I am reading Hahn-Banach theorem section from one functional analysis book. They have given corollary which says "Let $X$ be normed linear space and $\{x_1,…,x_n\}$ be linearly independent in $X$. Then there exist $\{f_1,…,f_n\}$ in $X^*$ such that $f_i(x_j)=\delta_{ij}$". The proof follows nicely from one of the version of Hahn-Banach theorem. Now immediately after this corollary, they have given exercise that "For every normed linear space $X$, $\dim X\leq \dim X^*$". Now I think somehow I have to use the above corollary to prove this. I know equality holds for finite dimensional case but suppose I assume $X$ infinite dimensional, I am not getting how to apply or use the corollary (or is there some different logic to be used?). May be I am missing some silly logic. Thanks.
For every normed linear space $X$, $\dim X\le \dim X^*$
functional-analysis
Best Answer
Assume that $\dim X\ > \dim X^*$. Since $ \dim X = \infty$, we have that $n:= \dim X^* < \infty.$
Now let $\{x_1,....,x_n, x_{n+1} \}$ be linearly independent in $X. $ By the corollary there is $\{f_1,...,f_{n+1}\}$ in $X^*$ such that $f_i(x_j)=\delta_{ij}.$ This last property gives that $\{f_1,...,f_{n+1}\}$ is linearly independent , hence $\dim X^* \ge n+1,$ a contradiction.