Existence of at least one continuous coordinate functional

Question

Motivation: I'm trying to prove the following

Suppose $X,Y$ are normed spaces and $X$ is nontrivial, then if $L(X,Y)$, the space of bounded linear operators endowed with the usual operator norm, is a Banach space, so is $Y$.

I suppose there must be some canonical proofs for this result, but before looking at them, I want to make one myself. And here it goes:

Given a Hamel basis $\{x_i\}$ of $X$, find one $x_i$ such that the corresponding coordinate functional $f$ is continuous (if existing). Then for any Cauchy sequence $\{y_n\}\subset Y$, define $g_n:=y_nf\in L(X,Y)$. Then, it is not hard to show that $\{g_n\}$ is Cauchy in $L(X,Y)$, and as a result $y_n\to (\lim g_n)(x_i)$.

So the only trouble is the existence of such a $f$. Of course, I know that I can never expect all of them to be continuous, but does there exist at least a continuous one, for each $X$ and each its Hamel basis?

EDIT As commented by @Marko, far less is actually required to prove the original theorem: any bounded functional $f$ which attains $1$ at some $x\ne 0$ is ok (existence by Hahn Banach Extension Theorem). However, I'm still curious to know about the existence of continuous coordinate functionals on a general Hamel basis.

Answer 1

That's a good question and I had never thought about it. This is what I managed to show:

If you are allowed to change the basis, there is a very straightforward positive answer:

Proposition 1: In every infinite dimensional normed space $X$ there exists a basis such that at least one coordinate functional is bounded.

Proof: Pick a nonzero $x_0\in X$ and an $f_0\in X^*$ such that $f(x_0)=1$. The kernel of $f_0$ has codimension $1$, so there exists a Hamel basis $B=\{x_i:i\in I\}\cup\{x_0\}$ of $X$ such that $x_i\in \ker f_0$, for every $i\in I$. This implies that that $f_0$ is the coordinate functional of $x_0$ and it is also bounded.

But you want to check whether this property hold for every basis. I believe the answer is negative in this case. Here is my attempt:

Lemma 1: Let $X$ be an infinite dimensional normed space. Then there exists a Hamel basis which is dense in $X$.

You can find a proof of Lemma 1, here. We need one more Lemma which is based on something you wrote at a previous question of yours (it was a clever idea indeed!). I omit the proof because I think it is relatively easy:

Lemma 2: Let $X$ be an infinite dimensional normed space, $\{b_i : i\in I\}$ a Hamel basis of $X$ and $\{b_i^{\#}: i\in I\}$ the coordinate functionals. Pick an $i_0\in I$. The following are equivalent:

1. $b_{i_0}\notin \overline{\langle b_i: i\in I\setminus\{i_0\}\rangle}$,

2. the functional $b_{i_0}^{\#}$ is bounded.

With these Lemmas we can show that

Proposition 2: In every infinite dimensional normed space there exists a basis such that none of its coordinate functionals are bounded.

Proof: By Lemma 1, there exists a Hamel basis $\{b_i: i\in I\}$ which is dense in $X$. Since $X$ has no isolated points, for every $i_0\in I$ the set $\{b_i: i\in I\setminus\{i_0\}\}$ is still dense in $X$. In particular, $b_{i_0}\in \overline{\langle b_i: i\in I\setminus\{i_0\}\rangle}$, for every $i_0\in I$ and by Lemma 2 this means that $b_{i_0}^{\#}$ is discontinuous for every $i_0\in I$.

Of course in a finite dimensional space you can't find any dense Hamel bases, so the previous argument doesn't work. After all, in finite dimensional spaces all of the coordinate functionals are bounded.