Let $V$ be a normed vector space over $\mathbb R$. For a normalized Hamel basis $\mathcal B$ of $V$, consider the linear functional $f_{\mathcal B}:V\to\mathbb R$ taking constant value $1$ on $\mathcal B$; explicitly, $f_{\mathcal B}\bigl(\sum_{v\in\mathcal B}c_vv\bigr)=\sum_{v\in\mathcal B}c_v\,.$
Is there some characterization of the spaces $V$ such that $f_{\mathcal B}$ is bounded for some $\mathcal B$? what if $V$ is also assumed to be a Banach space?.
This is true for all $V$. We may assume $\dim V>1$; then by Hahn-Banach there exists a bounded functional $f:V\to\mathbb{R}$ of norm greater than $1$. Let $H=f^{-1}(\{1\})$ and let $S$ be the unit sphere of $V$. It suffices to show that $S\cap H$ spans $V$, since then a maximal linearly independent subset of $S\cap H$ will be a normalized Hamel basis $\mathcal{B}$ with $f_{\mathcal{B}}=f$.
First, since $f$ has norm greater than $1$, there is some unit vector $u\in S$ such that $|f(u)|>1$. Let $z=u/f(u)$, so $z\in H$ and $\|z\|<1$. Now let $v\in H$ be distinct from $z$ and observe that the entire line $L=\{tv+(1-t)z:t\in\mathbb{R}\}$ is contained in $H$. As $t\to\pm\infty$, $\|tv+(1-t)z\|\to\infty$ since $v\neq z$. But also, when $t=0$, $\|tv+(1-t)z\|=\|z\|<1$. Thus by the intermediate value theorem, there are at least two points on $L$ of norm $1$, one for a negative value of $t$ and one for a positive value of $t$. The span of those two points then contains $L$, and in particular contains $v$. That is, $v$ is in the span of $S\cap H$.
We have thus shown that every element of $H$ except for $z$ is in the span of $S\cap H$. Since $\dim V>1$, $H$ contains more than one point and in particular contains an entire line through $z$. But then $z$ can be written as a linear combination of two other points on that line, so $z$ is also in the span of $S\cap H$. Thus the span of $S\cap H$ contains all of $H$.
Finally, suppose $v\in V$ is arbitrary. If $f(v)\neq 0$ then $v$ is a scalar multiple of an element of $H$ and thus is in the spane of $S\cap H$. If $f(v)=0$ then we can fix an element $w\in H$ and then $w$ and $w+v$ are both in $H$, and thus $v=(w+v)-v$ is in the span of $S\cap H$. Thus $S\cap H$ spans all of $V$.