Is it true that $\dim(X) \leq \dim(X^{\ast})$ for every infinite dimentional Banach space $X$?

Given an arbitrary infinite dimensional Banach space $X$, can we deduce that it's dimension $\dim(X)$ (the cardinality of one of its Hamel bases) is less or equal of the dimension $\dim(X^{\ast})$ of its dual space (the space of all continuous linear functionals $f:X\to\mathbb{R}$)?

This is an interesting question which has been unaddressed for a long time, so I'll give it a shot.

Lets denote the cardinality of a set $A$ by $|A|$. For a normed space $X$ we define its density character $d(X)$ as the smallest cardinality of its dense sets, that is $d(X)=\min\{|D|: D\subseteq X, \overline{D}=X\}$. In particular a separable normed space $X$ has $d(X)=\aleph_0$.

We need the following three lemmas:

Lemma 1: If $X$ is an infinite dimensional vector space over $\mathbb{R}$ and $\dim X\geq |\mathbb{R}|$, then $\dim X=|X|$.

A proof of Lemma 1 can be found here.

Lemma 2: Let $X$, $Y$ be infinite dimensional Banach spaces with $d(X)\leq d(Y)$. Then $|X|\leq |Y|.$

You can find a proof here.

For the last step, it is known from functional analysis that when $X^*$ is separable, then so is $X$. If you check the proof carefully you'll realise that what is actually proven is the following:

Lemma 3: Let $X$ be a normed space. Then $d(X)\leq d(X^*)$.

Proof. Let $\{x_a^*: a\in A\}$ be a dense subset of $S_{X^*}$ of cardinality $|A|=d(X^*)$. For every $a\in A$ we pick an $x_a\in B_X$ with $x_a^*(x_a)>\tfrac{1}{2}$ and set $Y= \langle \{x_a: a\in A\} \rangle$. The set $Y$ is dense in $X$: Otherwise, there would exist an $f \in X^*$ with $\|f\|=1$ such that $f(y)=0$ for every $y\in Y$. But then for every $a\in A$, \begin{eqnarray*} \|f-x_a^*\|\geq |f(x_a)-x_a^*(x_a)|=\frac{1}{2}, \end{eqnarray*} which implies that $f\notin \overline{\{x_a^*: a\in A\}}=S_{X^*}$, a contradiction. So $X$ contains a dense subset of cardinality $|Y|\leq |A|$, therefore $d(X)\leq |A|=d(X^*)$.

Combining the previous lemmas, we get an affirmative answer to remilt's question.

That is a great answer to a very interesting problem. Given your current solution, I would like to suggest a possible generalization to the case of normed spaces.

Since in the finite dimensional case everything is working well, let $X$ be an infinite dimensional normed space and consider the canonical embedding of $X$ to its double dual $X^{**}$ given by:

$T:X\rightarrow X^{**},\hspace{10pt} T(x)=T_x \hspace{5pt}$ where for $x\in X,$ we define $T_x:X^*\rightarrow\mathbb{R}$ such that $ \hspace{7pt} {T_x}(x^*)=x^*(x)$

Of course, T is an isometric embedding.

Consider the spaces:

$T(X)\hspace{5pt}$ (which is isometric to $X$ and obviously dense in $\overline{T(X)}$) and

$\overline{T(X)}\hspace{5pt}$ (which is a Banach space, as it is a closed subspace of the Banach space $X^{**}$)

(This process is of course standard when considering the completion of a normed space $X$).

Since $\overline{T(X)}$ is a Banach space, by your argument we must have that $\dim(\overline{T(X)})\leq \dim(({\overline{T(X)}})^*)$

But, since $T(X)$ is dense in $\overline{T(X)},$ we must have that the spaces ${(T(X))}^*$ and $({\overline{T(X)}})^*$ are isometric.

(Indeed, consider this result when stated in the following more general fashion:

Let $X$ be a normed space, $Z$ a dense subspace of $X$ and $Y$ a Banach space. Then, the spaces $\mathcal{B}(X,Y)$ and $\mathcal{B}(Z,Y)$ are isometric.

I will be glad to give hints to the proof of this fact, if anyone is interested).

Now combine all the previous arguments together to get:

$\dim(X)= \dim(T(X))\leq \dim({\overline{T(X)}})\leq \dim(({\overline{T(X)}})^*) = \dim(({T(X)})^*)= \dim(X^*)$

(Notice that we have also implicitly used the fact that if two spaces are isometric, then their duals must be isometric as well).