Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know the meaning of dual of linear map in inner product spaces, also it is defined in Banach space[ Rudin functional analysis].

What is the definition of Dual of linear map if vector space are Frechet space/ or more generally locally convex sapce which is not normable.

share|cite|improve this question
We can try the definition with duality bracket: if $T\colon X\to Y$ is linear, we should have $$\langle T^*f,x\rangle=\langle f,Tx\rangle$$ where $f\in Y'$ and $x\in X$ ($T^*\colon Y'\to X'$). – Davide Giraudo Aug 2 '12 at 12:01
up vote 3 down vote accepted

Definition of dual linear map in topological vector spaces over field $F$ is the same as in normable case. Given linear map $T:V\to W$ between topological vector spaces we define its adjoint as $$ T' : W'\to V':f\mapsto f\circ T $$ Motivation for this definition of dual linear operator from categorical point of view is the following. If you want to study some object (for our purposes linear operators between topological vector spaces) its useful to study maps from this object ($\mathrm{Hom}(V,W)$ in our case) into something simple (field of scalras $F$ will fit). This useful maps can be obtained by applying contravariant $\mathrm{Hom}(-,F)$ functor to the linear operators $T\in\mathrm{Hom}(V,W)$. Then you get $\mathrm{Hom}(T,F)=T'$

Unfortunately, this definition may not have much sense, because for general topological vector spaces its dual can be trivial. But if you restrict your interest to locally convex spaces this problem won't rise.

share|cite|improve this answer

Let $X$ and $Y$ two locally convex spaces, and $X'$, $Y'$ their respective duals. If $T\colon X\to Y$ is linear, we can define $T^*\colon Y'\to X'$ by $$\forall l\in Y', \forall x\in X, \quad \langle T^*l,x\rangle_{X',X}=\langle l,Tx\rangle_{Y',Y}.$$

share|cite|improve this answer
I see, indeed it seems it not used, since it seems well-defined in any case. Thanks! – Davide Giraudo Aug 2 '12 at 15:21
What local convexity (and hence separation of points) does give you is that taking the adjoint is injective from $L(X,Y)$ to $L(Y',X')$. You can of course put other topologies on the dual spaces than the weak ones but then you have to check some things. I think this is discussed in detail in all texts on locally convex spaces, e.g. Trèves, Kelley, Koethe, Schaefer etc. (but it may look far more complicated in some of them than it actually is...) – t.b. Aug 2 '12 at 15:30
@t.b. Thanks. It's indeed the confusion I made. I confused the condition for injectivity with the condition for well-definiteness. – Davide Giraudo Aug 2 '12 at 15:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.