is always complete (regardless of whether $X$ is complete or not) and points can be separated by elements of the dual space, which is often useful. A dual space which is particularly interesting is $\mathscr{D}'$: the space of distributions (here $\mathscr{D}$ is $C^\infty_0$ endowed with its canonical LF topology); it can be shown that if $P(\partial)$ is any linear constant coefficient partial differential operator, then there is a distribution $E$ (called a fundamental solution for $P(\partial)$) such that $P(\partial)E=\delta_0$ in the sense of distributions (this is the theorem of Malgrange-Ehrenpreis). This is a very important result because of the following corollary: if $f\in C^\infty_0$, then the PDE $P(\partial)u=f$ admits a $C^\infty$ solution explicitly given by $u=E\ast f$. In physics (say electrodynamics) this is used all the time to solve PDE's. Also, one often seeks weak solutions for the PDE $P(\partial)u=f$ (i.e. this equation is satisfied in the distribution sense) and then tries to prove that they are strong (usual) solutions; this is a very convenient method.

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The dual space is a concept that shows up in greater generality in Linear Algebra. If $V$ is a vector space then its dual space $V^*$ is the set of all linear functionals from $V$ to its base field $F$.

When $V$ is a finite dimensional vector space, then so is $V^*$ and $\text{dim } V^*=\text{dim }V$. Thus there is a linear isomorphism between them when they are taken over the same field. Why is this important? Such an isomorphism gives us an example of a unnatural isomorphism in category theory. Since $V^*$ is also a vector space, it also has a dual denoted $V^{**}$ called the double-dual of $V$. Unlike between $V$ and $V^*$ there is a natural isomorphism between $V$ and $V^{**}$.

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  • $\begingroup$ he is asking in the context of functional analysis. -2 $\endgroup$ – user88576 Feb 9 '14 at 23:33
  • $\begingroup$ I was under the impression that he has only heard of the dual space in the context of functional analysis. He said he didn't understand the full power of the concept, and I gave him an instance of how it useful elsewhere. $\endgroup$ – Robert Wolfe Feb 9 '14 at 23:37
  • $\begingroup$ I know the definition, but I didn't know why it is use full. Thank you. $\endgroup$ – Emin Feb 10 '14 at 16:32

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