# At most finitely many (Hamel) coordinate functionals are continuous - different proof

If $X$ is a vector space over $\mathbb R$ and $B=\{x_i; i\in I\}$ is a Hamel basis for $X$, then for each $i\in I$ we have a linear functional $a_i(x)$ which assigns to $x$ the $i$-th coordinate, i.e., the functions $a_i$ are uniquely determined by the conditions that $$x=\sum_{i\in I} a_i(x)x_i,$$ where only finitely many summands are non-zero.

If $X$ is a Banach space, then at most finitely many of them can be continuous.

I have learned the following argument from comments in this question.

Suppose that $\{b_i; i\in\mathbb N\}$ is an infinite subset of $B$ such that each $f_{b_i}$ is continuous. W.l.o.g. we may assume that $\lVert{b_i}\rVert=1$.

Let $$y:=\sum_{i=1}^\infty \frac1{2^i}b_i.$$ (Since $X$ is complete, the above sum converges.)

We also denote $y_n:=\sum_{i=1}^n \frac1{2^i}b_i$. Since $y_n$ converges to $y$, we have $f_{b_k}(y)=\lim\limits_{n\to\infty} f_{b_k}(y_n)=\frac1{2^k}$ for each $k\in\mathbb N$. Thus the point $x$ has infinitely many non-zero coordinates, which contradicts the definition of Hamel basis.

I have stumbled upon Exercise 4.3 in the book Christopher Heil: A Basis Theory Primer. Springer, New York, 2011. In this exercise we are working in an infinite-dimensional space $X$. Basically the same notation as I mentioned above is introduced, $a_i$'s are called coefficient functionals and then it goes as follows:

(a) Show by example that it is possible for some particular functional $a_i$ to be continuous.

(b) Show that $a_i(x_j) = \delta_{ij}$ for $i,j\in I$.

(c) Let $J = \{i\in I : a_i \text{ is continuous}\}$. Show that $\sup \{j\in J; \lVert a_j \rVert<+\infty\}$

(d) Show that at most finitely many functionals $a_i$ can be continuous, i.e., $J$ is finite.

(e) Give an example of an infinite-dimensional normed linear space that has a Hamel basis $\{x_i; i\in I\}$ such that each of the associated coefficient functionals $a_i$ for $i\in I$ is continuous.

The part (c) can be shown easily using Banach-Steinhaus theorem (a.k.a. Uniform boundedness principle). But I guess that the author of the book has in mind a different proof for part (d) from what I sketched above, since (c) is an easy consequence of (d) -- so he would probably not be put the exercises in this order. (But maybe I was just trying to read to much between the lines.)

Question: I was not able to find a proof od (d) which uses (c). Do you have some idea how to do this?

NOTE: My question is not about the parts (a), (b), (e). I've included them just for the sake of the completeness, in order to include sufficient context for the question.

How about this: If you have a Hamel basis $\{x_i\}$, and replace each $x_i$ by a nonzero scalar multiple of itself, then the result is still a Hamel basis. The corresponding functionals $a_i$ are of course replaced by nonzero scalar multiples of themselves (the multiplier for $a_i$ is the reciprocal of the multiplier for $x_i$). The new functional is continuous iff the original was continuous. If $J$ is infinite, then you can carry out this "replacement" by scalar multiples in such a way that the $\sup$ in (c) is infinite.

Speaking of different proofs, there is also a simple argument using Baire's Category Theorem.

Let $\{b_i:i\in I\}$ be a Hamel basis of $X$ and suppose that $(b_n^{\#})_{n\in\mathbb{N}}$ is a sequence of bounded coordinate functionals. Then $\bigcup_{n=1}^\infty \ker b_n^{\#}=X$.

This is easy to check since for each $x\in X$, $x=\sum_{i\in F}\lambda_ib_i$ for some finite $F\subseteq I$, so there exists an $n_0\in \mathbb{N}$ such that $b_{n_0}$ does not appear in this expression. Then $x\in \ker b_{n_0}^{\#}$.

Since $b_n^{\#}$'s are bounded, their kernels are closed, so by Baire's Theorem there exists an $n_0\in\mathbb{N}$ such that $\ker b_{n_0}^{\#}$ has non empty interior. This implies that $\ker b_{n_0}^{\#}=X$, which is a contradiction, since $b_{n_0}\notin \ker b_{n_0}^{\#}$.

• @JonMarkPerry My guess is that the OP indeed intended that to be a spoiler. Sep 17 '16 at 11:59
• rolled it back @MartinSleziak
– JMP
Sep 17 '16 at 12:35