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.

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    $\begingroup$ I could perhaps point out that clearly a countable infinity of the coordinate functions cannot be continuous. Suppose for basis elements $y_1,y_2,\ldots$ the corresponding coordinate functions $g_1,g_2,\ldots$ are continuous. Consider (invoking completeness) the point $y = \sum_{n=1}^\infty 2^{-n} y_n/\|y_n\|$. By continuity, all of $g_1(y),g_2(y),\ldots$ are nonzero - which is impossible. $\endgroup$ – Mike F Apr 16 '11 at 10:10
  • $\begingroup$ @Mike: I am afraid that your EDIT and comment are wrong. Take for example the sequence space $l^1$ and as $y_i$ the $i$-th unit vector. Clearly the evalution functionals at the $i$-th coordinate are continuous. $\endgroup$ – Vobo Apr 16 '11 at 16:23
  • $\begingroup$ @Vobo: But the $y_i$ are not a basis for $\ell^1$ in the algebraic sense since $(1/2,1/4,1/8,\ldots) \in \ell^1$ is not a finite linear combination of the $y_i$. $\endgroup$ – Mike F Apr 16 '11 at 18:22
  • $\begingroup$ @Mike: Sure, but they are linearly independent, so can be extended to a basis. $\endgroup$ – Vobo Apr 16 '11 at 18:28
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    $\begingroup$ @Vobo: It's more subtle than that. The coordinate function for $y_i$ as an element of the Hamel basis is not the same as the coordinate function for $y_i$ as an element of the Schauder basis. In fact the coordinate functions (call them $h_i$) for the Hamel basis must have the property that for any $x$, only finitely many $h_i(x)$ are nonzero. $\endgroup$ – Robert Israel Apr 29 '11 at 21:32

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:

  1. $\tau$ coincides with the norm topology;
  2. $H$ is finite.


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.

  • $\begingroup$ This doesn't seem to answer my question. I do know that all the coordinate functions cannot be continuous in the infinite-dimensional case (although I did not make this clear at the outset). Please see my edit above. $\endgroup$ – Mike F Apr 16 '11 at 11:01
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    $\begingroup$ @Mike: It does answer the original question and the new one with only little effort: Of course, finitely many coordinate functionals can be continuous. Simply split your Banach space as direct sum $X = F \oplus Y$ with $F$ finite dimensional. Then the dual space of $X$ splits as $F^\ast \oplus Y^\ast$ and then extend a basis of $F$ to a basis of $X$ by choosing a basis of $Y$. By what dissonance says the dual basis elements of $F$ will be continuous. $\endgroup$ – t.b. Apr 16 '11 at 11:26
  • $\begingroup$ (Of course, I assumed $Y$ to be closed in the above comment). $\endgroup$ – t.b. Apr 16 '11 at 11:44
  • $\begingroup$ @Theo: Oh! Thanks I guess that settles things. $\endgroup$ – Mike F Apr 16 '11 at 17:42
  • $\begingroup$ @Mike: You're welcome. But I think you should now reward dissonance for his/her efforts (i.e., accept & upvote :)) $\endgroup$ – t.b. Apr 16 '11 at 18:09

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