I'm trying to spot the error in the following argument, but without success. Can you help me out?
Let $Y$ be an infinite dimensional normed space on, say, $\mathbb{R}$, and let $f: Y \rightarrow \mathbb{R}$ be an unbounded linear map, which always exists (by using the existence of an algebraic basis of $Y$ and the axiom of choice).
By the homomorphism theorem we get $\bar{f}: Y/\ker f \rightarrow\mathbb{R}$ linear, bijective and such that $\pi\circ \bar{f}=f$, $\pi$ being the natural projection to the quotient. Therefore, since $f$ is unbounded and $\pi$ is boundend, then $\bar{f}$ is unbounded.
But $\bar{f}$ is a linear bijective map, so $Y/\ker f$ has dimension $1$ and I found a linear unbounded map between two finite dimensional spaces. What is wrong?
Best Answer
As $f$ is unbounded, $\ker(f)$ is not closed. That means that $Y / \ker(f)$ is not a normed space and you can't apply the homomorphism theorem.