Okay I have been working under the assumption that this is "obvious" for a while now, but it started to bug me and now I'm fumbling to prove it.

Suppose $X$ is a normed linear space (possibly infinite dimensional). Let $\mathcal{B}$ be a Hamel basis of $X$. Fix $b \in \mathcal{B}$. Is the coordinate projection $P_b : X \to \mathbb{C}$ or $\mathbb{R}$ defined by $P_b(b) = 1$ and $P_b(x)=0$ for $x \notin \mathrm{span}(\{b\})$ continuous?


1 Answer 1


No. With notation as in the problem, suppose $X$ is infinite-dimensional and complete, and let $b_1, b_2, ...$ be a countable subset of $B$. WLOG $\| b_i \| = 1$. Then the series

$$\sum_i \frac{b_i}{2^i}$$

converges to some element of $X$ which, by assumption, can be expressed as a finite sum $\sum_j c_j b_j$, where $b_j \in B$ and the sum ranges over some finite index set $J$. It follows that

$$\sum_{i=1}^n \frac{b_i}{2^i} - \sum_j c_j b_j \to 0$$

as $n \to \infty$. From here it follows that all but finitely many (at most $|J|$) of the projections $P_{b_i}$ fail to be continuous because they do not preserve the above limit.

You should at least require that $B$ is a Schauder basis, but even then I think the conclusion fails.

  • $\begingroup$ Start of an alternate argument: if this were true, then I think $X$ would be homeomorphic to $\mathbb{R}^B$ with the product topology, but I think if $B$ is uncountable then the latter fails to be metrizable. $\endgroup$ Apr 16, 2013 at 3:27

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