Discontinuous linear operator e and its core Let $T:E\to \mathbb{R}$ nonzero linear operator with $E$ vector space. So, show the following equivalence: $T$ is discontinuous if and only if $Ker(T)$ is dense in $E$.
 A: I was trying to come up with something elegant, but all I could do was get my hands dirty, so...
First, clearly the only continuous linear functional with a dense kernel is the zero functional. For the other direction, assume there exists some $v\in E$ and a balanced zero-neighborhood $U\subset E$ such that $(v+U)\cap\ker T=\emptyset$.
Note, then, that for all $u\in U$, one has $|Tu|<|Tv|$: if there exists some $u\in U$ with $|Tu|\geq |Tv|$ then $\left|-\frac{Tv}{Tu}\right|\leq 1$ hence $-\frac{Tv}{Tu}u\in U$, but then $v-\frac{Tv}{Tu}u\in\ker T$ in contradiction.
This implies that $T$ is continuous at zero (and due to linearity everywhere): for all $\epsilon>0$ there exists a zero neighborhood $V=\frac{\epsilon}{Tv}U$ such that $u\in V\implies |Tu-T0|<\epsilon$.

N.B.
You didn't indicate if $E$ has the structure of a normed vector space, so I didn't assume that. If you're unfamiliar with topological vector spaces, try to translate the argument to metric-space terminology (which amounts to choosing $U$ appropriately).
