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A linear operator $T\colon X\to Y$ between linear normed spaces is bounded if there exists a constant $M$ such that $$\|Tx\| \le M\|x\|\tag{*}$$ holds for every $x\in X$. Are there some situations when it is sufficient to verify that this condition is true for elements from some Hamel basis $B$?

Is there characterization of normed spaces here validity of $(*)$ for some Hamel basis implies boundedness? Are there at least some sufficient conditions? Is situation easier in Banach spaces? Can we say something at least for $Y=\mathbb R, \mathbb C$?

Or are there perhaps no pairs $(X,B)$ such that boundedness on $B$ implies boundedness on $X$?

My feeling about this question is that this might be somehow related to the "shape" of the unit ball. But I guess that convexity of unit ball is probably not enough.

This basically arose from a rather elementary discussion about a linear operator defined on $X=c_{00}$. (The space of sequences which have only finitely many non-zero terms with the sup norm. In this can write down an explicit basis $e^{(i)}=(0,0,\dots,0,1,0,\dots)$ and in the conversation I got an impression that the other person thinks that it suffices to verify $(*)$ for these vectors. I corrected them and said that we need to check this for each $x\in X$ - or at least for all elements of the unit ball - if we follow the definition they've learned. But I realized that the question whether this work at least sometimes might be quite complicated.)

I searched a bit to see whether similar question has been posted here in the past. I only found this question, which deals with orthonormal basis rather than Hamel basis: Linear Operator bounded on a basis.

Martin Sleziak
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2 Answers2

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I think for the case $T:\ell^2\to\mathbb R$ this is not possible.

First, we choose a Hamel basis $\{e_n : n\in \mathbb N\}\cup \{f_i : i\in I\}$, where $e_n$ are the (standard, orthogonal) unit vectors.

Now we can define $T$ via $$ T e_n = 1 \\ T f_i = 0 $$

and accordingly on linear combinations of these.

Now $\| Tx \| \leq \|x\|$ holds for this basis. On the other hand, it can be seen that $T$ is unbounded: Indeed, for $x_n := e_1+\dots+e_n$ we have $\| x_n \|=\sqrt{n}$, but $T x_n = n$, so $\frac1{\| x_n \|} Tx_n = \sqrt{n} \to \infty$.

supinf
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    Why is it true that $|x_n| = \sqrt n$? – sas Nov 24 '21 at 14:42
  • Because $e_n$ are orthogonal unit vectors. If you calculate the $2$-norm, you get $|x_n|= |(1,1,\dots,1)| = \sqrt{1+1+\cdots+1} = \sqrt{n}$. – supinf Nov 24 '21 at 15:51
  • @sas I added the word "orthogonal" in the answer, to clarify. – supinf Nov 24 '21 at 15:57
  • But Hamel basis cannot be orthogonal in infinite dimensions. Nvm, got it, only countable part here is orthonormal. – sas Nov 24 '21 at 17:25
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To extend supinf's result, in Banach spaces this is impossible: It is known that for every Hamel basis of them $B=\{b_i: i\in I\}$, at most finitely many of the coordinate functionals are bounded. On the other hand each coordinate functional $b_i^{\#}$ is bounded on $B$, as for $i\neq j$, $b_i^{\#}(b_j)=0 \leq \|b_j\|$ and $b_i^{\#}(b_i)=1 \leq \frac{1}{\|b_i\|}\|b_i\|$, so each $b_i^{\#}$ is bounded on the base with a constant $M_i=\frac{1}{\|b_i\|}$.

tree detective
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