Are the coordinate functions of a Hamel basis for an infinite dimensional Banach space discontinuous?
Solution 1:
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.