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Let $d$ be a metric on any non empty set $X$ then prove or give counter example of the following:

$d_1(x,y) = d^n(x,y)$ where $n \in \Bbb R$ is a metric.

My Attempt: I know that $d_1$ is a metric for $n \in \Bbb N$ and $n = \frac {1}{2}$ I think that $d_1(x,y) = d^n$ is a metric for $n \geq 0$ and it is not a metric for $n < 0$ it is because for negative value of $n$ , $d$ will be of the form $d_1(x,y) = \frac{1}{d(x,y)}$ and $d$ should be non zero.

Sangam Academy
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For $n<0$ it is not well defined.

For $n = 0$, it is clearly not a metric and for $n = 1$ it clearly is.

If $n \in (0, 1)$, it is indeed a metric. Symmetry and "$d(x, x) =0 \iff x=0$" are obvious. We know that in this case we have for $a, b \in (0, \infty)$: $$ (a+b)^n \leq a^n+b^n. $$ To prove this, observe that $g:(0, \infty)^2 \rightarrow \mathbb{R}$, $g(x, y) := (x+y)^n - x^n - y^n$ is decreasing in $x$ and $y$, i.e. $\partial_x g \leq 0$ and $\partial_y g \leq 0$.

The above gives us the triangle inequality if we do the following ($x, y, z \in X$): $$ d(x, z)^n \leq (d(x, y) + d(y, z))^n \leq d(x, y)^n + d(y, z)^n $$



If $n>1$ we simply can not be sure. Let $X := \mathbb{R}$ and $d:X \rightarrow [0, \infty)$ be the discrete metric $$ d(x, y) := \begin{cases} 1 &, \text{ if } x \neq y \\ 0 &, \text{ else}. \end{cases} $$ Then $d^n = d$ is still a metric. But if we choose $d(x, y) := \lvert x - y \rvert$, then $d^n$ is not a metric. Note that $$ d^n(x,z) > d^n(x, y) + d^n(y, z) $$ for $x=\frac{5}{4}$, $z = 0$ and $y=\frac{1}{4}$. So the triangle inequality is violated.

Hyperbolic PDE friend
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  • Here I'm confusing with the case that you have taken. I think that you have taken "Case (i) for $n < 0$, it is not well defined. Case (ii) for $n = 0$, $d$ is not a metric. Case (iii) for $n \in (0,1)$ $d$ is a metric. Case (iv) for $n = 1$ $d$ is a metric. Case(v) for $n \in (1,\infty)$ $d$ is not a metric." I think, last cases overlapped. Please make cases as I said. – Sangam Academy Jul 06 '21 at 02:27
  • Case (v) is in general not a metric. But it can be, that is what I said. You just can not a priori assume that it is. There are instances of $d$ where it is and where it is not. – Hyperbolic PDE friend Jul 06 '21 at 08:16