Summation functional on a Hamel basis

Question

Let $X$ be an infinite-dimensional Banach space. Is it possible to choose a Hamel basis $B$ of $X$ such that the linear functional defined by $f(b)=1$ ($b\in B$) was continuous?

Answer 1

My first thought was surely not. But actually the answer is yes.

Say $f$ is a non-zero bounded linear functional on $X$. Let $N$ be the null space of $f$. Now $X\setminus N$ certainly spans $X$, so there exists a Hamel basis $B\subset X\setminus N$.

So $f(b)\ne0$ for every $b\in B$; now modify $B$, replacing every element by an appropriate scalar multiple and you get $f(b)=1$.