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I've been thinking about the following question and have convinced myself it works only for $p \in (0,1]$.

"Find all values of $p \in \mathbb{R}$ for which $|x-y|^p$ defines a metric on $\mathbb{R}$."

To handle the case when $p$ is negative, it's enough to notice that the "metric" will violate the first law of a metric "$|x-y| ^p =0$ if and only if $y=x$". Every value of $p$ satisfies symmetry and the values of $p$ I mentioned above satisfy all three conditions. I'm convinced that the final values for $p$, those larger than $1$, will violate the triangle inequality, however, I have no way to verify this if, say, $p$ is irrational.

Do anyone have any suggestions or hints that might help me. Anything is appreciated! :)

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    did you check $d(x,z) \le d(x,y)+d(y,z)$ ? – 2017-02-12
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    I did for positive integers using the binomial theorem, but I'm not sure how to deal with say $p=\pi.$ – 2017-02-12

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The fact that $|x|^p$ is convex for $p>1$ suffices to show that the "metric" $|x-y|^p$ for $p>1$ violates the triangular inequality and hence is not a metric.

There's something called the $p$-norm (https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm, How to prove triangle inequality for $p$-norm?) which is a metric for $p>1$.

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    What has convexity to do with being not a metric? – 2017-02-12
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    @user251257: Write out the two inequalities for $|(x-y)+(y-z)|^p$, one from the convexity condition, one from the triangle inequality. They exactly contradict each other. – 2017-02-13
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    @user251257, Lee Mosher's comment answers your question. – 2017-02-13
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    @LeeMosher what is the inequality from the convexity condition involving 3 points not necessarily on a line? It is not that I doubt you. But, I have absolutely no idea what you mean. – 2017-02-13
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    Consider this counter example for the triangular inequality: Triangular inequality requires $|(x-y)+(y-z)|^p \leq |(x-y)|^p + |(y-z)|^p$. Consider $x=3,y=2$ and $z=0$, thus we require $3^p - 2^p \leq 1$. But since $|x|^p$ is convex for $p>1$, we have that $3^p - 2^p \geq 1$ (use this http://math.stackexchange.com/questions/1140679/how-to-prove-that-convex-function-has-an-increasing-slope). Thus arriving at a contradiction. – 2017-02-13
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    The question is about "a metric on $\mathbb{R}$", so $x,y,z \in \mathbb{R}$ and so the worry about "3 points not necessarily on a line" does not apply in the context of this question. – 2017-02-13