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In $\mathbb{R}^n$ , we can define $p$-norm by $\Vert{x}\Vert=\left ({\sum_{i=1}^{n}{\vert x_i\vert}^p}\right )^{1/p}$, where $ 1\le p <{\infty}$. And we know any Normed Linear space is also a metric space where the metric induced by the norm is $$$$ $d_p(x, y) =\Vert{x-y}\Vert_p$$$$$ But, if $0<p<1$, then I know Minkowski inequality for norm doesn't hold. But, my question is if we define a function for $0<p<1$ in the way that we define for $p-norm$, is it define a metric? How to prove triangle inequality for the metric (if it is!)? If the question is already answered or any other related question that can give some hints please refer.

SoG
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    Here is a solution that holds in a more general setting of integration theory. A solution to your specific problem follows by considering $\Omega={1,\ldots,n}$ and $\mu$ the counting measure. The kay inequality $(a+b)^p\leq a^p+b^p$, $a,b\geq0$ and $0<p<1$ can be obtained be Jensen's inequality for concave functions. – Mittens Nov 01 '21 at 01:09

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If $0<p<1$ you can define a $p$ norm by equality $$||x||_p =\sum_{j=1}^n |x_j |^p . $$ Since $$|u +v|^p \leq ||u|+|v||^p =\frac{|u|+|v|}{||u|+|v||^{1-p}} \leq \frac{|u|}{|u|^{1-p}} +\frac{|v|}{|v|^{1-p}} =|u|^p +|v|^p$$ for $u, v\in \mathbb{R}$ it is easy to show that the function $||\cdot ||_p$ is in the fact $p$-norm.

So we have that $$||x+y||_p \leq ||x||_p + ||y||_p $$ and since the other conditios are obviously satisfied the function $$d_p (x,y) =||x-y||_p $$ defines a metric on $\mathbb{R}^n .$

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    This is not a norm since it is not $1$-homogeneous. It does define a metric however. – Mittens Oct 31 '21 at 22:26
  • You don't need a homogenity of any functional $\varphi $ to define a metric by equation $$d(x,y) =\varphi (x-y).$$ It is enough if: $\varphi (0) =0 \leftrightarrow x=0,$ $\varphi (x) =\varphi (-x) ,$ $\varphi (x+y) \leq \varphi (x) +\varphi (y).$ –  Oct 31 '21 at 22:34
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    I know that, what I am saying is that $d_p$ ( in your notation) is not a norm. It is surely a metric… I did not downvote you by the way, I don’t like engaging in that kind of childish stuff. – Mittens Oct 31 '21 at 23:12