linear functional is continuous if and only if it is locally bounded at orgin

Question

Given a topological vector space $X$, a functional $f:X\rightarrow\mathbb{R}$ is continuous if and only if there exists an open neighbourhood $N$ of $0$ s.t. $|f(x)|\leq 1$ for all $x\in N$.

It easy to prove if $f$ is continuous then such $N$ exists. But I do not know how to prove the reverse direction. Any help, please?

Answer 1

$\epsilon N(=\{\epsilon x: x\in N\})$ is also a neighbourhood of $0$ and $|f(y)| \leq \epsilon$ for all $y $ in this neighborhood. So $f$ is continuous at $0$. For continuity at any other point $x$ use the fact that $|f(y)-f(x)|=|f(x-y)|$.