I have to prove that every finite dimensional Banach space is reflexive. The text suggests me to use the fact that every N-dimensional space is linearly isomorphic to $ \mathbb{R}^N $, and that if two spaces $ X,Y $ are linearly isomorphic, then if one is reflexive then the other one is reflexive. So I need to prove that $ (\mathbb{R}^N, ||\cdot||_2) $ is reflexive. In the following, $X^*$ is the space of the linear and bounded operators over $X$ and $X^{**} $ means $ (X^*)^*$. I tried the following way.
I have to prove that the canonical map: \begin{align} & \tau:\mathbb{R}^N \rightarrow (\mathbb{R}^N)^{**} \text{ s.t. } \\ & \tau(x)=\Lambda_x \text{ , where } \Lambda_xL=Lx \ \ \forall L\in (\mathbb{R}^N)^* \end{align}
is surjective. That is, I have to show that every $\Lambda \in (\mathbb{R}^N)^{**} $ is a $\Lambda_x$. So, fix $ \Lambda \in (\mathbb{R}^N)^{**} $ .
$ \text{Since } \mathbb{R}^N \text{is isomorphic to } (\mathbb{R}^N)^* \text{there exists a N-dimensional basis of } (\mathbb{R}^N)^*.\\ \text{So, I call this base } B=\{B_i\}_{i=1,..,N}. \text{Set } L \in (\mathbb{R}^N)^*. \text{I have } \Lambda L=\Lambda( \sum_{i=1}^{N} \alpha_iBi)=\sum_{i=1}^N\alpha_{i}\Lambda(B_i). \text{Let's call } b_\Lambda=(\Lambda(B_1),…,\Lambda(B_N)) \text{ and } \alpha_L=(\alpha_1,…,\alpha_n). \text{Since } L= \sum_{i=1}^N\alpha_{i}B_i, \text{ I have} Lx= \sum_{i=1}^N \alpha_{i} B_i(x).\text{I set } b_L(x)=(B_1(x),…,B_N(x)) \text{. Now I should prove that there exists an x s.t. } b_\Lambda=b_L(x). $
I have to say I'm a bit confused here. I don't know if this is the right approach. Besides I don't understand how the fact that I'm considering $ \mathbb{R}^N $ and not any other finite dimensional space should help me in the proof.
Best Answer
Hint Try instead to prove that $$\dim(\mathbb R^N)^*=N$$
Once you prove this, look at the picture: $$ \tau:\mathbb{R}^N \rightarrow (\mathbb{R}^N)^{**}$$ you have $\dim(\tau( \mathbb{R}^N))=N= \dim( (\mathbb{R}^N)^{**})$.
What can you say about a subspace of $\mathbb R^N$ of dimension $N$?
Hint 2: Let $v_1,..,v_N$ be a basis in $\mathbb R^N$. For each $1 \leq j \leq N$ show that there exists some $f_j \in (\mathbb R^N)^*$ such that $$\begin{align} f_j(v_j)=1 & \\ f_j(v_i)=0 & \forall i \neq j \end{align}$$
Prove that $f_1,.., f_N$ is a basis in $\mathbb R^N$.
Hint 3 If $V$ is a finite dimensional vector space, $W$ is a subspace of $V$ and $\dim(W)=\dim(V)$ show that $V=W$.
To do this use the fact that every basis of $W$ can be completed to a basis of $V$.
Now, use this with $\tau(\mathbb R^N) \subseteq (\mathbb{R}^N)^{**}$ to deduce that $$\tau(\mathbb R^N) = (\mathbb{R}^N)^{**}$$ meaning that $\tau$ is surjective, which is exactly what you need to show.