# Exponential norm?

Can a norm "grow exponentially"?

Let $||\cdot||_*: \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$ be a norm such that:

$$\lim_{|x| \rightarrow \infty } \frac{ ||x||_* }{ e^{|x|} } > 0$$

where $|\cdot| \mathbb{R}^n \rightarrow \mathbb{R}_{\geq 0}$ is a Euclidean norm.

Is that possible?

What about so called "Nagumo norms"?

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No, since all the norms in $\mathbb R^n$ are equivalent: let $N$ a norm; we can find $C>0$ such that $\lVert x\rVert_*\leq C\cdot |x|$ so $\frac{\lVert x\rVert_*}{e^{|x|}}\leq C\frac{|x|}{e^{|x|}}$ and $\lim_{|x|\to\infty}\frac{\lVert x\rVert_*}{e^{|x|}}=0$ for all norm $\lVert\cdot\rVert$.

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No, on $\mathbb{R}^n$ all norms are equivalent, and hence, there exists $c,C>0$ such that $c\|x\|_*\leq |x|\leq C\|x\|_*$.

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The other answers are better, since they show your property does not hold for any sequence $x_n$ with $|x_n|\rightarrow\infty$.

But one doesn't need much to show your property can't hold in general: Let $|x|=1$. Then $$\lim\limits_{\alpha\rightarrow\infty}{\Vert \alpha x\Vert_* \over e^{|\alpha x|}} =\lim\limits_{\alpha\rightarrow\infty}{\alpha\Vert x\Vert_* \over e^{ \alpha }}=0 .$$

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