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.

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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 $b_{n_0}\notin \ker b_{n_0}^{\#}$.