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I don't know whether the answer to this question will help answering this question.

Let $V$ be an infinite-dimensional vector space and $A \subset V$ a Hamel basis in $V$. What is the largest (and smallest) norm one can define on this space such that the given $A$ is normalised.

The norm $\|\cdot\|_N$ is larger than the norm $\|\cdot\|_M$, if we have $\forall x \in V :\|x\|_N \ge \|x\|_M$, or equivalently the unit ball of $\|\cdot\|_M$ contains that of $\|\cdot\|_N$

Attempt:

Since any $x \in V$ has a unique representation in $A$ of the form $x = \sum_{k=1}^n x_k a_k$, where $a_k \in A$, then by the axioms of the norm $$ \|x\| = \left\| \sum_{k=1}^n x_k a_k \right\| \le \sum_{k=1}^n |x_k| \|a_k\| = \sum_{k=1}^n |x_k| $$ I thought about the $p$-norm and there order relation on their unit balls but I don't know if it is possible to define a norm (wrt $A$) as $$ x \mapsto \|x\|_p := \left(\sum_{k=1}^n |x_k|^p\right)^{1/p}. $$ This seems to me a norm since it should act exactly as that in $\mathbb R^n$, or even $\ell^p(\mathbb N)$. We know also that $\|\cdot\|_q < \|\cdot\|_p$, for $1 \le p < q < \infty$.

So does it make sense to say that the $1$-norm is the largest (and the $\infty$-norm is the smallest) norm thus defined on any vector space ?

Physor
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  • These are all indeed norms and your calculation does show that $|x|\le|x|1$ for all $x\in V$. I am a little skeptical about $|\cdot|\infty$ being the least normalized norm, it is definitely dominated by all the $|\cdot|_p$ norms but this doesn't say much – Just dropped in Nov 21 '21 at 21:45
  • So do you mean that the 1-norm is larger than any norm (not only the other $p$-norms) if the have $A$ normalised ? Could you please explain that in a proof ? – Physor Nov 21 '21 at 21:53
  • But you just did: take $x=\sum x_ke_k$, this is a finite sum since ${e_k}$ is a Hamel basis. Thus $|x|=|\sum x_ke_k|\le\sum |x_k||e_k|=\sum|x_k|=|x|_1$, all we used is the triangle inequality – Just dropped in Nov 21 '21 at 21:59
  • It would be nice to see whether $V$ can be a Banach space with this 1-norm. – Lorenzo Pompili Nov 22 '21 at 00:19

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