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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.

Sourav Ghosh
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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