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Let $(E,N)$ be a normed space, and $M$ be another norm. Can you find a necessary and sufficient condition on $M$ so that $\sqrt{NM}$ is also a norm on $E$?

I hink the condition is that $M$ is proportional to $N$.

Of course, the only thing to investigate is the triangle inequality: we want to prove that if for all $x, y \in E$ there holds $$ \sqrt{N(x+y)M(x+y)} \leq \sqrt{N(x)M(x)} + \sqrt{N(y)M(y)} $$ then there is $\lambda >0$ such that $M=\lambda N$.

Does anoyone know?

I tried to prove a "reverse" triangle inequality without success: $$ N(x+y)M(x+y) \geq (N(x)-N(y))(M(x)-M(y)) = N(x)M(x)+N(y)M(y) - N(x)M(y)-M(x)N(y) $$ wheras $(\sqrt{N(x)M(x)} + \sqrt{N(y)M(y)})^2 = N(x)M(x) + N(y)M(y) +2\sqrt{N(x)M(x)N(y)M(y)}$ resembles it, but I'm unable to finish.

EDIT: I tried (assuming that $\sqrt{NM}$ is a norm) that for all $x, y \in E$ there holds $(\sqrt{N(x)M(y)} - \sqrt{N(y)M(x)})^2 \leq 0$, which then shows that $N$ and $M$ are proportional.

J.Mayol
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  • You can construct other examples using $p$ norms, $M,N$ need not be proportional – LPZ Apr 30 '24 at 12:27
  • @LPZ You could develop your comment into an answer if you'd like, or at least give a bit more details on the counterexample you're thinking of so that someone else may do so. – Bruno B Apr 30 '24 at 14:47
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    If we assume that the vector space $E$ is defined over $\mathbb{R}$. Then since $A(x)=\sqrt{N(x)M(x)}$ satisfies:

    \begin{equation} (i),,,, A(x)=0 \text{ iff }x=0 \end{equation}

    \begin{equation} (ii),,,, A(\lambda x)=|\lambda|A(x) \text{ for any }\lambda \in \mathbb{R} \end{equation}

    it suffices to show that the set $S:={A(x)\leq 1}$ is convex (it is also necessary).

     – Ragno May 06 '24 at 14:36

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