Unbounded linear functional $f(x)=\|x\|$

Question

Assume $f$ is an unbounded linear functional on a real linear normed space $X$. How to prove that there exists $x\in X, x\neq 0$ such that

$$f(x)=\|x\|$$

Answer 1

As $f$ is unbounded, there exists $v\in X$ such that $f(v) > \Vert v \Vert$. Choose $w\in X \setminus \langle v \rangle$ (this is possible as the the space needs to be infinite dimensional to admit an unbounded functional) such that $f(w)<0 < \Vert w \Vert$. Then the function $$g: [0,1] \rightarrow \mathbb{R}, \ g(t) = f(tv + (1-t)w) - \Vert tv + (1-t)w \Vert$$

is continuous (the restriction of $f$ to the finite-dimensional space generated by $v,w$ is continuous) and thus by the intermediate value theorem exists a $t_0\in (0,1)$ such that $g(t_0)=0$. Then $x= t_0v + (1-t_0)w$ does the job.