I have been thinking if the following statement is true:
Let $X$ be a normed space. Then $X$ is reflexive if and only if for every set $M\subset X^*$ such that $\bigcap_{\phi\in M} Ker\;\phi=\{0\} $ is necessarily true that $\overline{span(M)}=X^*$. Here $span(M)$ denotes the smallest linear supspace of $X^*$ containing $M$.
Q1: Is this statement true ?
Q2: If answer to Q1 is positive can we then say that for $X$ to be reflexive, there just cannot exist too many functionals in $X^*$ ?
Q3: If answer to Q1 is positive does the statement have any geometrical meaning ?
Thanks for any help.
Q1. Your statement is true. Let's denote by $\tau$ the canonical map of $X$ into $X^{**}$.
Assume that $X$ is reflexive. Let $M \subset X^*$ be a set s.t. $\bigcap\limits_{\phi \in M} \ker \phi = \{0\}$. Then if $0 \ne \alpha \in X^{**}$ then $\alpha$ cannot be zero on $M$ since $\alpha = \tau(x)$ for some $0 \ne x \in X$ and there exists $\phi \in M$ s.t. $\phi(x) = \alpha(\phi) \ne 0$. Therefore if an element $\alpha \in X^{**}$ vanishes on all $M$ then in is zero and therefore $\overline{span M} = X^{*}$ due to Hahn-Banach theorem.
Now assume that $X$ is not reflexive and take arbitrary $\alpha \in X^{**} \setminus \tau(X)$. Let $M = \ker \alpha$. Then $M$ is closed and $M \ne X^{*}$. Assume that there exists $0 \ne x \in X$ s.t. $\phi(x) = 0$ for all $\phi \in M$. Then $M \subset \ker \tau(x)$ and therefore $\alpha = c \tau(x)$ for some constant $c$ (since they have the same kernel). Hence $\bigcap\limits_{\phi \in M} \ker \phi = \{0\}$ and $M = \overline{span M} \ne X^*$.
Q2 and Q3. I don't think it is correct (or important) to treat reflexivity as "there is not too many functionals". It seems to be connected with the amount of functionals; for example there is a statement that if $X$ is reflexive and separable then $X^*$ is also separable, but it is not the point. And also if you consider not only normed spaces you can see that a "small" reflexive space can have a really "big" dual; the example is the space $\mathscr{D}(U)$ of test functions and its dual $\mathscr{D}'(U)$ of distributions. $\mathscr{D}$ is reflexive but the space of distributions is much "bigger" than it.
I think that the reason (and correct interpretation) for you statement is the connection between weak and $*$-weak topology on $X^*$ (if you are familiar with them). Reflexivity for normed spaces is equivalent to coincidence of weak and $*$-weak topology on $X^*$ (since $X^{**}$ is the space of all weakly continuous functionals on $X^*$ and $X$ is the space of only $*$-weakly continuous functionals). Now consider following well-known facts.
Let set $M \subset X^*$ satisfy a condition $\bigcap\limits_{\phi \in M} \ker \phi = \{0\}$. Then closure of $span M$ in $*$-weak topology is $X^*$.
A subspace in arbitrary normed space is weakly closed iff it is closed wrt norm.
Here your statement becomes more clear since for reflexive spaces $X$ and $S \subset X^*$ it is true that closures of $X$ wrt weak and $*$-weak topologies are the same. But also if $S$ is a subspace then also closure wrt norm and weak topology are also the same.
It appears that my interpretation of reflexivity is more topological rather than geometrical.