Example of $\alpha$-norm.

Question

We say that $\mathcal N$ is a $\alpha$-norm if $\mathcal N(xà=0\iff x=0$, $\mathcal N$ is sub-linear and $\mathcal N(\lambda x)=\lambda ^\alpha \mathcal N(x)$. Do you have an example of such a norm ? I was wondering for example of $\mathcal N(x)=x^2$ being a $2$-norm, but it doesn't look sub-additive since I can't do better that $$\mathcal N(x+y)=|x+y|^2=2(|x|^2+|y|^2)=2(\mathcal N(x)+\mathcal N(y)). $$

Answer 1

Note that an $\alpha$-norm does not exist for $\alpha>1$ as we find $$2\mathcal{N}(x)<2^{\alpha}\mathcal{N}(x)=\mathcal{N}(2x)\leq2\mathcal{N}(x)$$ which is a contradiction.

Would you consider $\mathcal{N}:\mathbb{R}\rightarrow\mathbb{R}_{\geq0}$ given by $$\mathcal{N}(r)=\sqrt{|r|}$$ to be an example of an $\alpha$-norm? Clearly we have $\mathcal{N}(x)=0$ if and only if $x=0$, we find for all $a,b\in\mathbb{R}$ that $$\sqrt{|a+b|}\leq\sqrt{|a|+|b|}\leq\sqrt{|a|}+\sqrt{|b|}$$ and for all $\lambda,x\in\mathbb{R}$ we have $\mathcal{N}(\lambda x)=|\lambda|^{\frac{1}{2}}\mathcal{N}(x)$.