if a point is a root of all linear functionals on a normed space, then it's zero I've been trying to do some work as literal and detailed as possible in order to see I know my analysis for every detail.
I tried to explain my self why if there is $x_0\in X$ ($X$ is a normed space) such that $\forall\varphi\in X^*:\:\varphi(x_0)=0$ then $x_0=0$.
I could justify it using some high theorems, like consequences of the hahn-banach theorem, but I really don't feel like this is necessary and I would love a farily easy explanation if anyone is familiar with. If you feel like all the reasoning for this claim do require these kind of theorem please let me know that.
Thanks a lot
 A: The point is that for every $x\in X-\{0\}$ there is $\phi\in X^\ast$ with $\phi(x)\neq 0$. I do think that one does need the Hahn-Banach theorem (more precisely a corollary, which allows one to extend continuous linear maps), as you need a continuous linear form. There is a certain analog in linear algebra, which does not need the Hahn-Banach theorem, but does need the existence of a basis for every vector space: if $V$ is a vector space over a field $k$ and $x\in V-\{0\}$ then there is $\phi\in V^\ast$ (here this is the ordinary, non-topological dual space): let $B$ be a basis of $V$ containing $x$, let $\phi$ be the linear extension of the map $B\rightarrow k$ taking $x$ to $1$ and all other elements of $B$ to $0$.
I think the result in question is too general to admit an easy proof just using the definitions. You see, your result includes the one from linear algebra as a special case (which is also somewhat nontrivial), and simply using a basis you cannot expect the extension to be continuous.
