Connection between linear mappings and continous mappings i'm looking for a non-continuous linear mapping f:$V \to \mathbb{R}$ with $V$ a normed linear space.
I found that I needed to use a infinite-dimensional V because if V is finite-dimensional then f would automaticaly be continuous.
I maybe thought to this one: 1 if $x \le 0$ and 0 if $x>0$ but i'm not sure if i used that V is normed linear space.
As followed i wanted to prove that if f is continuous in zero that f is continuous overall. The implication of "f is continuous" from "f is continuous in zero" is trivial. But I'm not sure if i can prove the reverse implication because i don't understand the problem with the linear mapping, so i need some help there.
I also thought that if f is uniformly continuous that f is normal continuous and so on but the other implication needs a counter example. I'm just a little stuck that i need to work with linear mappings because i know the implication for a normal function. So maybe someone can help me understanding the connection between continuous and linear mappings?
Thanks a lot.
 A: Example: Let $V=\bigoplus^\infty\mathbb{R}=c_{00}(\mathbb{R})$ be the space of sequences of reals with only finitely many nonzero terms (i.e., eventually constant 0 sequences), equipped with, e.g. the norm
$$
\lVert (x_n)\rVert = \sqrt{\sum_{n=1}^\infty x_n^2}
$$
and let $f\colon V\to\mathbb{R}$ be
$$
f((x_n))=\sum_{n=1}^\infty nx_n.
$$
Then $f$ is linear, but discontinuous since $f((0,0,\dots,0,\epsilon,0,0,\dots))=n\epsilon\to\infty$ as $n\to\infty$ but $\lVert(0,0,\dots,0,\epsilon,0,0,\dots)\rVert=\epsilon$.
Note: A linear map $f$ is uniformly continuous iff continuous iff continuous at $0$.  That is just $f(x)-f(y)=f(x-y)=f(x-y)-f(0)$ so if you can control $f$ on the ball of radius $r$ near $0$, then you have the same control on how much $f$ can vary on the ball of radius $r$ near any other points.  This proves the only nontrivial implication "continuous at $0$ $\Rightarrow$ uniformly continuous" (the other two: "uniformly continuous $\Rightarrow$ continuous" and "continuous $\Rightarrow$ continuous at $0$" are trivial).
