Let $X$ be a Banach space and let $\hat{X}$ be a dense subset of $X$. If $p$ is a seminorm on $X$ such that $p(x) =0 $ for all $x \in \hat{X}$, does $p(x) =0$ for all $x\in X$ (is $p$ the trivial seminorm)? In other words, if $p$ is uniformly zero on a dense subset of $X$, is it zero for all $X$?  

For instance, as a special case, are there any seminorms $p$ on $\mathbb{R}$ such that $p(x)=0$ for all $x\in \mathbb{Q}$ but there exists $r$ such that $p(r) \ne 0$?  

If the seminorm is continuous this follows immediately, but this is now always the case.



Seminorm which is zero on dense subset
It depends on whether $\hat{X}$ spans $X$ (in the algebraic sense, i.e. finite linear combinations).


If it does, then for every $x \in X$, we can write $x = a\_1 x\_1 + \dots + a\_n x\_n$ for some $x\_1, \dots, x\_n \in \hat{X}$ and some $a\_1, \dots, a\_n \in \mathbb{R}$. Then if $p$ is any seminorm that vanishes on $\hat{X}$ we have
$$p(x) \le |a\_1| p(x\_1) + \dots + |a\_n| p(x\_n) = 0.$$


If $\hat{X}$ does not span $X$, by Zorn's lemma we can find a (necessarily discontinuous) nontrivial linear functional $f : X \to \mathbb{R}$ that vanishes on $\hat{X}$. Set $p(x) = |f(x)|$.


Note that if $X$ is finite dimensional, only the first case can happen, since all proper subspaces of a finite dimensional normed space are closed. In particular your example with $X=\mathbb{R}$, $\hat{X}=\mathbb{Q}$ is impossible.


