I have to prove the triangle inequality

$(|x_1 - z_1|^p + |x_2 - z_2|^p)^{1/p} \leq (|x_1 - y_1|^p + |x_2 - y_2|^p)^{1/p} + (|y_1 - z_1|^p + |y_2 - z_2|^p)^{1/p}$ for $p \geq 1$ on $\mathbb{R}^2$.

Two things are confusing me:

1) How we manipulate the expression, seeing as we can't use the standard triangle inequality $|x - z| \leq |x - y| + |y - z|$. I am used to using a $|x - z| = |x - z + y - y | < |x - y| + |y - z|$ method, however is this still valid given the powers of $p$?

2) Why this is an invalid statement for $0 < p < 1$

Sorry if these questions are a little basic, I'm just looking for a little guidance not a solution. Any help would be appreciated.

Thanks, Ash.

  • 2
    $\begingroup$ For 2), consider $(x_1,x_2)=(1,0)$, $(z_1,z_2)=(0,1)$, and $(y_1,y_2)=(0,0)$. $\endgroup$ – Jonas Meyer Jan 15 '11 at 23:09

This is where Minkowski's Inequality comes in handy. See this (a question about Minkowski's inequality which uses Hölder's inequality in the proof).

  • $\begingroup$ If I am understanding correctly, using Minkowski's Inequality we can set $f=x-y, g=y-z$ and recover $||x-z||_p \leq ||x-y||_p + ||y-z||_p$, proving the statement? $\endgroup$ – Ash Jan 15 '11 at 23:37
  • 1
    $\begingroup$ @Ash: Essentially, yes. $\endgroup$ – Arturo Magidin Jan 16 '11 at 3:22

(1) It is not in general true that $|x+y|^p \leq |x|^p + |y|^p$ when $p\gt 1$: for example, take $x=y=1$, $p=2$. Then $|x+y|^p = |1+1|^2 = 4$, whereas $|x|^p + |y|^p = 1^2 + 1^2 = 2$. So, no, you cannot use such a "fact" for the proof, because it's not true. You really do need to prove the inequality that is given to you.

If $x_1=z_1$ and $x_2=z_2$, there is nothing to do, so we may assume that the left hand side is nonzero.

What you want to show is essentially Mikowski's inequality, much along the lines of the answer Trevor has linked to.

As to why there is such an essential difference between the case $p\geq 1$ and the case $0\lt p\lt 1$, the reason is that while the function $y=x^p$ is convex when $p\gt 1$ (and technically also for $x=1$) it is not so for $0\lt p\lt 1$. This is what hides behind the problems for $0\lt p\lt 1$.


In our class, for $p > 1$, we used Hölder's inequality $(\ast)$ $$ \sum_{k = 1}^{n} | a_k b_k| \le \left( \sum_{k = 1}^{n} | a_k |^p \right)^{\frac{1}{p}} \left( \sum_{k = 1}^{n} | b_k |^p \right)^{\frac{1}{p}}. $$ Let $q := \frac{p}{p - 1} > 1$ and $a_i, b_i \in \mathbb{R}$ for all $i \in \{1, \ldots, n\}$. Then we have $\frac{1}{p} + \frac{1}{q}$. $p$ and $q$ are said to be Hölder conjugates. \begin{align*} \sum_{i=1}^n|a_{i}+b_{i}|^p & =\sum_{i=1}^n|a_{i}+b_{i}|\,|a_{i}+b_{i}|^{p-1} \\& \le \sum_{i=1}^n|a_{i}|\,|a_{i}+b_{i}|^{p-1} +\sum_{i=1}^n|b_{i}|\,|a_{i}+b_{i}|^{p-1} \\ & \overset{(*)}{\le} \left(\sum_{i=1}^n|a_{i}|^p\right)^{\frac 1p} \left(\sum_{i=1}^n|a_{i}+b_{i}|^{q(p-1)}\right)^{\frac 1q} + \left(\sum_{i=1}^n|b_{i}|^p\right)^{\frac 1p} \left(\sum_{i=1}^n|a_{i}+b_{i}|^{q(p-1)}\right)^{\frac 1q} \\& =\left( \left(\sum_{i=1}^n|a_{i}|^p\right)^{\frac 1p}+ \left(\sum_{i=1}^n|b_{i}|^p\right)^{\frac 1p}\right) \left(\sum_{i=1}^n|a_{i}+b_{i}|^{p}\right)^{\frac 1q}, \end{align*} because we have $(p - 1)q = p$. Dividing by $\left(\sum_{i=1}^n|a_{i}+b_{i}|^{p}\right)^{\frac 1q}$ and using $1-\frac 1q=\frac 1p$ gets us \begin{align} \left(\sum_{i=1}^n|a_{i}+b_{i}|^{p}\right)^{\frac 1p}\le \left(\sum_{i=1}^n|a_{i}|^p\right)^{\frac 1p}+ \left(\sum_{i=1}^n|b_{i}|^p\right)^{\frac 1p}, \end{align} The triangle equality is trivially true, when $\sum_{i=1}^n|a_{i}+b_{i}|^{p}=0$, so we can ignore this here. Now set $a_i := y_i-x_i$ and b_i := z_i-y_i$ to obtain the desired result.


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