Let $(X,\|\;\|)$ a normed vector space over $K$ s.t. $\dim_K(X)=\infty$. I want to prove that:

$$ \dim_K(X^*)=\infty \\ \Big(X^*=\{f:X\to K\;\;|\;\;f\text{ is linear and bounded}\}=\mathcal B(X,K)\big) $$ My attempt goes like this:

Since $\dim_K(X)=\infty$, lets consider $\{e_i\}_{i\in\Bbb N}\subseteq X$ linearly independent ($l.i.$)

Let $m\in\Bbb N$ fixed, then clearly $\{e_1,e_2,.\ldots,e_m\}\subseteq X$ is $\;l.i.$, and let $\;f_1,\ldots,f_m\in X^*$.

Lets define

$$M_k= \langle \{e_j\}_{j\ne k}^m\rangle \;\;\forall k\in\{1,2,\ldots,m\}$$

so, $$\forall k\in\{1,\ldots,m\}\;\;e_k\in X\setminus M_k$$ then, by Hahn-Banach,

$$ \forall k\in\{1,\ldots,m\}\;\exists f_k\in X^*\text{ s.t. } \; f_k \big|_{M_k}=0, \;\; f_k(e_k)>0\;\land\;\;\|f_k\|=1 $$ Now, lets take a linear combination of the cero functional $$ \sum_{i=1}^m\lambda_i f_i = f_0,\text{ where }\;\lambda_i\in K $$ so, $\forall k\in\{1,\ldots,m\}$ we get that $$ 0=f_0(e_k)=\sum_{i=1}^m\lambda_if_i(e_k)=\lambda_kf_k(e_k) $$ and, since $f_k(e_k)>0$, it follows that $\lambda_k=0$. Thus $\{f_1,\ldots,f_m\}$ is $l.i.$ and, since $m\in\Bbb N$ is arbitrary, thus $\{f_1,\ldots,f_m\}$ is $l.i.\;\;\forall m\in\Bbb N$. Thereby $\dim_K(X^*)=\infty$.

What do you think? Did I miss something? Would appreciate your comments or suggestions.


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