0
$\begingroup$

I am currently studying for my Functional Analysis test and then started thinking about the following and figured it is true (if it is not true, please do tell me - I am just thinking about this to try get as deep an understanding of the work as possible):

Let $X$ be a finite dimensional normed space. Then the algebraic dual space $X^*$ and the dual space $X'$ coincide.

My reasoning for why this is true:

$X' \subseteq X^*$ is trivial. We thus need to show that $X^* \subseteq X'$ also holds.

Since $X$ is a finite-dimensional normed space, we know that every linear operator on $X$ is bounded.

Now consider a functional $f$ on $X$. It makes sense to speak of functionals on $X$ since $X$, as a normed space, is also a topological vector space (and hence a vector space). That is, $f \in X^*$.

Now we know that $f$ is also a linear operator (since functionals are linear operators from $X$ onto $\mathbb C$) and, since $X$ is finite dimensional, $f$ is bounded, that is, $f \in X'$. We thus have that $X^* \subseteq X' $.

that is, $$X^* = X'.$$

Is this argument valid? Also, what happens when $X$ is not finite-dimensional? My intuition tells me that this will then not be true - otherwise there would be no real reason to distinguish between the definitions of $X^*$ and $X'$. But can somebody maybe show me why? :)

$\endgroup$
4
$\begingroup$

Yes. Your argument is valid. In a nutshell, your argument is this:

Take any $f \in X^*$. Since $X$ is finite dimensional, $f$ is bounded. So, $f \in X'$. So, $X^* \subseteq X'$, as desired.

When $X$ is infinite dimensional, we can always construct an unbounded linear function $f$ on $X$. Finding an explicit construction of such an $X$ is tricky. However, it suffices to find an example that works on a vector space of countably infinite dimension (it is non-trivial to see that this is indeed enough; you'll need to inject the axiom of choice appropriately).

To that end: let $\{e_k\}_{k \in \Bbb N}$ be a (Hamel-)basis for $X$ consisting of unit-vectors. We can the define a linear functional by $$ f(e_n) = n $$ Hopefully you can see that this must be unbounded.

$\endgroup$
  • $\begingroup$ Could you explain a little bit why it suffices to consider only countably infinite dimension? Thank you very much. $\endgroup$ – Jiaqi Li Oct 3 '17 at 22:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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