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I have to prove that if $\dim(X)=\infty$ then $\dim(X^*)=\infty$. But this question is from the section of excercise of Hahn-Banach extension theorem of my professor. I really don't know how to start this, please if you can give me a couple of tips it would be very helpfull. Thanks.

TeemoJg
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    Just because it appears in that section it doesn't mean that you have to use H-B Theorem. This result is very elementary. Prove it by contradiction. – Kavi Rama Murthy Oct 27 '20 at 23:53
  • @KaviRamaMurthy ... in ZF it cannot be proved that an infinite-dimensional vector space has any linear functional at all. So I do not understand why you say that you do not need to use Hahn-Banach. – GEdgar Oct 28 '20 at 00:37

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Lemma 1: If $F \in X^{*}$ is nonzero and $\mathcal{N}=\{x \in X: F(x)=0\},$ then there exists a one-dimensional subspace $\mathcal{M}$ of $X$ such that $$ X=\mathcal{N}\oplus\mathcal{M}=\{x+y: \quad x \in \mathcal{N} \quad \text { and } \quad y \in \mathcal{M}\} \quad \text { and } \quad \mathcal{N} \cap \mathcal{M}=\{0\}. $$

Lemma 2: Given $x\neq 0$, then there is an $F \in X^*$ such that $$||F||=1 \qquad \text{ and } \qquad F(x)=||x||_X $$

Both lemmas are direct consequences from Hahn-Banach Theorem and can be easily verified.

Now, let's suppose (by contradiction) that $\dim (X^*)=n\leq \infty$, then we can consider a base $\mathcal{B}=\{F_1,\dots,F_n \}$ of $X^*$ and apply the lemma 1 for each $F_i$, this gaves the one dimentional subspaces $\mathcal{M}_i$ and the $\mathcal{N}_i$ with the above mentioned properties. Note that by the definition of the $\mathcal{M}_i$, $X=\mathcal{M}_1\oplus \mathcal{M}_2\oplus \dots \oplus \mathcal{M}_n \oplus X_0$, where $X_0$ is the infinitely dimensional space given by $\bigoplus \mathcal{N}_i/\bigoplus M_i$.

So, consider $x$ an arbitrary element of $X_0$, and $F\in X^*$, note that: $$ F(x)=F_1(x)+F_2(x)+\dots+F_n(x)=0,$$ where the first equality is due to the fact that $F_i$ is a base for $X^*$, and the second come from the fact that $X_0\cap M_i = \{0 \}$, note that is for any linear functional and any $x\in X_0$. In particular given an $x\in X_0$ we can consider the $F$ given by lemma 2, as $F(x)=0=||x||$, we can conclude that $||x||=0$ for all $x\in X_0$, therefore $x=0$ and hence $X_0=\{0\}$ which is a contradiction to the fact that $X_0$ is infinitely dimensional.

With this we have that $\dim X^* = \infty$.

TeemoJg
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