Let $A : E \rightarrow E$ be a linear operator, and $E$ a (typically infinite-dimensional) Banach space. Suppose that $S$ is a collection of norm 1 elements of $E$ spanning a dense subspace of $E$.
- Does $||Ax|| \leq C ||x||$ for all $x \in S$ imply that $A$ is bounded ?
2.(EDITED) Does $||x|| \leq c ||Ax||$ for all $x \in S$ and $A$ bounded, injective imply that $A$ is bounded below ?
Clearly this would be true if $S$ were dense itself, but since it only spans a dense subspace I'm not so sure.
Reason for the edit : Without those hypotheses, the answer is clearly no to point 2 : as in David's answer, take a Hamel basis $e_n$ and let $Ae_n=e_n$ and let $Ax=0$ for any $x \notin span(e_n)$.
The answer is "no" to your first question:
In $\ell_2$, extend the unit vectors, $(e_i)_{i=1}^\infty$, to a Hamel basis $(f_\alpha)_{\alpha\in I}$ of $\ell_2$. For $i\in\Bbb N$, define $Ae_i=e_i$. Define $A$ on the other elements of $(f_\alpha)_{\alpha\in I}$ so that it is unbounded (for instance, take $(h_i)_{i=1}^\infty$ a sequence from $(f_\alpha)_{\alpha\in I}$ disjoint from $(e_i)_{i=1}^\infty$ and map $h_i$ to $i\cdot h_i $). Note $I$ must be uncountable; hence this can be done.
The answer to your second question is "no" as well, as Daniel Fischer's comment in the original post shows.