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I have an exercise that states:

a) Prove that for $0<p<1$, $d_1(x,y) =(|x_1-y_1|^p+|x_2-y_2|^p)^{1/p}$ is not a metric on $\mathbb{R}^2$ and

b) Prove that for $0<p<1$, $d_2(x,y) = |x_1-y_1|^p+|x_2-y_2|^p$ is a metric on $\mathbb{R}^2$

Now obviously the only problem is the triangle inequality. For the first question I found a counter example by taking the points $(1,0)$, $(0,1)$ and $(1,1)$. So this is fine.

I am stuck in the second. How do I prove the triangle inequality for b? I tried to approach it by taking a Taylor series approximation but I got lost very quickly. My second idea was to use the Minkowski inequality but I couldn't apply it somehow. Thanks in advance.

EDIT: corrected the equations

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    $\begingroup$ b) does not depend on $p$, type? Do you mean $\left(\cdots\right)^p$? $\endgroup$ – user127.0.0.1 Jan 29 '14 at 2:03
  • $\begingroup$ Sorry for the typo, I corrected it. $\endgroup$ – rom Jan 29 '14 at 4:20
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I am assuming that for $b)$ you mean:

Prove that for $0 < p < 1,\,$ $d_2(x,y) = \left(|x_1 - y_1| + |x_2 - y_2|\right)^p$ is a metric on $\mathbb{R}^2$.

There are two things you need here:

  • $|x - y| \leq |x| + |y|$.
  • $(a + b)^p \leq a^p + b^p$, for $a,b \geq 0$

This first one is the standard triangle inequality that you are most likely familiar with. To prove the second, you can take the fact that the function $f(x) = x^p$ is convex. What properties you need are:

  • $f(0) = 0$
  • $f'(x) \geq 0$
  • $f''(x) \leq 0$

The first and third of these properties will give you a triangle like inequality:

$$f(a + b) \leq f(a) + f(b).$$

What is this true? Take the difference:

$$ g(x) = f(x + a) - f(x) - f(a) $$

Note that $g'(x) = f'(x + a) - f'(x)$. Since $f''(x) \leq 0$ you must have that $f'$ is decreasing and so $g'(x) \leq 0$. Now take a look at $g(0) = f(a) - f(0) - f(a) = f(0) = 0$. So what you get is $g(x) \leq 0$ for all $x \geq 0$. That is:

$$f(a + b) \leq f(a) + f(b).$$

Now to prove your propositon given these two triangle inequalities is not hard anymore:

Show: $d(x,z) \leq d(x,y) + d(y,z)$. First note that since $f'(x) \geq 0$ you have $f(x) \leq f(y)$ for $x \leq y$. We need this for $(*)$ in the inequality chain. Now get:

$$ \begin{align} d(x,z) & = \left(|x_1 - z_1| + |x_2 - z_2|\right)^p \\ & = \left(|x_1 - y_1 + y_1 - z_1| + |x_2 - y_2 + y_2 - z_2|\right)^p \\ & \leq \left(|x_1 - y_1| + |x_2 - y_2| + |y_1 - z_1| + |y_2 - z_2|\right)^p (*)\\ & \leq \left(|x_1 - y_1| + |x_2 - y_2|\right)^p + \left(|y_1 - z_1| + |y_2 - z_2|\right)^p\\ & = d(x,y) + d(y,z) \end{align} $$

And you are done.

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  • $\begingroup$ An easier way to prove your proposition on $(a + b)^p \le a^p + b^p$ is to normalize to $a + b = 1$, then we get that $a^p \ge a$ and $b^p \ge b$ which immediately gives the result. $\endgroup$ – Deven Ware Jan 29 '14 at 3:09
  • $\begingroup$ @DevenWare That's also a great way to go about it, and is in fact how you would go about proving the (harder) Minkowski's inequality. Though, I would say that knowing this technique is helpful for many other triangle inequality problems, especially ones that don't have a homogeneity property. $\endgroup$ – breeden Jan 29 '14 at 3:15
  • $\begingroup$ There was a typo that I corrected. Sorry for that. A question, for $0<p<1$ is $f(s)=x^$ convex? Isn't it concave? But I think that you've actually provided me with the solution. Thanks. $\endgroup$ – rom Jan 29 '14 at 4:25
  • $\begingroup$ @rom, right, sorry. I get to two confused sometimes :) Concave - because it looks more like a cave ;) $\endgroup$ – breeden Jan 29 '14 at 5:04
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I assume that you mean:

Prove that for $p\geq 1$ $$d_2(x,y)=\left(|x_1−y_1|^{p}+|x_2−y_2|^{p}\right)^{1/p}$$ is a metric on $\mathbb{R}^2$.

So, the statement of the triangle inequality then becomes: $$ \left(|x_1−y_1|^{p}+|x_2−y_2|^{p}\right)^{1/p}+\left(|y_1−z_1|^{p}+|y_2−z_2|^{p}\right)^{1/p}\geq \left(|x_1−z_1|^{p}+|x_2−z_2|^{p}\right)^{1/p}.$$

Put $\mathbf{u}=\langle x_1-y_1,x_2-y_2\rangle$ and $\mathbf{v}=\langle y_{1}-z_1,y_2-z_2\rangle$, and try rewriting this expression in terms of $p$-norms: $$ \Vert\mathbf{x}\Vert_{p} = \Vert \langle x_{i} ,\ldots x_{n}\rangle\Vert_{p} = \left(\sum_{i=1}^{n}\vert x_{i}\vert^{p} \right)^{1/p}.$$

This should help from here: http://en.wikipedia.org/wiki/Minkowski_inequality

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  • $\begingroup$ No, I don't mean this and the Minkowski inequality holds for $p\ge1$, it cannot be applied here directly. $\endgroup$ – rom Jan 29 '14 at 4:19

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