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? :)


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.

  • $\begingroup$ 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

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