Are the coordinate functions of a Hamel basis for an infinite dimensional Banach space discontinuous? The question is in the title really, but I suppose I could at least fix some notation here.
Let $X$ be an infinite dimensional Banach space - over the reals for the sake of concreteness. Use choice to produce a Hamel basis $(x_i)_{i \in I}$ for $X$. I have the impression that this basis should necessarily interact badly with the topology and, therefore, not be of much (any?) use for doing analysis on $X$, but I've never really thought about why this should be the case.
Is there a rigorous sense in which we can say that this basis is "useless"?
One way to make this precise (although I'd be interested in other points of view) might be to consider the corresponding linear functionals $(f_i)_{i \in I}$ - uniquely determined by taking $f_i(x_j)$ equal to $1$ or $0$ according as $i=j$ or not. I would be very surprised if it were possible for any of these functionals to be continuous, but this conviction is based only on my vague notion that a Hamel basis for $X$ must be somehow "pathological" and not on any sound reasoning.
Because I have no idea how to approach this question (admittedly also because I don't think this is a particularly constructive thing to be dwelling on at this time of year...) I thought I would appeal to the hard-earned wisdom of the good people of Mathematics Stack Exchange.
Edit for clarity: I actually already know at most finitely many of the coordinate functions can be continuous (see comment below) which is, now that I think about it, a fairly critical failure in and of itself and probably enough to justify the word "useless" above. My question is whether they must all be discontinuous.
 A: Your impression is right. Hamel bases and topology interact badly in infinite-dimensional spaces. An easy example would be the space of real polynomial functions in one variable over the interval $[0, 1]$ equipped with norm 
$$\lVert p \rVert = \int_0^1\lvert p(x)\rvert\, dx.$$
Then clearly $\{1, x, x^2, \ldots\}$ is a Hamel basis. We then have a family of linear functionals
$$P_j(a_0+a_1x+\ldots+a_nx^n)=\begin{cases} a_j & j \le n \\ 0 & \text{otherwise}\end{cases}$$
which we may call projectors but only in the algebraic sense. In fact they are not continuous. 
Take $(x-1)^n$. Then
$$\int_0^1 \left\lvert (x-1)^n \right\rvert\, dx = \frac{(-1)^{n+1}}{n+1}\to 0,$$
but $P_0\big( (x-1)^n \big)=(-1)^n$, that is, $P_0$ maps a convergent sequence into a non-regular one. We can find similar examples for the other $P_j$. 

Of course the previous space was not complete. This is in the nature of things: a consequence of Baire cathegory theorem is that a Hamel basis in a Banach space is necessarily uncountable and thus it needs be an object much more complicated. There is something we can say in this case too, though. In fact, I recall having seen this as an exercise somewhere: 
Exercise Let $(V, \lVert \cdot \rVert)$ be a normed space and $H\subset V$ an algebraic basis. Then for all $x \in V$ we have 
$$x=P_{h_1}(x)h_1 + \ldots P_{h_k}(x)h_k$$
for uniquely determined $h_1 \ldots h_k \in H$. The mappings $P_h$ thus defined are linear. Call $\tau$ the coarsest topology on $V$ s.t. those mappings are all continuous. Then the following are equivalent: 


*

*$\tau$ coincides with the norm topology; 

*$H$ is finite. 


EDIT
Here's a proof for $1 \Rightarrow 2$. Suppose $H$ is not finite. For each $h_1 \ldots h_k \in H$ and $\varepsilon >0$, let 
$$B(h_1 \ldots h_k, \varepsilon) = \{x \in V \mid \lvert P_{h_j}(x) \rvert < \varepsilon,\ j=1 \ldots k \}.$$
Those sets form a fundamental system of neighborhoods of the origin for the topology $\tau$ and so for the norm topology too. Fix $h_1 \ldots h_k$ and $\varepsilon$. Since $H$ is infinite we can find $h \in H, h \ne h_1 \ldots h_k$. Call $r_h=\{\lambda h \mid \lambda \in \mathbb{K}\}$. This set clearly is unbounded. We have $r_h \subset B(h_1 \ldots h_k, \varepsilon)$ and so $B(h_1 \ldots h_k, \varepsilon)$ is unbounded too.
We have thus shown that every $\lVert \cdot \rVert$-neighborhood of the origin of $V$ needs be unbounded. This is a contradiction.
